如何将uint16数组转换为uint8数组

问题描述

将具有5个元素的uint16数组转换为uint8数组

[edit]
从OP注释到解决方案之一(更详尽的解释):
感谢答谢，但我想将uint16的两个字节分开，然后我将得到的总共10个字节应该放在uint8的数组中"
[/edit]

to convert array of uint16 having 5 elements to array of uint8

[edit]
From OP in comment to one of the solutions (explains more thoroughly):
"thanx for replying but i want to separate both bytes of uint16 and then the total 10 bytes which , i will get should be in array of uint8"
[/edit]

推荐答案

uint16 arr1[5] = {2, 3, 20, 90, 3857};
uint8  arr2[10] = {0};

uint8* ptr = (uint8 *) &arr1; //cast the 16bit pointer to an 8bit pointer
for(int i=0; i<10; i++)
{
 arr2[i] = *ptr; //pass data to other array
 ptr++;          //move your pointer
}

for(int i = 0; i < 10; i++)
{
  cout << ptr[i] << endl;
}

最好的问候
Espen Harlinn

Best regards
Espen Harlinn

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