Question

这有点奇怪，但是在将数据从较小的数组复制到较大的数组时，我似乎在执行strncpy时会丢失数据。我很遗憾为什么会这样。请帮忙！这是代码（已从更大的代码中删除）

#include <stdlib.h>     
#include <stdio.h>      
#include <string.h>

#define SIZE_UNSIGNED_LONG_LONG 8
#define DATALEN 2048

int main(int argc, char *argv[])
{

    printf("size of unsigned long long is %d \n", sizeof(unsigned long long));

    //unsigned long long to unsigned char
    unsigned long long a = 3;
    unsigned char cval[8];
    cval[7] = (a & 0xFF);
    cval[6] = ((a >> 8) & 0xFF);
    cval[5] = ((a >> 16) & 0xFF);
    cval[4] = ((a >> 24) & 0xFF);
    cval[3] = ((a >> 32) & 0xFF);
    cval[2] = ((a >> 40) & 0xFF);
    cval[1] = ((a >> 48) & 0xFF);
    cval[0] = ((a >> 56) & 0xFF);

    unsigned char data[2048];
    strncpy((char *)data,(char *)cval, SIZE_UNSIGNED_LONG_LONG);//destination buffer

    //unsigned char to unsigned long long
    unsigned long long counter = 0;
    counter = counter |(data[7]);
    counter = counter |(data[6]<< 8);
    counter = counter |(data[5]<< 16);
    counter = counter |(data[4]<< 24);
    counter = counter |(data[3]<< 32);
    counter = counter |(data[2]<< 40);
    counter = counter |(data[1]<< 48);
    counter = counter |(data[0]<< 56);
    printf("Trial1: counter is = %llu\n", counter);
    counter = 0;
    counter = counter |(data[0]);
    counter = counter |(data[1]<< 8);
    counter = counter |(data[2]<< 16);
    counter = counter |(data[3]<< 24);
    counter = counter |(data[4]<< 32);
    counter = counter |(data[5]<< 40);
    counter = counter |(data[6]<< 48);
    counter = counter |(data[7]<< 56);
    printf("Trial2: counter is = %llu\n", counter);

    return 0;
}

Answer 1

您应该使用 memcpy（），而不是使用 strncpy（）。如果找到null（'\ 0'）字符，则 strncpy（）将剩余字节填充为零。

Answer 2

也许你的问题不存在：

counter = counter |(data[3]<< 32);

发出警告：

left shift count >= width of type

使用strncpy从较小到较大的数据阵列复制时丢失的数据

2 个答案: