What happens if you try to compile and run this program? #include
int main(void) {
int i, t[4];
t[3] = 0;
for (i = 1; i >= 0; i--)
ti] =
= -131 ii
prints ("*", +(11);
return 0;

Your question is poorly formatted

#include <stdio.h>

int main(void) {

int i, t[4];

t[3] = 0;

for (i = 1; i >= 0; i--)

t[i] = t[3] * i;

printf("%d", +t[1]);

return 0;

}

It outputs 0

Explanation:

I'll start my explanation from the third line

This line declares an integer variable i and an array of 4 elements of type integer

int i, t[4];

This line initialize the 3rd index element to 0

t[3] = 0;

The next two lines is an iteration;

The first line of the iteration iterates the value of i in descending order from 1 to 0

for (i = 1; i >= 0; i--)

This line of the iteration calculates t[1] as t[3] * i and t[0] as t[3] * i; Since t[3] is 0; both t[1] and t[0] will be 0

t[i] = t[3] * i;

This line prints t[1] which is 0

printf("%d", +t[1]);

answer; //alcohol hampers your vision greatly, making it harder to perceive things in detail and causing you to be more detached from your surroundings.