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*“.

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