Question

my code is:

#include<stdio.h>

main() {

short int i=0;

for(i<=5 && i>=-1; ++i; i>0)
    printf("%d, ",i);

return 0;
}

OUTPUT:-

i don’t know from where it starts but end in the sequence

..., -4, -3, -2, -1

can u help me understand the working of this code snippet?

Answer 1

for(i<=5 && i>=-1; ++i; i>0)

is equivalent to:

for(; ++i;)

because i<=5 && i>=-1 and i > 0 expressions have no side-effects.

Now the controlling expression is ++i, it means the loop is executed until ++i is evaluated to 0.

i is a short object so ++i is equivalent to i = (int) i + 1.

When (int) i + 1 is converted to short object i and the value is not representable in a short, the conversion is implementation defined (see C99, 6.3.1.3p3).

In your implementation the behavior is that when the value is not representable in a short, it just wraps around and becomes a huge negative value (SHRT_MIN). The loop is executed repeatedly until the ++i controlling expression is 0.