为什么在函数定义中对于函数自变量没有强制提到“ int”数据类型？

Question

Recently I went through some code similar to this: (The code is proprietary, and hence adding a similar one)

#include<stdio.h>

void test_it(var)
{
    printf("%d\n",var);
}

int main()
{
    test_it(67);
    return 0;
}

The arguments of test_it do not have datatype mentioned.

I compiled it as gcc test_it.c ... : Surprisingly No Warnings/Error

Again I compiled using: gcc -Wall test_it.c ... : No Warnings/Error yet again

(Getting more aggressive now...)

I compiled it again using: gcc -Wall -Wextra test_it.c ... :
warning: type of 'var' defaults to 'int' finally I got the warning.

I tried using multiple arguments as:

void test_it(var1, var2)
{
    printf("%d\n%d\n",var1, var2);
}

int main()
{
    test_it(67,76);
    return 0;
}

Same beahavior!!

Also I tried this:

void test_it(var)
{
    printf("%d\n",var);
}

main()   // Notice that no `int` there
{
    test_it(67);
    return 0;
}

This code gave warning with -Wall option only.

So my question is why the int datatype is not mandatory for function arguments in function definition?

EDIT:

Rewording the question:

Why gcc doesn't give warning with -Wall in the case of omitting datatype of function arguments, but gives warning for omitting the function return type? Why does it ignore it in the first case?

Answer 1

In C89, the default type is assumed to be int . This ( is valid in C89 ), however the default type rule has been abandoned in C99 . See the difference:

C89 - Compiles fine

C99 prog.c:3:6: error: type of 'var' defaults to 'int'

Try to compile with the -std=c99 flag.

Answer 2

By default the functions arguements are of type int in C89. So the code executes fine.
You can go through this related question.