反转c中的数组 [英] Reverse an array in c

问题描述

我检查了一些用C语言反转数组的视频，因此我知道我的逻辑是正确的.这是我的代码:

I checked some videos of reversing an array in C therefore I know that my logic is correct. Here's my code:

#include <stdio.h>
void reverse( int arr[], unsigned int len )
{
int i=0;
int n=len;
int j=len-1;
int temp;

while (i<n) 
    {
        temp=arr[i];
        arr[i]=arr[j];
        arr[j]=temp;
        i++;
        j--;
    }
}

void reverse( int arr[], unsigned int len ); 

int main( void )
{
  int a[] = {11, 19, 13};
  unsigned int len = 3;

  reverse( a, len );

  int k=0;
  for( k=0; k<len; k++ )
  {   
     printf( "%d ", a[k] );
  }
  printf( "\n" ); 

  return 0;
}

它输出相同的数组.我找不到问题所在.我应该为reverse函数返回一些东西吗?

It outputs the same array. I couldn't find where the problem is. Should I return something for the reverse function?

推荐答案

您将阵列反转了两次.这就是为什么您拥有原始数组.

You are reversing the array twice; that's why you have the original array.

如何?以一个10元素的数组为例.第一步，将交换第0个和第9个元素.最后一步，您将交换第9个和第0个元素.最终结果:没有变化.

How? Take for an example a 10-element array. As your first step, you'll swap 0th and 9th element. As your last step, you'll swap 9th and 0th element. Net result: no change.

您不想将i最多运行到n.您想将i最多运行到j，这样就永远不会将元素交换回来(即，将条件更改为while (i < j) { ... })

You don't want to run i up to n. You want to run i up to j, so that you never swap elements back (i.e. change the condition to while (i < j) { ... })

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