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C++ includes useful generic functions like?std::for_each?and?std::transform, which can be very handy. Unfortunately they can also be quite cumbersome to use, particularly if the?functor?you would like to apply is unique to the particular function.
#include <algorithm>
#include <vector>
namespace {
struct f {
void operator()(int) {
// do something
}
};
}
void func(std::vector<int>& v) {
f f;
std::for_each(v.begin(), v.end(), f);
}
If you only use?f?once and in that specific place it seems overkill to be writing a whole class just to do something trivial and one off.
In C++03 you might be tempted to write something like the following, to keep the functor local:
void func2(std::vector<int>& v) {
struct {
void operator()(int) {
// do something
}
} f;
std::for_each(v.begin(), v.end(), f);
}
however this is not allowed,?f?cannot be passed to a?template?function in C++03.
The new solution
C++11 introduces lambdas allow you to write an inline, anonymous functor to replace the?struct f. For small simple examples this can be cleaner to read (it keeps everything in one place) and potentially simpler to maintain, for example in the simplest form:
void func3(std::vector<int>& v) {
std::for_each(v.begin(), v.end(), [](int) { /* do something here*/ });
}
Lambda functions are just syntactic sugar for anonymous functors.
Return types
In simple cases the return type of the lambda is deduced for you, e.g.:
void func4(std::vector<double>& v) {
std::transform(v.begin(), v.end(), v.begin(),
[](double d) { return d < 0.00001 ? 0 : d; }
);
}
however when you start to write more complex lambdas you will quickly encounter cases where the return type cannot be deduced by the compiler, e.g.:
void func4(std::vector<double>& v) {
std::transform(v.begin(), v.end(), v.begin(),
[](double d) {
if (d < 0.0001) {
return 0;
} else {
return d;
}
});
}
To resolve this you are allowed to explicitly specify a return type for a lambda function, using?-> T:
void func4(std::vector<double>& v) {
std::transform(v.begin(), v.end(), v.begin(),
[](double d) -> double {
if (d < 0.0001) {
return 0;
} else {
return d;
}
});
}
"Capturing" variables
So far we've not used anything other than what was passed to the lambda within it, but we can also use other variables, within the lambda. If you want to access other variables you can use the capture clause (the?[]?of the expression), which has so far been unused in these examples, e.g.:
void func5(std::vector<double>& v, const double& epsilon) {
std::transform(v.begin(), v.end(), v.begin(),
[epsilon](double d) -> double {
if (d < epsilon) {
return 0;
} else {
return d;
}
});
}
You can capture by both reference and value, which you can specify using?&?and?=?respectively:
[&epsilon]?captures by reference[&]?captures all variables used in the lambda by reference[=]?captures all variables used in the lambda by value[&, epsilon]?captures variables like with [&], but epsilon by value[=, &epsilon]?captures variables like with [=], but epsilon by reference
转发地址:https://stackoverflow.com/questions/7627098/what-is-a-lambda-expression-in-c11
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