为什么定义没有返回类型的main()不会出错而进行编译? [英] Why does defining main() with no return type compile with no error?

问题描述

我只是通过使用GCC(g++)进行编译来测试此代码；

I was just testing this code by compiling it with GCC (g++);

#include <stdio.h>

main(int argc, char **argv){

    printf("something");

}

它似乎可以构建并运行良好，只是发出警告；

It seems to build and run fine with just a warning saying;

ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]

尽管我定义了main()，没有返回类型，并且没有返回任何值.
然后，我决定通过使用Visual C ++编译器进行编译来测试相同的代码，但是它有效地抛出了错误:

Though I defined main() with no return type and not returning any value to it.
I then decided to test the same code by compiling it with Visual C++ compiler but it efficiently threw the error:

missing type specifier - int assumed. Note: C++ does not support default-int

并且构建没有成功.

我期望GCC抛出错误并停止编译，而不是发出警告，并使其成功构建.

I was expecting GCC to throw an error and stop compiling instead of a warning and letting it build successfully.

为什么GCC赋予main()这样的特权，即不让其知道其返回类型?

Why does GCC give main() such privilege of not letting it know its return type?

推荐答案

程序格式错误. C ++标准不允许省略返回类型.

The program is ill-formed. Omitting the return type is not permitted by the C++ standard.

编译器不将其视为致命错误的原因是历史性的.在1999年标准之前，C确实允许省略函数的返回类型.它默认为int. C ++源自C，因此早期的(预标准)C ++具有相同的规则.

The reason the compiler doesn't treat it as a fatal error is historical. Prior to the 1999 standard, C did permit the return type of a function to be omitted; it would default to int. C++ is derived from C, so early (pre-standard) versions of C++ had the same rule.

在现代C ++中，省略返回类型是一个错误.需要编译器诊断这样的错误，但是不需要将其视为致命错误.通过显示警告，编译器就标准而言已经完成了工作.

In modern C++, omitting the return type is an error. The compiler is required to diagnose such an error, but it's not required to treat it as fatal. By printing a warning, the compiler has done its job as far as the standard is concerned.

不要忽略警告.

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