为什么即使没有任何类声明也需要原型？ [英] Why prototype is required even without any class declaration?

问题描述

如果我只是这样做：
Ex1：

#include <iostream>

int main()
{
    //try to call doSomething function
    doSomething();
}

void doSomething()
{
    std::cout << "Call me now!" << std::endl;
}

我得到编译错误！因为编译不知道是什么是doSomething。

I get compilation error! Because the compile doesn´t know what is "doSomething".

但是如果我改变doSomething的位置在第一位，程序编译成功。
Ex2：

But if I change the position of doSomething to come in first place, the program compiles successfully. Ex2:

#include <iostream>

void doSomething()
{
    std::cout << "Call me now!" << std::endl;
}

int main()
{
    //try to call doSomething function
    doSomething();
}

我可以声明原型如下：
Ex3：

#include <iostream>

void doSomething(void);

int main()
{
    //try to call doSomething function
    doSomething();
}

void doSomething()
{
    std::cout << "Call me now!" << std::endl;
}

但是为什么第一个例子不工作？为什么我甚至必须声明一个原型或最后调用函数和主函数？

But why the first example does not work? Why I even have to declare a prototype or call functions first and main function at last?

谢谢！

推荐答案

如果编译器没有看到定义或声明，就不能调用函数，首先是简单的。原型或实际定义必须出现在调用之前。

You can't call a function without the compiler having seen either the definition or a declaration, first -- simple as that. A prototype or the actual definition must appear before a call.

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