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

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  • Difficulty Level : Medium
  • Last Updated : 11 Oct, 2021

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;
}

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 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“.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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