将int转换为字节并有效地切换endian

I have to do some int -> byte conversion and switch to big endian for some MIDI data I'm writing. Right now, I'm doing it like:

我需要进行一些int ->字节转换，并切换到big endian，以获取我正在编写的一些MIDI数据。现在，我这样做:

int tempo = 500000;
char* a = (char*)&tempo;

//reverse it
inverse(a, 3);

[myMutableData appendBytes:a length:3];

and the inverse function:

逆函数:

void inverse(char inver_a[],int j)
{
    int i,temp;
    j--;
    for(i=0;i<(j/2);i++)
    {
      temp=inver_a[i];
      inver_a[i]=inver_a[j];
      inver_a[j]=temp;
       j--;
     }
}

It works, but it's not real clean, and I don't like that I'm having to specify 3 both times (since I have the luxury of knowing how many bytes it will end up).

它是有效的，但它不是真正的干净，而且我不喜欢我必须同时指定3次(因为我有足够的空间知道它最终会有多少字节)。

Is there a more convenient way I should be approaching this?

有更方便的方法来处理这个问题吗?

3 个解决方案

#1

Use the Core Foundation byte swapping functions.

使用核心基础字节交换函数。

int32_t unswapped = 0x12345678;
int32_t swapped = CFSwapInt32HostToBig(unswapped);
char* a = (char*) &swapped;
[myMutableData appendBytes:a length:sizeof(int32_t)];

#2

This should do the trick:

这应该能达到目的:

/*
  Quick swap of Endian.    
*/

#include <stdio.h>

int main(){

  unsigned int number = 0x04030201;
  char *p1, *p2;
  int i;

    p1 = (char *) &number;
    p2 = (p1 + 3);


  for (i=0; i<2; i++){
      *p1 ^= *p2;
      *p2 ^= *p1;
      *p1 ^= *p2;
  }


return 0;   
}

You can pack it into a function in whatever way you want to use it. The bitwise swap should compile into some pretty neat assembly :)

你可以用任何你想用的方式将它打包成一个函数。位交换应该编译成一些非常整洁的程序集:)

Hope it helps :)

希望它能帮助:)

#3

int tempo = 500000;

//reverse it
inverse(&tempo);

[myMutableData appendBytes:(char*)tempo length:sizeof(tempo)];

and the inverse function:

逆函数:

void inverse(int *value)
{
    char inver_a = (char*)value;
    int j = sizeof(*value); //or u can put 2
    int i,temp;
    // commenting this j--;
    for(i=0;i<(j/2);i++)
    {
      temp=inver_a[i];
      inver_a[i]=inver_a[j];
      inver_a[j]=temp;
       j--;
     }
}