C ++在64位平台中传递char数组地址 [英] c++ pass char array address in 64bit platform

问题描述

我有func（），必须带unsigned int参数。

i have func() which must have to take unsigned int parameter.

有一个原因使func（unsigned int val）无效，

there is a reason for void func(unsigned int val),

我必须将各种类型（unsigned char，unsigned short，unsigned int）传递给func参数。

i have to pass various type(unsigned char, unsigned short, unsigned int) to func parameter.

我也必须通过将char数组转换为func（），我的当前解决方案如下面的代码。

and i have to pass char array to func(), my current solution like below code.

编辑：是否有简单的方法在64位平台上移植此代码？

is there easy way to port this code in 64bit platform?

char test_str[128] = { 0 };
void func(unsigned int val)
{
  memcpy(test_str, (char *)val, 128); //current my solution
  printf("%s\n", test_str);
}

int main()
{
  char str[128] = "hello world";
  func((unsigned int)(char *)&str); //current my solution
  return 0;
}

注意：intptr_t

note: intptr_t

推荐答案

您的代码有很多问题，但至少对我来说，编译器首先抱怨的是：

There are a lot of problem with your code, but at least for me the one the compiler first complains about is:

错误：从指针转换为较小类型'unsigned int'会丢失
信息│偏移量：4个字节：
0x7ffe4b93ddd4内容：9 func（（unsigned int）（char *） str）;

error: cast from pointer to smaller type 'unsigned int' loses information │Offset: 4 byte: 0x7ffe4b93ddd4 contents:9 func((unsigned int)(char *)str);

我假设您正在尝试将char数组的字面地址潜入unsigned int参数中。但是，无符号int只能容纳4个字节（在我的平台上），这不足以容纳完整地址，因为所有指针都需要8个字节（同样在我的平台上）。

I assume that you're trying to sneak in the literal address of the char array into the unsigned int parameter. However an unsigned int can only hold 4 bytes (in my platform), that is not enough to hold the full address, since all pointers require 8 bytes (again, in my platform). See for yourself.

#include <stdio.h>

int main (int argc, char  *argv[])
{
    char str[128] = "hello world";
    unsigned int *address = (unsigned int *)str;

    printf("address: %p \t contains: %s \t pointer size: %lu \n", address, (char *)address, sizeof(char *));
    // address: 0x7ffcf9492540          contains: hello world          pointer size: 8
    printf("Size of address in pointer: %lu \n", sizeof(long long *));
    // will print the same size as the past sizeof() operation
    printf("Size of unsigned int variable: %lu \n", sizeof(unsigned int));
    // 4 bytes, not enough to fit in the necessary 8 bytes

    return 0;
}

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