assignment works as a condition - c

Question

Consider the following Code,
int i;
while(i=0)
printf("Hello");
Now Generally speaking i=0 is an assignment and not a condition for while to check.
But the GCC compiler lets it go with a warning and even evaluates it correctly (does not execute the print statement).
Why? I usually would do with parenthesis for the truth value but my juniors feel that I am wrong and there is no real reason for the parenthesis in this!
EDIT: Zeroing down on the 'actual' doubt, Please consider the following test case
int callme(){
return 0;
}
int main(int argc,char*argv[]){
int c;
while(c = callme()){
printf("Calling...\n");
}
return 0;
}

Answer 1

The expression i = 0 does 2 things:
Has the side effect of storing o in i
Yields the value 0
I usually would do with parenthesis for the truth value but my juniors
feel that i am wrong and there is no real reason for the parenthesis
in this
It's usually a hint to the compiler meaning "I actually want this, I didn't forget a =, shut up".
For your specific case there's no reason to write if (i = 0): you already know what if (0) does. But it's pretty useful when used as:
if ((i = some_function()))
...

Answer 2

i=0 is always an assignment (unless you have it as part of int i = 0; where it is an initialization). But any non-void expression may appear inside the condition of a while loop and if it evaluates to non-zero, the body of the loop will be executed, and if it is zero, the body of the loop will not be executed.
The notation:
while (i = 0)
printf("Hello\n");
is always equivalent to:
i = 0;
There is very little justification for writing the loop at all.
People do write other expressions:
while (c = getchar())
...process EOF or a non-null character...
But that's usually a bug. It is more likely that you should be writing:
while ((c = getchar()) != EOF)
...process a character - possibly null...
or even:
while ((c = getchar()) != EOF && c != '\0')
...process a non-null character...
The first getchar() loop gets a warning from GCC; the latter two do not because of the the explicit test of the value from the assignment.
The people who write a condition like this:
while ((c = getchar()))
really annoy me. It avoids the warning from GCC, but it is not (IMNSHO) a good way of coding.

Answer 3

When you use an assignment operator such as
a=0;
You assign the value to 'a', and still return the number 0.
To test your question, I tried these lines of codes:
int a;
printf("%d", a=0);
and these lines displayed 0.
Then, I tested another set of codes:
int b;
printf("%d", b=15);
Here, the lines displayed 15.
So, if you do:
while(a=0)
{
printf("zero");
}
The (a=0) statement would return false, thus not displaying anything.
But if you do:
while(a=15)
{
printf("fifteen");
}
The "fifteen" will be displayed endlessly, because the statement (a=15) will return a non zero value, or 15, which is not false, not zero, thus it is true. :)

Answer 4

As cnicutar has told above the assignment also yields the value zero.
Some additional info:
It is a common coding mistake for people to omit an extra '=' whereby the comparison becomes an assignment.
An easy way to avoid this is to write the comparison as below, in which case even if a '=' is missed compiler will give an error
while(0 == i)
{
prinf("Hello");
}