I would like to use a lambda as a parameter for a C++ function, but I don't know which type to specify in the function declaration. What I would like to do is this:
void myFunction(WhatToPutHere lambda){
//some things
}
I have tried void myFunction(auto lambda)
and void myFunction(void lambda)
but none of these codes compiled. In case it matters, the lambda doesn't return anything.
How can I use a lambda as a parameter in a C++ function?
You have 2 ways: make your function template:
template <typename F>
void myFunction(F&& lambda)
{
//some things
}
or erase type (with std::function
for example):
void
myFunction(const std::function<void()/*type of your lamdba::operator()*/>& f)
{
//some things
}
You have two choices, basically.
Make it a template:
template<typename T>
void myFunction(T&& lambda){
}
or, if you do not want (or can't) do that, you can use type-erased std::function
:
void myFunction(std::function<void()> const& lambda){
}
Conversely, your attempt with auto
would've been correct under the concepts TS as currently implemented in gcc, where it'd be an abbreviated template .
// hypothetical C++2x code
void myFunction(auto&& lambda){
}
or with a concept:
// hypothetical C++2x code
void myFunction(Callable&& lambda){
}
If this is an inline
function, prefer a template, as in
template<typename Func>
void myFunction(Func const&lambda)
{
//some things
}
because it binds to anything that makes sense (and will cause compiler error for anything else), including lambdas, instances of named classes, and std::function<>
objects.
On the other hand, if this function is not inline
, ie implemented in some compilation unit, you cannot use a generic template but must use a specified type, which is best taken a std::function<>
object and passed via reference.
Pass it as you'd pass a simple function. Just give it a name with auto
#include <iostream>
int SimpleFunc(int x) { return x + 100; }
int UsingFunc(int x, int(*ptr)(int)) { return ptr(x); }
auto lambda = [](int jo) { return jo + 10; };
int main() {
std::cout << "Simple function passed by a pointer: " << UsingFunc(5, SimpleFunc) << std::endl;
std::cout << "Lambda function passed by a pointer: " << UsingFunc(5, lambda) << std::endl;
}
Output:
Simple function passed by a pointer: 105
Lambda function passed by a pointer: 15
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