反向字节为64位值 [英] Reverse bytes for 64-bit value

问题描述

我正在尝试为分配分配64位地址指针的字节，并使用以下代码：

I'm trying to reverse the bytes for a 64 bit address pointer for an assignment and have this code:

char swapPtr(char x){
x = (x & 0x00000000FFFFFFFF) << 32 | (x & 0xFFFFFFFF00000000) >> 32;
x = (x & 0x0000FFFF0000FFFF) << 16 | (x & 0xFFFF0000FFFF0000) >> 16;
x = (x & 0x00FF00FF00FF00FF) << 8  | (x & 0xFF00FF00FF00FF00) >> 8;
return x;
}

但是，这只会弄乱一切。但是，类似的功能可以在64位长的情况下完美运行。指针有什么需要做的吗？

But, it just messes everything up. However, a similar function works perfectly for a 64bit long. Is there something different that needs to be done for pointers?

我进行函数调用的方式是否会引起问题？

Could the way I'm making the function call be an issue?

对于指针：

*(char*)loc = swapPtr(*(char*)loc);

很长一段时间：

*loc = swapLong(*loc);

推荐答案

您不能使用 char x 为指针！！！！ char 只有一个字节长。

You cannot use char x for a pointer!!!! A char is only a single byte long.

您至少需要

unsigned long int swapPtr(unsigned long int x) {

或者更好，使用指针的类型

Or better, use the type of the pointer

void* swapPtr(void* x) {

当您开始移动指针时，编译器很可能会抱怨；在这种情况下，最好将参数显式转换为无符号的64位整数：

Quite likely your compiler will complain when you start bit shifting pointers; in that case you're better off explicitly casting your argument to an unsigned 64 bit integer:

#include <stdint.h>
uint64_t x;

还请注意，您必须使用变量地址进行调用，因此您必须使用

Note also that you have to call with the address of a variable, so you call with

result = swapLong(&loc);

不是 * loc （在 loc 指向的地方-值，而不是地址）。

not *loc (which looks at the place where loc is pointing - the value, not the address).

完整程序：

#include <stdio.h>
#include <stdint.h>

uint64_t swapLong(void *X) {
  uint64_t x = (uint64_t) X;
x = (x & 0x00000000FFFFFFFF) << 32 | (x & 0xFFFFFFFF00000000) >> 32;
x = (x & 0x0000FFFF0000FFFF) << 16 | (x & 0xFFFF0000FFFF0000) >> 16;
x = (x & 0x00FF00FF00FF00FF) << 8  | (x & 0xFF00FF00FF00FF00) >> 8;
return x;
}

int main(void) {
  char a;
  printf("the address of a is 0x%016llx\n", (uint64_t)(&a));
  printf("swapping all the bytes gives 0x%016llx\n",(uint64_t)swapLong(&a));
}

输出：

the address of a is 0x00007fff6b133b1b
swapping all the bytes gives 0x1b3b136bff7f0000

编辑，您可以使用类似的

#include <inttypes.h>

printf("the address of a is 0x%016" PRIx64 "\n", (uint64_t)(&a));

其中，宏 PRIx64 扩展为您需要以十六进制格式打印64位数字的格式字符串。比上面的清洁一点。

where the macro PRIx64 expands into "the format string you need to print a 64 bit number in hex". It is a little cleaner than the above.

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