问题描述
我在使用lambda时遇到问题,我想将它们保存在指针中,但不使用常量变量来创建它们.
i have a problem with lambdas and is that i want to save them in pointers but create them using no constant variables.
// Example program
#include <iostream>
#include <string>
template<short n>
void speak(){
std::cout << "speak:[" << n << "]" << std::endl;
}
int main()
{
typedef void(*func)(void);
func ptr[8];
const short j = 10;
for(short i = 0; i < 4; i++){
ptr[i] = [=](void)->void{
std::cout << "lamda:[" << j << "]" << std::endl;
};
/*ptr[i] = [=](void)->void{
std::cout << "lamda:[" << i << "]" << std::endl;
};*/
ptr[i + 4] = speak<j>;
//ptr[i + 4] = speak<i>;
}
for(short i = 0; i < 8; i++){
ptr[i]();
}
return 0;
}
有可能用常数var"j"定义一个lambda,而没有用"i"定义一个lambda.有没有办法使用i来做到这一点,所以就没有必要这样做:
It's posible to define a lambda with a constant var "j", but no with "i". Is there any way to do it using i, so there'll be no need to do:
ptr[0] = [=](void)->void{
std::cout << "lamda:[" << 0 << "]" << std::endl;
};
ptr[1] = [=](void)->void{
std::cout << "lamda:[" << 1 << "]" << std::endl;
};
ptr[2] = [=](void)->void{
std::cout << "lamda:[" << 2 << "]" << std::endl;
};
ptr[3] = [=](void)->void{
std::cout << "lamda:[" << 3 << "]" << std::endl;
};
模板化函数是我实现它的另一种尝试,但没有成功
当我尝试使用"i"时,得到了:错误:无法转换'main()::< lambda()>'"赋值到'func {aka void()()}'""*
推荐答案
您使用 std :: function
代替了指针.只需添加标头< functional>
,然后更改
You use std::function
instead of a pointer. Just include the header <functional>
, and change
typedef void(*func)(void);
到
typedef std::function<void(void)> func;
一切都会正常!
不确定为什么()
也能工作时为什么喜欢写(void)
.
Not sure why you like to write (void)
when ()
works too.
之所以不能使用函数指针,是因为函数指针仅指向代码.它没有任何捕获的变量. std :: function
是一个也可以保存捕获的变量的类.
The reason you can't use a function pointer is because a function pointer just points to code; it doesn't have any captured variables. std::function
is a class which is also able to hold the captured variables.
这篇关于它有什么方法可以将带有捕获参数的lambda存储在指针中的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!