i have a doubt in c

Question

why does this code outputs 1
int main(void)
{
int c=0;
unsigned int a = 0;
for(a=0; a<10; a--)
c++;
printf("%d", c);
return 0;
}

i know that we have unsigned int here that cant go below 0 so it will go highest integer limit and end the loop there but when i do this :

int main(void)
{
unsigned int a = 0;
a--;
printf("%d", a);
return 0;
}
this prints -1 . well what is going on i think i am missing something
HELP HERE......

Answer 1

ohh man ..........how can i forget that .......thank u bro, so much .

Answer 2

You need `printf("%u")` if you want to print unsigned integers, %d is for signed numbers.

A signed -1 is `1111 1111` in binary which happens to be how the biggest unsigned number looks in binary aswell.

So basically you are doing a "conversion" from unsigned to signed before printing.

Answer 3

The loop condirion in first snippet defines that loop goes as long as <a> is less than 10, and during loop <c> will be incremented. As you decrement <a> from 0 to 4294967295 the condition becomes false, and the loop only goes around once. That is how you got 1 in output ...