为什么可以用具有不同返回类型的lambda构造std :: function? [英] Why can std::function be constructed with a lambda with a different return type?
问题描述
以下编译正常:
#include <functional>
int main()
{
std::function<const int&()> f = []() -> int {return 1;};
const int& r = f(); // r is a dangling reference
return 0;
}
怎么可能将具有const int&
返回类型的std::function
设置为具有int
返回类型的lambda?允许这种类型的转换隐式发生而没有警告是恕我直言.
How come it's possible to set an std::function
with a const int&
return type to a lambda with an int
return type? Allowing this sort of cast to happen implicitly and with no warning is a gotcha IMHO.
推荐答案
您可以使用任何带有相关参数可调用且其返回值可隐式转换为std::function
返回值的对象来构造std::function
. int
可隐式转换为const int&
,因此符合规则.
You can construct a std::function
with any object which is callable with the relevant arguments and whose return value is implicitly convertible to the std::function
return. int
is implicitly convertible to const int&
, so the rules are met.
编译器可以对此进行警告,但是对于一个特殊的角落情况,似乎需要做很多工作.
A compiler could feel free to warn about this, but it seems like a lot of work for a particularly corner-y corner case.
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