Pre/Post Increment Operator

Q1:

int arr[] = {10, 20, 30, 40, 50, 60};
int* ptr = arr;
*(ptr++) += -1; // 9, 20, 30, 40, 50, 60
*(++ptr) += -2; // 9, 20, 28, 40, 50, 60
ptr += 2;
*ptr += -3;// 9, 20, 28, 40, 47, 60

Ans:

9, 20, 28, 40, 47, 60

Demo:

#include <stdio.h>

int main(){
    int arr[] = {10, 20, 30, 40, 50, 60};
    int* ptr = arr;
    *(ptr++) += -1; // 9, 20, 30, 40, 50, 60
    *(++ptr) += -2; // 20, 20, 28, 40, 50, 60
    ptr += 2;
    *ptr += -3;// 20, 20, 40, 40, 47, 60
    
    for(int i = 0; i < sizeof(arr) / sizeof(arr[0]); i++){
        printf("arr[%d] = %d \n", i, arr[i]);
    }

    return 0;
}

Explanation:

int* ptr = arr;

     ptr V  
arr[] = {10, 20, 30, 40, 50, 60};

*(ptr++) += -1;

++在 ptr 後面, 他會先賦值再移動指標 Step 1. *ptr --> 10 Step 2. *ptr = *ptr + (-1) --> arr[0] = 9 Step 3. ptr++ --> this pointer has shifted to arr[1]

*(++ptr) += -2;

++在 ptr 前面, 他會先移動再賦值 Step 1. ++ptr --> this pointer has shifted to arr[2] Step 2. *(++ptr) == arr[2] = 30 Step 3. *(++ptr) = *(++ptr) + (-2) --> arr[2] = arr[2] -2 --> arr[2] = 28

ptr += 2

ptr += 2 --> this pointer has shifted to arr[4]

*ptr += -3

*ptr += -3 --> arr[4] = arr[4] -3 --> arr[4] = 47

Summary:

(1)
a = ptr++
ptr的值先給 a, ptr再加 1

a = ++ptr
ptr先加 1, ptr的值再傳給 a


(2)
*(ptr++)
++在 ptr 後面, 他會先賦值再移動指標

*(++ptr)
++在 ptr 前面, 他會先移動再賦值