将 char 指针转换为 unsigned char 数组

Question

我想将一个 char 指针转换为一个 unsigned char var，我以为我可以通过强制转换来做到这一点，但它不起作用:

char * pch2;
//Code that puts something in pc2
part1 = (unsigned char) pch2;

我有代码:

result.part1 = (unsigned char *) pch2;
printf("STRUCT %s\n",result.part1);

result 只是一个带有 unsigned char 数组的结构。

编辑:

            pch2 = strtok( ip, "." );

            while( pch2 != NULL ){
                printf( "x %d x: %s\n", i, pch2 );
                pch2[size-1] = '\0';

                if(i == 1)
                    result.part1 = (unsigned char *) pch2;
                if(i == 2)
                    result.part2 = (unsigned char *) pch2;
                if(i == 3)
                    result.part3 = (unsigned char *) pch2;
                if(i == 4)
                    result.part4 = (unsigned char *) pch2;
                i++;
                pch2 = strtok (NULL,".");
            }   
            printf("STRUCT %c\n",result.part1);

结构:

typedef struct
{
    unsigned char part1;
    unsigned char part2;
    unsigned char part3;
    unsigned char part4;
} res;

Answer 1

您转换为 unsigned char 而不是 unsigned char* 您忘记了 *

part1 = (unsigned char*) pch2;

如果 pch2 不是 null 终止程序将崩溃，如果你幸运的话，当你使用 strlen 时，所以你需要在使用打印之前先将它 null 终止pch2，试试这个:

pch2[size-1] = '\0';  /* note single quote */
result.part1 = (unsigned char *) pch2;

更新:像这样定义你的结构:

typedef struct
{
    const char *part1;
    const char *part2
    const char *part3;
    const char *part4;
} res;

并在完全不强制转换的情况下分配给它:

result.part1 = pch2;