c - 为什么没有参数的函数(与实际函数定义相比)可以编译？

Question

我刚刚看到某人的 C 代码，我很困惑为什么要编译它。有两点不太明白。

1. 与实际函数定义相比，函数原型(prototype)没有参数。

2. 函数定义中的参数没有类型。

<小时/>

#include <stdio.h>

int func();

int func(param)
{
    return param;
}

int main()
{
    int bla = func(10);    
    printf("%d", bla);
}

为什么这有效？我已经在几个编译器中测试了它，它工作得很好。

Answer 1

所有其他答案都是正确的，但仅限于 completion

A function is declared in the following manner:

  return-type function-name(parameter-list,...) { body... }

return-type is the variable type that the function returns. This can not be an array type or a function type. If not given, then int is assumed.

function-name is the name of the function.

parameter-list is the list of parameters that the function takes separated by commas. If no parameters are given, then the function does not take any and should be defined with an empty set of parenthesis or with the keyword void. If no variable type is in front of a variable in the paramater list, then int is assumed. Arrays and functions are not passed to functions, but are automatically converted to pointers. If the list is terminated with an ellipsis (,...), then there is no set number of parameters. Note: the header stdarg.h can be used to access arguments when using an ellipsis.

为了完整起见，再次强调。 From C11 specification 6:11:6 (第179页)

The use of function declarators with empty parentheses (not prototype-format parameter type declarators) is an obsolescent feature.