g ++“调用”函数无括号(不是f(),但f;)。为什么总是返回1? [英] g++ "Calling" function without parenthesis ( not f() but f; ). Why always returns 1?
问题描述
c ++(GNU GCC g ++)。 callingfunction without()
函数不工作,但编译成功。
c++ (GNU GCC g++ ). "Calling" function without () The function is not working, but compiles ok.
更令人惊讶的是,这样的代码总是返回1 ...
More surprisingly, such code always returns 1...
有什么解释吗?
我希望函数名称只是一个常规指针,但似乎有点不同。
I expected function name to be just a regular pointer, but seems it's a bit different...
I got all 1's only by chance?
include< iostream>
using namespace std;
void pr()
{
cout<< sth;
}
int main()
{
pr;
cout<< pr; // output:1
cout<< * pr; // output:1
cout<< & pr; // output:1
}
#include <iostream>
using namespace std;
void pr ()
{
cout << "sth";
}
int main()
{
pr;
cout << pr; // output: 1
cout << *pr; // output: 1
cout << ≺ // output: 1
}
推荐答案
你实际上不是在代码中调用 pr
,而是将函数指针传递给 cout
。 pr
在传递给 cout $ c $时会转换为
bool
c>。如果你把 cout<< boolalpha
,您将输出 true
,而不是 1
。
You're not actually calling pr
in your code, you're passing the function pointer to cout
. pr
is then being converted to a bool
when being passed to cout
. If you put cout << boolalpha
beforehand you will output true
instead of 1
.
EDIT:
对于C ++ 11,您可以写入以下重载:
With C++11 you can write the following overload:
template <class RType, class ... ArgTypes>
std::ostream & operator<<(std::ostream & s, RType(*func)(ArgTypes...))
{
return s << "(func_ptr=" << (void*)func << ")(num_args="
<< sizeof...(ArgTypes) << ")";
}
这意味着调用 cout < pr
将打印(func_ptr =< pr的地址)(num_args = 0)
。函数本身可以做任何你想要的,这只是为了证明,使用C ++ 11的可变参数模板,你可以匹配任意arity的函数指针。这仍然不能用于重载的函数和函数模板,而不指定你想要的重载(通常通过转换)。
which means the call cout << pr
will print (func_ptr=<address of pr>)(num_args=0)
. The function itself can do whatever you want obviously, this is just to demonstrate that with C++11's variadic templates, you can match function pointers of arbitrary arity. This still won't work for overloaded functions and function templates without specifying which overload you want (usually via a cast).
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