它是可移植的定义和声明函数后调用它在main中的C？ [英] Is it portable to define and declare the function after calling it in main in C?

问题描述

我遇到了与C和C ++之间的函数定义/声明有关的以下差异。

I came across the following difference related to function definition/declaration between C and C++.

在C ++中，如果我在声明和定义函数之前调用函数，

In C++ if I call a function before declaring and defining it,I get the error message 'function name is invalid identifier' which is fine for me as it sounds reasonable.

在Visual Studio编译器的C中，当我编译下面的程序时，我得到了一个错误消息函数名是无效的标识符错误消息：

In C with Visual Studio compiler when I compiled the following program I get the error message:

error C2371: 'fun' : redefinition; different basic types

，在gcc-4.3.4中，它成功执行，只有这个警告：

and in gcc-4.3.4 it executed successfully with just this warning:

warning: conflicting types for ‘fun’

这里是程序：

#include <stdio.h>
int main(){
    fun();
    return 0;
}
void fun(){
    printf("It is fun");
}

所以在C中调用一个函数，它？！

So is it fine in C to just call a function and later bother about defining it?! And why the compilers behave differently?

推荐答案

在C89中调用没有原型的函数时（免责声明：我不知道关于C99），你隐式声明为

When you call the function without a prototype in C89 (disclaimer: I don't know about C99), you are implicitly declaring it as

int fun（）;

int fun();

除了在C ++中，空的parantheses只是意味着你没有指定它需要什么参数，而不是它需要零个参数。

Note that in C, other than in C++, the empty parantheses just mean that you haven't specified what arguments it takes, not that it takes zero arguments.

当你重新声明函数为返回void，你得到警告，但没有错误。如果你返回一些非平凡的（structs，float，...）而不是void或者如果你尝试使用函数的int结果，那么你可能在运行时遇到了麻烦。

When you redeclare the function as returning void, you get the warning, but no error. If you return something non-trivial (structs, floats, ...) instead of void or if you try to use the int result of the function, then you might be in deep trouble at runtime.

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