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The value of $j$ at the end of the execution of the following C program

int incr (int i)
{
static int count = 0;
count = count + i;
return (count);
}
main () {
int i, j;
for (i = 0; i <= 4; i++)
j = incr (i);
}



is

1. $10$
2. $4$
3. $6$
4. $7$
edited | 959 views

At $i=0, j=0$
At $i=1, j=1$
At $i=2, j=3$
At $i=3, j=6$
At $i=4, j=10$

answered by Boss (34.1k points)
edited by
0
$\frac{(n)*(n+1)}{2}$ Put n = 4.
0
This is because count being a static variable is not reinitialized at every call post the first call of incr(i)
ans 10
answered by Active (5.1k points)