关于c函数指针的2个问题 [英] 2 questions about function pointer in c

问题描述

我有两个问题。

1.我看到

int (*Ptr)(int,int);
Ptr=someOtherFuncion;

它不应该是这样的？

Its not should be like that?

Ptr=&someOtherFuncion;

2.我知道函数指针的回调
像

2.I learn that function pointer its for callback like that

    someOtherFunction(functionPointer)

是什么产生差异，如果我会把未指针常规的函数？

what is the differeence if I will put a regular function that is not pointer?

推荐答案

函数的名称几乎立刻衰变到一个函数指针，所以 someOtherFunction 衰减到该同一指针和放大器; someOtherFunction 为您提供了明确。事实上，操作数的地址运算符（＆安培; ）是少数地方之一是衰减的不的发生

The name of a function decays almost immediately to a pointer to the function, so someOtherFunction decays to the same pointer that &someOtherFunction gives you explicitly. In fact, the operand of the address-of operator (&) is one of the few places were the decay doesn't happen.

这有有趣的结果：即使你取消引用函数指针，它再次衰减的时候了。所以下面都是等价的：

This has amusing consequences: Even if you dereference the function pointer, it decays again right away. So the following are all equivalent:

someOtherFunction(1, 2);
(*someOtherFunction)(1, 2);
(**someOtherFunction)(1, 2);
(***someOtherFunction)(1, 2);

所以，如果你感觉不适分配给一个函数指针没有明确的地址的，通过各种手段把＆安培; 在那里，但你没有到

要解决这个问题的第二部分：一个功能的总是的通过函数指针调用，但是因为上面提到的瞬间衰减的，正常的功能可以被称为只是方式与功能相同指针。

To address the second part of the question: A function is always called through a function pointer, but because of the above-mentioned instant decay, normal functions can be called just the same way as function pointers.

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