Question

here is my snippet of code:

float square_root(x)
float x;
{
 .......
}

int main(){
    printf("Square_root 2 = %f\n", square_root(4));
}

When I pass number 4.0 to the square_root() function, x parameter inside the function is 4.0000000 so its ok.
But when I pass just 4 (like in example), x variable inside the function becomes 1.976262583365e-323#DEN

Why does that happen?

Answer 1

You’re using the old style of function declaration, where the argument types are listed separately. See C function syntax, parameter types declared after parameter list

As you are passing an int, it’s not being converted to float by default. Your function, though, is interpreting the argument as a float, which gives undesirable results.

The modern approach is to include the argument type in the function declaration, which will allow your int argument to be automatically converted to a float.

float square_root(float x)
{
     .......
}