c - 对指针进行类型转换以更改其大小(以字节为单位)

Question

我正在尝试使用指针在充满十六进制值的内存区域中移动。我使用 uint8_t 指针来确保它一次只指向 1 个字节。我想转到一个特定位置(从起始位置偏移字节数)并存储一组长度为 1、2、4 或 8 字节的字节。然后我想将该值解释并打印为有符号或无符号十进制(由枚举参数确定)。

我认为我可以简单地将指针从 1 字节 uint8_t 更改为字节数的正确大小(例如 uint64_t 为 8 btyes)然后存储该值(因为我不能只打印它一次一个字节，我需要评估整个字节/字节组)。这是我目前所拥有的:

void showValueAtOffset(FILE* Out, const uint8_t* const baseAddr, uint32_t Offset, Sign Sgn, uint8_t nBytes) {//------------------doesn't work----------------
   uint8_t *p = baseAddr;//create a pointer that points to the first byte of memory in the array
   for(int i = 0; i < Offset; i++){//use a for loop to move p to the location indicated by Offset 
       p++;
   }
   if(nBytes == 1){
       //pointer p already has the correct typecast for the number of bytes
       uint8_t N = *p;//store the value of the byte
       if(Sgn == SIGNED){
           int8_t result = N;
           fprintf(Out, "%d  ", result);//print the value
       }
       else{//if UNSIGNED
           fprintf(Out, "%u  ", N);//print the value
       }
   }
   else if(nBytes == 2){
       uint16_t q = (uint16_t) p;//create the pointer q with the correct typecast for the number of bytes
       uint16_t N = *q;//store the value of the bytes
       if(Sgn == SIGNED){
           int16_t result = N;
           fprintf(Out, "%d  ", result);//print the value
       }
       else{//if UNSIGNED
           fprintf(Out, "%u  ", N);//print the value
       }
   }
   else if(nBytes == 4){
       uint32_t q = (uint32_t) p;//create the pointer q with the correct typecast for the number of bytes
       uint32_t N = *q;//store the value of the bytes
       if(Sgn == SIGNED){
           int32_t result = N;
           fprintf(Out, "%d  ", result);//print the value
       }
       else{//if UNSIGNED
           fprintf(Out, "%u  ", N);//print the value
       }
   }
   else if(nBytes == 8){
       uint64_t q = (uint64_t) p;//create the pointer q with the correct typecast for the number of bytes
       uint64_t N = *q;//store the value of the bytes
       if(Sgn == SIGNED){
           signed int result = (signed int) N;
           fprintf(Out, "%d  ", result);//print the value
       }
       else{//if UNSIGNED
            unsigned int result = (unsigned int) N;
            fprintf(Out, "%u  ", result);//print the value
       }
   }
   else{
       //this should not happen according to the preconditions
   }
   fprintf(Out, "\n");
}

这行不通，我收到了几个错误，例如“无效的类型争论‘unary *’(有‘uint32_t’)”和“警告:从指针转换为不同大小的整数”。

我做错了什么？

Answer 1

错误的直接来源是您应该使用uint16_t *q = (uint16_t *) p;，而不是uint16_t q = (uint16_t) p;，即将一个指针类型强制转换为另一个指针类型，然后再解析它。

仍然，修改后的代码依赖于机器架构，存在大/小端和对齐的问题。简而言之，该代码将在 x86 机器上运行，但可能会由于未对齐的内存访问而崩溃，或者在其他体系结构上产生错误的数字。一种可移植的方式是这样的:

uint32_t to_uint32_be(const void* num)
{
    const uint8_t *p = num;
    uint32_t res = 0U;
    int i;
    for (i = 0; i < 4; i++)
        res = (res << 8) + p[i];
    return res;
}

在上面的代码中，'be'代表big-endian，即数据以网络字节顺序存储，最高有效字节在前。