Question

我想从unsigned int获取unsigned char *的值，我的逻辑给出了错误的值

unsigned int Address = 0x0a5b0644; // this value is getting ss fun parameter
m_Address        = (unsigned char*)Address;  // this logic wrote in side C++ file

unsigned char * m_Address; // this is declared inside H file

这里m_Address不是gettting值0x0a5b0644

我可以知道这样做吗

Answer 1

从整数类型转换为指针将是实现定义。

如果您确实需要无符号整数类型来存储指针，请在uintptr_t中使用intptr_t（或<cstdint>表示有符号整数类型）。它们作为可选功能在 C ++ 11 中引入。他们能够转换为void *类型并转换回来。

Answer 2

您可以尝试以下操作假设unsigned int是4个字节，{0x0a，0x5b，0x06,0x44}

unsigned char bytes[11];// the number of charactoer in the string + a null terminating
sprintf(bytes, "%#04x", Address); //<here it will print it

将无符号整数转换为unsigned char *

2 个答案: