How to conditionally define a lambda?
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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:
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++
|
show 5 more comments
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:
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++
3
"my code with this lambda does not compile" - What is the error you get?
– Suma
1 hour ago
You are trying to reference local variablegenerator
without capturing it. If second lambda was capture-free it would compile.
– VTT
1 hour ago
4
Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?
– Artyer
1 hour ago
1
Is a conditional really a performance issue here? Have you profiled the code?
– Jesper Juhl
1 hour ago
1
@Artyer Done well, the call to the lambda can be inlined.
– Angew
1 hour ago
|
show 5 more comments
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:
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++
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:
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++
c++
edited 1 hour ago
Artificial Hairless Armpit
asked 1 hour ago
Artificial Hairless ArmpitArtificial Hairless Armpit
1,2941535
1,2941535
3
"my code with this lambda does not compile" - What is the error you get?
– Suma
1 hour ago
You are trying to reference local variablegenerator
without capturing it. If second lambda was capture-free it would compile.
– VTT
1 hour ago
4
Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?
– Artyer
1 hour ago
1
Is a conditional really a performance issue here? Have you profiled the code?
– Jesper Juhl
1 hour ago
1
@Artyer Done well, the call to the lambda can be inlined.
– Angew
1 hour ago
|
show 5 more comments
3
"my code with this lambda does not compile" - What is the error you get?
– Suma
1 hour ago
You are trying to reference local variablegenerator
without capturing it. If second lambda was capture-free it would compile.
– VTT
1 hour ago
4
Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?
– Artyer
1 hour ago
1
Is a conditional really a performance issue here? Have you profiled the code?
– Jesper Juhl
1 hour ago
1
@Artyer Done well, the call to the lambda can be inlined.
– Angew
1 hour ago
3
3
"my code with this lambda does not compile" - What is the error you get?
– Suma
1 hour ago
"my code with this lambda does not compile" - What is the error you get?
– Suma
1 hour ago
You are trying to reference local variable
generator
without capturing it. If second lambda was capture-free it would compile.– VTT
1 hour ago
You are trying to reference local variable
generator
without capturing it. If second lambda was capture-free it would compile.– VTT
1 hour ago
4
4
Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?
– Artyer
1 hour ago
Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?
– Artyer
1 hour ago
1
1
Is a conditional really a performance issue here? Have you profiled the code?
– Jesper Juhl
1 hour ago
Is a conditional really a performance issue here? Have you profiled the code?
– Jesper Juhl
1 hour ago
1
1
@Artyer Done well, the call to the lambda can be inlined.
– Angew
1 hour ago
@Artyer Done well, the call to the lambda can be inlined.
– Angew
1 hour ago
|
show 5 more comments
2 Answers
2
active
oldest
votes
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); }, /* ... */)
}
}
Why not write a lambda that takes a lambda instead?
– Yakk - Adam Nevraumont
5 mins ago
add a comment |
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.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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); }, /* ... */)
}
}
Why not write a lambda that takes a lambda instead?
– Yakk - Adam Nevraumont
5 mins ago
add a comment |
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); }, /* ... */)
}
}
Why not write a lambda that takes a lambda instead?
– Yakk - Adam Nevraumont
5 mins ago
add a comment |
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); }, /* ... */)
}
}
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); }, /* ... */)
}
}
edited 58 mins ago
answered 1 hour ago
Vittorio RomeoVittorio Romeo
59.7k17165309
59.7k17165309
Why not write a lambda that takes a lambda instead?
– Yakk - Adam Nevraumont
5 mins ago
add a comment |
Why not write a lambda that takes a lambda instead?
– Yakk - Adam Nevraumont
5 mins ago
Why not write a lambda that takes a lambda instead?
– Yakk - Adam Nevraumont
5 mins ago
Why not write a lambda that takes a lambda instead?
– Yakk - Adam Nevraumont
5 mins ago
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered 1 hour ago
AngewAngew
135k11261354
135k11261354
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3
"my code with this lambda does not compile" - What is the error you get?
– Suma
1 hour ago
You are trying to reference local variable
generator
without capturing it. If second lambda was capture-free it would compile.– VTT
1 hour ago
4
Seems like a premature optimisation. Surely function call overhead would be greater than ternary overhead?
– Artyer
1 hour ago
1
Is a conditional really a performance issue here? Have you profiled the code?
– Jesper Juhl
1 hour ago
1
@Artyer Done well, the call to the lambda can be inlined.
– Angew
1 hour ago