Pre/Post Increment Operator

Search…

Q1:
1
int arr[] = {10, 20, 30, 40, 50, 60};
2
int* ptr = arr;
3
*(ptr++) += -1; // 9, 20, 30, 40, 50, 60
4
*(++ptr) += -2; // 9, 20, 28, 40, 50, 60
5
ptr += 2;
6
*ptr += -3;// 9, 20, 28, 40, 47, 60
Copied!
Ans:
9, 20, 28, 40, 47, 60
Demo:
1
#include <stdio.h>
2
​
3
int main(){
4
    int arr[] = {10, 20, 30, 40, 50, 60};
5
    int* ptr = arr;
6
    *(ptr++) += -1; // 9, 20, 30, 40, 50, 60
7
    *(++ptr) += -2; // 20, 20, 28, 40, 50, 60
8
    ptr += 2;
9
    *ptr += -3;// 20, 20, 40, 40, 47, 60
10
    
11
    for(int i = 0; i < sizeof(arr) / sizeof(arr[0]); i++){
12
        printf("arr[%d] = %d \n", i, arr[i]);
13
    }
14
​
15
    return 0;
16
}
17
​
Copied!
Explanation:
int* ptr = arr;
1
     ptr V  
2
arr[] = {10, 20, 30, 40, 50, 60};
Copied!
*(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
(1)
2
a = ptr++
3
ptr的值先給 a, ptr再加 1
4
​
5
a = ++ptr
6
ptr先加 1, ptr的值再傳給 a
7
​
8
​
9
(2)
10
*(ptr++)
11
++在 ptr 後面, 他會先賦值再移動指標
12
​
13
*(++ptr)
14
++在 ptr 前面, 他會先移動再賦值
Copied!

Basic Concept in C - Previous

Understand about pointer

Next - Basic Concept in C

string and constant charaters

Last modified 4mo ago

Copy link

Contents

Q1:

Explanation:

Summary: