Question

The below code runs in an infinte loop.
‘i’ has been intialised with the value 1 and then compared with 0.

So printf() stmt should execute once but it runs infinetly.

unsigned int i = 1;
for (; i >= 0; i--) {
    printf("Hello: %u\n",i);
}

please explain this behaviour.

Answer 1

As other answers said, it’s because it’s unsigned and all. I will tell you an elegant way to do what you want to do with an unsigned integer.

unsigned int i=10;
while(i --> 0) printf("Hello:%u\n", i+1);

This --> is referred to sometimes as the goes to operator. But it’s in fact just -- and >. If you change the spacing, you’ll get

while( i-- > 0 )

My2c