How to conditionally define a lambda? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Data science time! April 2019 and salary with experience The Ask Question Wizard is Live!Are functors actually faster than pointers to functions?How do you set, clear, and toggle a single bit?How do I iterate over the words of a string?How to convert a std::string to const char* or char*?How can I profile C++ code running on Linux?Why are #ifndef and #define used in C++ header files?Undefined, unspecified and implementation-defined behaviorC++11 introduced a standardized memory model. What does it mean? And how is it going to affect C++ programming?What is a lambda expression in C++11?What is an undefined reference/unresolved external symbol error and how do I fix it?Average values of a MAT channel

Why is Nikon 1.4g better when Nikon 1.8g is sharper?

Is there hard evidence that the grant peer review system performs significantly better than random?

How often does castling occur in grandmaster games?

What is "gratricide"?

If Windows 7 doesn't support WSL, then what does Linux subsystem option mean?

Chinese Seal on silk painting - what does it mean?

How could we fake a moon landing now?

Project Euler #1 in C++

How to write the following sign?

How to react to hostile behavior from a senior developer?

Why do early math courses focus on the cross sections of a cone and not on other 3D objects?

Putting class ranking in CV, but against dept guidelines

An adverb for when you're not exaggerating

What is the meaning of 'breadth' in breadth first search?

Significance of Cersei's obsession with elephants?

How fail-safe is nr as stop bytes?

Trademark violation for app?

How do living politicians protect their readily obtainable signatures from misuse?

Should I use a zero-interest credit card for a large one-time purchase?

Why do we need to use the builder design pattern when we can do the same thing with setters?

AppleTVs create a chatty alternate WiFi network

Why wasn't DOSKEY integrated with COMMAND.COM?

Time to Settle Down!

Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?



How to conditionally define a lambda?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!Are functors actually faster than pointers to functions?How do you set, clear, and toggle a single bit?How do I iterate over the words of a string?How to convert a std::string to const char* or char*?How can I profile C++ code running on Linux?Why are #ifndef and #define used in C++ header files?Undefined, unspecified and implementation-defined behaviorC++11 introduced a standardized memory model. What does it mean? And how is it going to affect C++ programming?What is a lambda expression in C++11?What is an undefined reference/unresolved external symbol error and how do I fix it?Average values of a MAT channel



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








9















The following function will randomly "sprinkle salt" on a loaded image. For the sake of boosting performance, the conditional statement



uint j = rows == 1 ? 0 : randomRow(generator);


should not be inside the loop.



Instead, I want to define a lambda getJ before the loop as



auto getJ = rows == 1 ? []() return 0; : []() return randomRow(generator); ;


However, my code with this lambda does not compile with the following red squiggled text:



enter image description here



Question



How to conditionally define such a lambda?



void salt_(Mat mat, unsigned long long n)

const uchar channels = mat.channels();
uint cols = mat.cols;
uint rows = mat.rows;

if (mat.isContinuous())

cols *= rows;
rows = 1;


default_random_engine generator;
uniform_int_distribution<uint> randomRow(0, rows - 1);
uniform_int_distribution<uint> randomCol(0, cols - 1);



// auto getJ = rows == 1 ? []() return 0; : []() return randomRow(generator); ;


uchar * const data = mat.data;

for (unsigned long long counter = 0; counter < n; counter++)

uint i = randomCol(generator);
uint j = rows == 1 ? 0 : randomRow(generator);
//uint j = getJ();

uint index = channels * (cols * j + i);
for (uchar k = 0; k < channels; k++)
data[index + k] = 255;











share|improve this question



















  • 3





    "my code with this lambda does not compile" - What is the error you get?

    – Suma
    2 days ago











  • You are trying to reference local variable generator without capturing it. If second lambda was capture-free it would compile.

    – VTT
    2 days ago







  • 5





    Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?

    – Artyer
    2 days ago






  • 2





    Is a conditional really a performance issue here? Have you profiled the code?

    – Jesper Juhl
    2 days ago






  • 1





    @Artyer Done well, the call to the lambda can be inlined.

    – Angew
    2 days ago

















9















The following function will randomly "sprinkle salt" on a loaded image. For the sake of boosting performance, the conditional statement



uint j = rows == 1 ? 0 : randomRow(generator);


should not be inside the loop.



Instead, I want to define a lambda getJ before the loop as



auto getJ = rows == 1 ? []() return 0; : []() return randomRow(generator); ;


However, my code with this lambda does not compile with the following red squiggled text:



enter image description here



Question



How to conditionally define such a lambda?



void salt_(Mat mat, unsigned long long n)

const uchar channels = mat.channels();
uint cols = mat.cols;
uint rows = mat.rows;

if (mat.isContinuous())

cols *= rows;
rows = 1;


default_random_engine generator;
uniform_int_distribution<uint> randomRow(0, rows - 1);
uniform_int_distribution<uint> randomCol(0, cols - 1);



// auto getJ = rows == 1 ? []() return 0; : []() return randomRow(generator); ;


uchar * const data = mat.data;

for (unsigned long long counter = 0; counter < n; counter++)

uint i = randomCol(generator);
uint j = rows == 1 ? 0 : randomRow(generator);
//uint j = getJ();

uint index = channels * (cols * j + i);
for (uchar k = 0; k < channels; k++)
data[index + k] = 255;











share|improve this question



















  • 3





    "my code with this lambda does not compile" - What is the error you get?

    – Suma
    2 days ago











  • You are trying to reference local variable generator without capturing it. If second lambda was capture-free it would compile.

    – VTT
    2 days ago







  • 5





    Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?

    – Artyer
    2 days ago






  • 2





    Is a conditional really a performance issue here? Have you profiled the code?

    – Jesper Juhl
    2 days ago






  • 1





    @Artyer Done well, the call to the lambda can be inlined.

    – Angew
    2 days ago













9












9








9


1






The following function will randomly "sprinkle salt" on a loaded image. For the sake of boosting performance, the conditional statement



uint j = rows == 1 ? 0 : randomRow(generator);


should not be inside the loop.



Instead, I want to define a lambda getJ before the loop as



auto getJ = rows == 1 ? []() return 0; : []() return randomRow(generator); ;


However, my code with this lambda does not compile with the following red squiggled text:



enter image description here



Question



How to conditionally define such a lambda?



void salt_(Mat mat, unsigned long long n)

const uchar channels = mat.channels();
uint cols = mat.cols;
uint rows = mat.rows;

if (mat.isContinuous())

cols *= rows;
rows = 1;


default_random_engine generator;
uniform_int_distribution<uint> randomRow(0, rows - 1);
uniform_int_distribution<uint> randomCol(0, cols - 1);



// auto getJ = rows == 1 ? []() return 0; : []() return randomRow(generator); ;


uchar * const data = mat.data;

for (unsigned long long counter = 0; counter < n; counter++)

uint i = randomCol(generator);
uint j = rows == 1 ? 0 : randomRow(generator);
//uint j = getJ();

uint index = channels * (cols * j + i);
for (uchar k = 0; k < channels; k++)
data[index + k] = 255;











share|improve this question
















The following function will randomly "sprinkle salt" on a loaded image. For the sake of boosting performance, the conditional statement



uint j = rows == 1 ? 0 : randomRow(generator);


should not be inside the loop.



Instead, I want to define a lambda getJ before the loop as



auto getJ = rows == 1 ? []() return 0; : []() return randomRow(generator); ;


However, my code with this lambda does not compile with the following red squiggled text:



enter image description here



Question



How to conditionally define such a lambda?



void salt_(Mat mat, unsigned long long n)

const uchar channels = mat.channels();
uint cols = mat.cols;
uint rows = mat.rows;

if (mat.isContinuous())

cols *= rows;
rows = 1;


default_random_engine generator;
uniform_int_distribution<uint> randomRow(0, rows - 1);
uniform_int_distribution<uint> randomCol(0, cols - 1);



// auto getJ = rows == 1 ? []() return 0; : []() return randomRow(generator); ;


uchar * const data = mat.data;

for (unsigned long long counter = 0; counter < n; counter++)

uint i = randomCol(generator);
uint j = rows == 1 ? 0 : randomRow(generator);
//uint j = getJ();

uint index = channels * (cols * j + i);
for (uchar k = 0; k < channels; k++)
data[index + k] = 255;








c++






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago







Artificial Hairless Armpit

















asked 2 days ago









Artificial Hairless ArmpitArtificial Hairless Armpit

1,3091535




1,3091535







  • 3





    "my code with this lambda does not compile" - What is the error you get?

    – Suma
    2 days ago











  • You are trying to reference local variable generator without capturing it. If second lambda was capture-free it would compile.

    – VTT
    2 days ago







  • 5





    Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?

    – Artyer
    2 days ago






  • 2





    Is a conditional really a performance issue here? Have you profiled the code?

    – Jesper Juhl
    2 days ago






  • 1





    @Artyer Done well, the call to the lambda can be inlined.

    – Angew
    2 days ago












  • 3





    "my code with this lambda does not compile" - What is the error you get?

    – Suma
    2 days ago











  • You are trying to reference local variable generator without capturing it. If second lambda was capture-free it would compile.

    – VTT
    2 days ago







  • 5





    Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?

    – Artyer
    2 days ago






  • 2





    Is a conditional really a performance issue here? Have you profiled the code?

    – Jesper Juhl
    2 days ago






  • 1





    @Artyer Done well, the call to the lambda can be inlined.

    – Angew
    2 days ago







3




3





"my code with this lambda does not compile" - What is the error you get?

– Suma
2 days ago





"my code with this lambda does not compile" - What is the error you get?

– Suma
2 days ago













You are trying to reference local variable generator without capturing it. If second lambda was capture-free it would compile.

– VTT
2 days ago






You are trying to reference local variable generator without capturing it. If second lambda was capture-free it would compile.

– VTT
2 days ago





5




5





Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?

– Artyer
2 days ago





Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?

– Artyer
2 days ago




2




2





Is a conditional really a performance issue here? Have you profiled the code?

– Jesper Juhl
2 days ago





Is a conditional really a performance issue here? Have you profiled the code?

– Jesper Juhl
2 days ago




1




1





@Artyer Done well, the call to the lambda can be inlined.

– Angew
2 days ago





@Artyer Done well, the call to the lambda can be inlined.

– Angew
2 days ago












2 Answers
2






active

oldest

votes


















9















my code with this lambda does not compile with the following red squiggled text




You cannot use randomRow inside the body of the lambda expression without capturing it beforehand, as the generated closure object needs to have access to it.



Even if you were to use [&randomRow], the code would still fail to compile as every lambda expression produces a closure of unique type, even if the lambda expressions are exactly the same.



You can turn the problem on its head to avoid any overhead and achieve what you want - create a function that takes the lambda you want to invoke:



template <typename F>
void saltImpl(F&& getJ, /* ... */)

uchar * const data = mat.data;

for (unsigned long long counter = 0; counter < n; counter++)

uint i = randomCol(generator);
uint j = rows == 1 ? 0 : randomRow(generator);
//uint j = getJ();

uint index = channels * (cols * j + i);
for (uchar k = 0; k < channels; k++)
data[index + k] = 255;




Usage example:



void salt_(Mat mat, unsigned long long n)

const uchar channels = mat.channels();
uint cols = mat.cols;
uint rows = mat.rows;

if (mat.isContinuous())

cols *= rows;
rows = 1;


default_random_engine generator;
uniform_int_distribution<uint> randomRow(0, rows - 1);
uniform_int_distribution<uint> randomCol(0, cols - 1);

if (rows == 1)

saltImpl([] return 0; , /* ... */);

else

saltImpl([&] return randomRow(generator); , /* ... */)







share|improve this answer




















  • 1





    Why not write a lambda that takes a lambda instead?

    – Yakk - Adam Nevraumont
    2 days ago


















3














Why this fails is because the lambdas are of a different type. That's natural, their operator() have different definitions. Which means you want your following code to work with two different types. And the C++ way of making code work with different types is using templates.



Convert the code using getJ to a function template (it can be local to your implementation file), like this:



template <class G>
void salt_impl_(Mat mat, unsigned long long n, default_random_engine &generator, G getJ)

const uchar channels = mat.channels();
uint cols = mat.cols;
uint rows = mat.rows;

if (mat.isContinuous())

cols *= rows;
rows = 1;


uchar * const data = mat.data;

uniform_int_distribution<uint> randomCol(0, cols - 1);

for (unsigned long long counter = 0; counter < n; counter++)

uint i = randomCol(generator);
uint j = getJ();

uint index = channels * (cols * j + i);
for (uchar k = 0; k < channels; k++)
data[index + k] = 255;




void salt_(Mat mat, unsigned long long n)

const uchar channels = mat.channels();
uint cols = mat.cols;
uint rows = mat.rows;

if (mat.isContinuous())

cols *= rows;
rows = 1;


default_random_engine generator;
uniform_int_distribution<uint> randomRow(0, rows - 1);

if (rows == 1)
salt_impl_(mat, n, generator, []() return 0; );
else
salt_impl_(mat, n, generator, [&]() return randomRow(generator); );



Feel free to reduce the initial-part duplication between the function and the template by passing more parameters, making them members of a class, or something similar.



Also note that the non-trivial lambda must capture the variables which it accesses (randomRow and generator). I did this using the universal by-reference capture [&] in the code above.






share|improve this answer























    Your Answer






    StackExchange.ifUsing("editor", function ()
    StackExchange.using("externalEditor", function ()
    StackExchange.using("snippets", function ()
    StackExchange.snippets.init();
    );
    );
    , "code-snippets");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "1"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55708242%2fhow-to-conditionally-define-a-lambda%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9















    my code with this lambda does not compile with the following red squiggled text




    You cannot use randomRow inside the body of the lambda expression without capturing it beforehand, as the generated closure object needs to have access to it.



    Even if you were to use [&randomRow], the code would still fail to compile as every lambda expression produces a closure of unique type, even if the lambda expressions are exactly the same.



    You can turn the problem on its head to avoid any overhead and achieve what you want - create a function that takes the lambda you want to invoke:



    template <typename F>
    void saltImpl(F&& getJ, /* ... */)

    uchar * const data = mat.data;

    for (unsigned long long counter = 0; counter < n; counter++)

    uint i = randomCol(generator);
    uint j = rows == 1 ? 0 : randomRow(generator);
    //uint j = getJ();

    uint index = channels * (cols * j + i);
    for (uchar k = 0; k < channels; k++)
    data[index + k] = 255;




    Usage example:



    void salt_(Mat mat, unsigned long long n)

    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())

    cols *= rows;
    rows = 1;


    default_random_engine generator;
    uniform_int_distribution<uint> randomRow(0, rows - 1);
    uniform_int_distribution<uint> randomCol(0, cols - 1);

    if (rows == 1)

    saltImpl([] return 0; , /* ... */);

    else

    saltImpl([&] return randomRow(generator); , /* ... */)







    share|improve this answer




















    • 1





      Why not write a lambda that takes a lambda instead?

      – Yakk - Adam Nevraumont
      2 days ago















    9















    my code with this lambda does not compile with the following red squiggled text




    You cannot use randomRow inside the body of the lambda expression without capturing it beforehand, as the generated closure object needs to have access to it.



    Even if you were to use [&randomRow], the code would still fail to compile as every lambda expression produces a closure of unique type, even if the lambda expressions are exactly the same.



    You can turn the problem on its head to avoid any overhead and achieve what you want - create a function that takes the lambda you want to invoke:



    template <typename F>
    void saltImpl(F&& getJ, /* ... */)

    uchar * const data = mat.data;

    for (unsigned long long counter = 0; counter < n; counter++)

    uint i = randomCol(generator);
    uint j = rows == 1 ? 0 : randomRow(generator);
    //uint j = getJ();

    uint index = channels * (cols * j + i);
    for (uchar k = 0; k < channels; k++)
    data[index + k] = 255;




    Usage example:



    void salt_(Mat mat, unsigned long long n)

    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())

    cols *= rows;
    rows = 1;


    default_random_engine generator;
    uniform_int_distribution<uint> randomRow(0, rows - 1);
    uniform_int_distribution<uint> randomCol(0, cols - 1);

    if (rows == 1)

    saltImpl([] return 0; , /* ... */);

    else

    saltImpl([&] return randomRow(generator); , /* ... */)







    share|improve this answer




















    • 1





      Why not write a lambda that takes a lambda instead?

      – Yakk - Adam Nevraumont
      2 days ago













    9












    9








    9








    my code with this lambda does not compile with the following red squiggled text




    You cannot use randomRow inside the body of the lambda expression without capturing it beforehand, as the generated closure object needs to have access to it.



    Even if you were to use [&randomRow], the code would still fail to compile as every lambda expression produces a closure of unique type, even if the lambda expressions are exactly the same.



    You can turn the problem on its head to avoid any overhead and achieve what you want - create a function that takes the lambda you want to invoke:



    template <typename F>
    void saltImpl(F&& getJ, /* ... */)

    uchar * const data = mat.data;

    for (unsigned long long counter = 0; counter < n; counter++)

    uint i = randomCol(generator);
    uint j = rows == 1 ? 0 : randomRow(generator);
    //uint j = getJ();

    uint index = channels * (cols * j + i);
    for (uchar k = 0; k < channels; k++)
    data[index + k] = 255;




    Usage example:



    void salt_(Mat mat, unsigned long long n)

    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())

    cols *= rows;
    rows = 1;


    default_random_engine generator;
    uniform_int_distribution<uint> randomRow(0, rows - 1);
    uniform_int_distribution<uint> randomCol(0, cols - 1);

    if (rows == 1)

    saltImpl([] return 0; , /* ... */);

    else

    saltImpl([&] return randomRow(generator); , /* ... */)







    share|improve this answer
















    my code with this lambda does not compile with the following red squiggled text




    You cannot use randomRow inside the body of the lambda expression without capturing it beforehand, as the generated closure object needs to have access to it.



    Even if you were to use [&randomRow], the code would still fail to compile as every lambda expression produces a closure of unique type, even if the lambda expressions are exactly the same.



    You can turn the problem on its head to avoid any overhead and achieve what you want - create a function that takes the lambda you want to invoke:



    template <typename F>
    void saltImpl(F&& getJ, /* ... */)

    uchar * const data = mat.data;

    for (unsigned long long counter = 0; counter < n; counter++)

    uint i = randomCol(generator);
    uint j = rows == 1 ? 0 : randomRow(generator);
    //uint j = getJ();

    uint index = channels * (cols * j + i);
    for (uchar k = 0; k < channels; k++)
    data[index + k] = 255;




    Usage example:



    void salt_(Mat mat, unsigned long long n)

    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())

    cols *= rows;
    rows = 1;


    default_random_engine generator;
    uniform_int_distribution<uint> randomRow(0, rows - 1);
    uniform_int_distribution<uint> randomCol(0, cols - 1);

    if (rows == 1)

    saltImpl([] return 0; , /* ... */);

    else

    saltImpl([&] return randomRow(generator); , /* ... */)








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 2 days ago

























    answered 2 days ago









    Vittorio RomeoVittorio Romeo

    59.8k17165309




    59.8k17165309







    • 1





      Why not write a lambda that takes a lambda instead?

      – Yakk - Adam Nevraumont
      2 days ago












    • 1





      Why not write a lambda that takes a lambda instead?

      – Yakk - Adam Nevraumont
      2 days ago







    1




    1





    Why not write a lambda that takes a lambda instead?

    – Yakk - Adam Nevraumont
    2 days ago





    Why not write a lambda that takes a lambda instead?

    – Yakk - Adam Nevraumont
    2 days ago













    3














    Why this fails is because the lambdas are of a different type. That's natural, their operator() have different definitions. Which means you want your following code to work with two different types. And the C++ way of making code work with different types is using templates.



    Convert the code using getJ to a function template (it can be local to your implementation file), like this:



    template <class G>
    void salt_impl_(Mat mat, unsigned long long n, default_random_engine &generator, G getJ)

    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())

    cols *= rows;
    rows = 1;


    uchar * const data = mat.data;

    uniform_int_distribution<uint> randomCol(0, cols - 1);

    for (unsigned long long counter = 0; counter < n; counter++)

    uint i = randomCol(generator);
    uint j = getJ();

    uint index = channels * (cols * j + i);
    for (uchar k = 0; k < channels; k++)
    data[index + k] = 255;




    void salt_(Mat mat, unsigned long long n)

    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())

    cols *= rows;
    rows = 1;


    default_random_engine generator;
    uniform_int_distribution<uint> randomRow(0, rows - 1);

    if (rows == 1)
    salt_impl_(mat, n, generator, []() return 0; );
    else
    salt_impl_(mat, n, generator, [&]() return randomRow(generator); );



    Feel free to reduce the initial-part duplication between the function and the template by passing more parameters, making them members of a class, or something similar.



    Also note that the non-trivial lambda must capture the variables which it accesses (randomRow and generator). I did this using the universal by-reference capture [&] in the code above.






    share|improve this answer



























      3














      Why this fails is because the lambdas are of a different type. That's natural, their operator() have different definitions. Which means you want your following code to work with two different types. And the C++ way of making code work with different types is using templates.



      Convert the code using getJ to a function template (it can be local to your implementation file), like this:



      template <class G>
      void salt_impl_(Mat mat, unsigned long long n, default_random_engine &generator, G getJ)

      const uchar channels = mat.channels();
      uint cols = mat.cols;
      uint rows = mat.rows;

      if (mat.isContinuous())

      cols *= rows;
      rows = 1;


      uchar * const data = mat.data;

      uniform_int_distribution<uint> randomCol(0, cols - 1);

      for (unsigned long long counter = 0; counter < n; counter++)

      uint i = randomCol(generator);
      uint j = getJ();

      uint index = channels * (cols * j + i);
      for (uchar k = 0; k < channels; k++)
      data[index + k] = 255;




      void salt_(Mat mat, unsigned long long n)

      const uchar channels = mat.channels();
      uint cols = mat.cols;
      uint rows = mat.rows;

      if (mat.isContinuous())

      cols *= rows;
      rows = 1;


      default_random_engine generator;
      uniform_int_distribution<uint> randomRow(0, rows - 1);

      if (rows == 1)
      salt_impl_(mat, n, generator, []() return 0; );
      else
      salt_impl_(mat, n, generator, [&]() return randomRow(generator); );



      Feel free to reduce the initial-part duplication between the function and the template by passing more parameters, making them members of a class, or something similar.



      Also note that the non-trivial lambda must capture the variables which it accesses (randomRow and generator). I did this using the universal by-reference capture [&] in the code above.






      share|improve this answer

























        3












        3








        3







        Why this fails is because the lambdas are of a different type. That's natural, their operator() have different definitions. Which means you want your following code to work with two different types. And the C++ way of making code work with different types is using templates.



        Convert the code using getJ to a function template (it can be local to your implementation file), like this:



        template <class G>
        void salt_impl_(Mat mat, unsigned long long n, default_random_engine &generator, G getJ)

        const uchar channels = mat.channels();
        uint cols = mat.cols;
        uint rows = mat.rows;

        if (mat.isContinuous())

        cols *= rows;
        rows = 1;


        uchar * const data = mat.data;

        uniform_int_distribution<uint> randomCol(0, cols - 1);

        for (unsigned long long counter = 0; counter < n; counter++)

        uint i = randomCol(generator);
        uint j = getJ();

        uint index = channels * (cols * j + i);
        for (uchar k = 0; k < channels; k++)
        data[index + k] = 255;




        void salt_(Mat mat, unsigned long long n)

        const uchar channels = mat.channels();
        uint cols = mat.cols;
        uint rows = mat.rows;

        if (mat.isContinuous())

        cols *= rows;
        rows = 1;


        default_random_engine generator;
        uniform_int_distribution<uint> randomRow(0, rows - 1);

        if (rows == 1)
        salt_impl_(mat, n, generator, []() return 0; );
        else
        salt_impl_(mat, n, generator, [&]() return randomRow(generator); );



        Feel free to reduce the initial-part duplication between the function and the template by passing more parameters, making them members of a class, or something similar.



        Also note that the non-trivial lambda must capture the variables which it accesses (randomRow and generator). I did this using the universal by-reference capture [&] in the code above.






        share|improve this answer













        Why this fails is because the lambdas are of a different type. That's natural, their operator() have different definitions. Which means you want your following code to work with two different types. And the C++ way of making code work with different types is using templates.



        Convert the code using getJ to a function template (it can be local to your implementation file), like this:



        template <class G>
        void salt_impl_(Mat mat, unsigned long long n, default_random_engine &generator, G getJ)

        const uchar channels = mat.channels();
        uint cols = mat.cols;
        uint rows = mat.rows;

        if (mat.isContinuous())

        cols *= rows;
        rows = 1;


        uchar * const data = mat.data;

        uniform_int_distribution<uint> randomCol(0, cols - 1);

        for (unsigned long long counter = 0; counter < n; counter++)

        uint i = randomCol(generator);
        uint j = getJ();

        uint index = channels * (cols * j + i);
        for (uchar k = 0; k < channels; k++)
        data[index + k] = 255;




        void salt_(Mat mat, unsigned long long n)

        const uchar channels = mat.channels();
        uint cols = mat.cols;
        uint rows = mat.rows;

        if (mat.isContinuous())

        cols *= rows;
        rows = 1;


        default_random_engine generator;
        uniform_int_distribution<uint> randomRow(0, rows - 1);

        if (rows == 1)
        salt_impl_(mat, n, generator, []() return 0; );
        else
        salt_impl_(mat, n, generator, [&]() return randomRow(generator); );



        Feel free to reduce the initial-part duplication between the function and the template by passing more parameters, making them members of a class, or something similar.



        Also note that the non-trivial lambda must capture the variables which it accesses (randomRow and generator). I did this using the universal by-reference capture [&] in the code above.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 days ago









        AngewAngew

        135k11261354




        135k11261354



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55708242%2fhow-to-conditionally-define-a-lambda%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Get product attribute by attribute group code in magento 2get product attribute by product attribute group in magento 2Magento 2 Log Bundle Product Data in List Page?How to get all product attribute of a attribute group of Default attribute set?Magento 2.1 Create a filter in the product grid by new attributeMagento 2 : Get Product Attribute values By GroupMagento 2 How to get all existing values for one attributeMagento 2 get custom attribute of a single product inside a pluginMagento 2.3 How to get all the Multi Source Inventory (MSI) locations collection in custom module?Magento2: how to develop rest API to get new productsGet product attribute by attribute group code ( [attribute_group_code] ) in magento 2

            Category:9 (number) SubcategoriesMedia in category "9 (number)"Navigation menuUpload mediaGND ID: 4485639-8Library of Congress authority ID: sh85091979ReasonatorScholiaStatistics

            Magento 2.3: How do i solve this, Not registered handle, on custom form?How can i rewrite TierPrice Block in Magento2magento 2 captcha not rendering if I override layout xmlmain.CRITICAL: Plugin class doesn't existMagento 2 : Problem while adding custom button order view page?Magento 2.2.5: Overriding Admin Controller sales/orderMagento 2.2.5: Add, Update and Delete existing products Custom OptionsMagento 2.3 : File Upload issue in UI Component FormMagento2 Not registered handleHow to configured Form Builder Js in my custom magento 2.3.0 module?Magento 2.3. How to create image upload field in an admin form