用默认参数解引用函数-C ++ 14 vs C ++ 11 [英] Dereferencing a function with default arguments - C++14 vs C++11

问题描述

以下代码无法在带有选项-std=c++1y的g ++版本5.4.0中进行编译:

Following code can't be compiled with g++ version 5.4.0 with option -std=c++1y:

void f(int=0) ;

int main() {
    f(); // ok
    (*f)(2);// ok
    (*f)();// ok c++11; error with c++14: too few arguments to function
    return 0;
}

该函数声明为具有默认参数，那么这里出了什么问题? 谢谢你的帮助.

The function declared to have default argument, so what is wrong here? thanks for help.

为什么g++ -c -std=c++11可以编译?

推荐答案

接受(*f)()有效是GCC错误.该标准的字母表示使用带有一元*的函数名称应导致该函数名称衰减为指针.然后应该取消指针的引用，以获取调用表达式的函数地址.

Accepting (*f)() as valid is a GCC bug. The letter of the standard indicates that using a function name with unary * should cause the function name to decay into a pointer. The pointer should then be dereferenced to obtain the functions address for the call expression.

但是，海湾合作委员会似乎很聪明，并且忽略了上述行为.它简单地将(*f)视为f.调用f可以使用默认参数完成.

But GCC seems clever, and omits the above behavior. It treats (*f) simply as f. And calling f can be done with default arguments.

但是，可以强迫GCC进行衰减.应用于函数名称的一元+会强制将其衰减为指针.所以以下两个:

However, one can force GCC to preform the decay. Unary + applied on the function name will decay it to a pointer forcefully. So the following two:

(+f)();
(*+f)();

在两个 GCC 7.2 的任何一个标准修订版中，导致GCC发出error: too few arguments to function.和 GCC 6.3 .

Cause GCC to emit error: too few arguments to function on either standard revision, in both GCC 7.2 and GCC 6.3.

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