) What will be the result of following expression?
int p=20, k=10;
p=p*++k;
System.out.print (p="+p)
segment?
Answers
Answer:
Predict the output of following C programs.
// PROGRAM 1
#include <stdio.h>
int main(void)
{
int arr[] = {10, 20};
int *p = arr;
++*p;
printf("arr[0] = %d, arr[1] = %d, *p = %d",
arr[0], arr[1], *p);
return 0;
}
// PROGRAM 2
#include <stdio.h>
int main(void)
{
int arr[] = {10, 20};
int *p = arr;
*p++;
printf("arr[0] = %d, arr[1] = %d, *p = %d",
arr[0], arr[1], *p);
return 0;
}
// PROGRAM 3
#include <stdio.h>
int main(void)
{
int arr[] = {10, 20};
int *p = arr;
*++p;
printf("arr[0] = %d, arr[1] = %d, *p = %d",
arr[0], arr[1], *p);
return 0;
}
The output of above programs and all such programs can be easily guessed by remembering following simple rules about postfix ++, prefix ++ and * (dereference) operators
1) Precedence of prefix ++ and * is same. Associativity of both is right to left.
2) Precedence of postfix ++ is higher than both * and prefix ++. Associativity of postfix ++ is left to right.
(Refer: Precedence Table)
The expression ++*p has two operators of same precedence, so compiler looks for assoiativity. Associativity of operators is right to left. Therefore the expression is treated as ++(*p). Therefore the output of first program is “arr[0] = 11, arr[1] = 20, *p = 11“.
The expression *p++ is treated as *(p++) as the precedence of postfix ++ is higher than *. Therefore the output of second program is “arr[0] = 10, arr[1] = 20, *p = 20“.
The expression *++p has two operators of same precedence, so compiler looks for assoiativity. Associativity of operators is right to left. Therefore the expression is treated as *(++p). Therefore the output of third program is “arr[0] = 10, arr[1] = 20, *p = 20“.