数组赋值在C使用指针运算 [英] Array Assignments in C Using Pointer Arithmetic

问题描述

我怎样才能改变一个数组的值，当我使用指针算法访问一个特定元素？

 的#include＆LT;＆stdio.h中GT;诠释主（）{
  诠释一个[3] = {1，1，1}，B [3] = {2，2,2};  一个++ = B ++; //我怎样才能得到这个工作，所以[1] = B [1]？  返回0;
}
 

解决方案

数组不是指针。重复此三次;的数组不是指针的

您不能增加一个数组，它不是一个分配的值（即，可以发生变异的话）。当然你也可以索引到它得到一个值返回：

  A [1] = B [1];
 

其次，当前的code试图递增，然后分配一个新值的阵列本身的，当你的意思是要分配给数组的元素。在需要时数组降解为指针，所以这个工程太：

 为int * a_ptr = A;
为int * B_PTR = B;
* ++ a_ptr = * ++ B_PTR;
//或者更好的...
a_ptr [1] = B_PTR [1];
 

这就是你的意思做。我和，往往不是preFER版本1，使用索引的指针，以及因为它往往是更容易阅读。

How can I change the value in an array when I access a particular element using pointer arithmetic?

#include <stdio.h>

int main() {
  int a[3] = {1, 1, 1}, b[3] = {2, 2, 2};

  a++ = b++; // How can I get this to work so a[1] = b[1]?

  return 0;
}

解决方案

Arrays are not pointers. Repeat this three times; arrays are not pointers.

You cannot increment an array, it is not an assignable value (i.e., you cannot mutate it). You can of course index into it to get a value back:

a[1] = b[1];

Secondly, your current code is attempting to increment and then assign a new value to the array itself, when you meant to assign to an element of the array. Arrays degrade to pointers when required, so this works too:

int *a_ptr = a;
int *b_ptr = b;
*++a_ptr = *++b_ptr;
// or, better...
a_ptr[1] = b_ptr[1];

Which is what you meant to do. I prefer version 1 and, more often than not, use indexing with pointers as well because it is often easier to read.

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