Difference between ++*p, *p++ and *++p

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 second program is “arr[0] = 10, arr[1] = 20, *p = 20“.