C 程序错误代码混淆（比较值）

Question

Just trying to compare two values in C, and find the smaller value. Accidently did it in a really bad style, but I don't understand why C still let it run... the program didn't include the branch that two values are equal. But when it's given two equal values, it gives the program gives out the same value... may I ask what did C complier do? Thank you! (the example below returns 3...)

int min(int x, int y);

int main(void)
{
    int a = 3, b = 3;
    printf("%d", min(a,b));
}

int min(int x, int y)
{
    if(x > y)
    return y;
}
    else if(x < y)
{
    return x;
}

Answer 1

Accidently did it in a really bad style, but I don't understand why C still let it run..

The program should not compile because this code

int min(int x, int y)
{
    if(x > y)
    return y;
}
    else if(x < y)
{
    return x;
}

is syntactically incorrect.

This else statement

    else if(x < y)
{
    return x;
}

is placed outside any function including min .

It seems you mean

int min(int x, int y)
{
    if(x > y)
    {
        return y;
    }
    else if(x < y)
    {
        return x;
    }
}

But when it's given two equal values, it gives the program gives out the same value... may I ask what did C complier do?

The program has undefined behavior because the function min returns nothing when its two arguments are equal each other.

From the C Standard (6.9.1 Function definitions)

12 If the } that terminates a function is reached, and the value of the function call is used by the caller, the behavior is undefined.

It should be defined for example the following way

int min(int x, int y)
{
    return y < x ? y : x;
}