Difference between ++*p, *p++ and *++p (original) (raw)

Last Updated : 23 Jul, 2025

Predict the output of following C programs.

C `

// 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; }

C

// 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; }

C

// 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; }

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The output of the 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 associativity. 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 associativity. 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".