main() { int i; printf("%d", &i)+1; scanf("%d", i)-1

main()
{
int i;
printf("%d", &i)+1;
scanf("%d", i)-1;
}

Question Posted / guest

main() { int i; printf("%d", &i)+1; scanf("%d", i)-1; } ..

Answer / abhishek marshetty

Explanation: printf( ) prints address/garbage of i, scanf() dont have & sign, so scans address for i +1, -1 dont have any effect on code.

Is This Answer Correct ?

11 Yes

0 No

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