试图在c中将十六进制格式化为小端

Question

所以我正在尝试获取缓冲区中值的总和，但是，需要重新排列这些值才能这样做。本质上，我如何让 C 将此数组中的值转换为：uint16_t inp[] = {0xFF,0x81,0xFD,0x00,0x00,0x00,0x00,0x08,0x00,0x00}; 看起来像此数组：uint16_t buff[] = {0x81FF,0x00FD,0x0000,0x0800,0x0000}

Answer 1

感谢@kaylum，我明白了：

 #include <stdio.h> 
 #include <stdint.h>
 
uint16_t inp[] = {0xFF,0x81,0xFD,0x00,0x00,0x00,0x00,0x08,0x00,0x00};
uint16_t buff[sizeof(inp)/2];


int main(){
    int sum = 0;
    
    for(int i = 0; i < sizeof(buff);i=i+2){
        buff[i] = ((inp[i+1] | 0x00) << 8) | (inp[i] | 0x00);
        //printf("%.4X\n",buff[i]);
    }
    
    for(int i = 0; i < sizeof(buff);i++){
        sum = sum += buff[i];
    }
    printf("%u\n",sum);
    return 0;
}

试图在c中将十六进制格式化为小端

trying to format hex into little endian in c

c

hex

endianness