如何使用C将长值(32位)拆分为四个char变量(8位)？

Question

我有一个32位长的变量CurrentPosition,我想分成4位,8位字符.我如何在C中最有效地做到这一点？我正在使用8位MCU,8051架构.

unsigned long CurrentPosition = 7654321;
unsigned char CP1 = 0;
unsigned char CP2 = 0;
unsigned char CP3 = 0;
unsigned char CP4 = 0;
// What do I do next? 

我应该用一个指针引用CurrentPosition的起始地址然后再添加两个地址四次的地址吗？

这是小恩迪安.

我还希望CurrentPosition保持不变.

Answer 1

    CP1 = (CurrentPosition & 0xff000000UL) >> 24;
    CP2 = (CurrentPosition & 0x00ff0000UL) >> 16;
    CP3 = (CurrentPosition & 0x0000ff00UL) >>  8;
    CP4 = (CurrentPosition & 0x000000ffUL)      ;

您也可以通过指针访问字节,

unsigned char *p = (unsigned char*)&CurrentPosition;
//use p[0],p[1],p[2],p[3] to access the bytes.

Answer 2

我认为你应该考虑使用一个联盟:

union {
   unsigned long position;
   unsigned char bytes[4];
} CurrentPosition;

CurrentPosition.position = 7654321;

这些字节现在可以访问为:CurrentPosition.bytes [0],...,CurrentPosition.bytes [3]