C问题：为什么char实际上占用内存中的4个字节？

问题描述

我写了一个小程序，以检查在我的内存中占用多少字节，并且它显示char实际上占用内存中的4个字节。我理解这主要是因为字对齐，并没有看到char只有1个字节的优势。为什么不使用4字节的char？

  int main（void）
 {
 int a; 
 char b; 
 int c; 
 a = 0; 
 b ='b'; 
 c = 1; 
 printf（％p\\\
，& a）; 
 printf（％p\\\
，& b）; 
 printf（％p\\\
，& c）; 
 return 0; 
} 
  

输出：
0x7fff91a15c58
0x7fff91a15c5f
0x7fff91a15c54

更新：
我不相信malloc只会为char分配1个字节，即使sizeof（char）作为参数传递，因为malloc包含一个标题将确保标题是字对齐的。任何意见？

Update2：
如果被要求有效地使用内存而不填充，唯一的办法是创建一个特殊的内存分配器？或者是否可以禁用填充？

解决方案

你有int，char，int

请参见为什么限制字节对齐？

 字节0字节1字节2字节3 
 0x1000 
 0x1004 X0 X1 X2 X3 
 0x1008 
 0x100C Y0 Y1 Y2 
  

如果将它们存储在4字节，字节和4字节的形式，它将采取2个cpu周期来检索 int c 和一些位移，以获得c的实际值正​​确对齐，用作int 。

I wrote a small program to check how many bytes char occupies in my memory and it shows char actually occupies 4 bytes in memory. I understand it's mostly because of word alignment and don't see advantage of a char being only 1 byte. Why not use 4 bytes for char?

int main(void)
{
  int a;
  char b;
  int c;
  a = 0;
  b = 'b';
  c = 1;
  printf("%p\n",&a);
  printf("%p\n",&b);
  printf("%p\n",&c);
  return 0;
}

Output: 0x7fff91a15c58 0x7fff91a15c5f 0x7fff91a15c54

Update:I don't believe that malloc will allocate only 1 byte for char, even though sizeof(char) is passed as argument because, malloc contains a header will makes sure that header is word aligned. Any comments?

Update2:If you are asked to effectively use memory without padding, is the only way is to create a special memory allocator? or is it possible to disable padding?

解决方案

You have int, char, int

See the image here under "Why Restrict Byte Alignment?"http://www.eventhelix.com/realtimemantra/ByteAlignmentAndOrdering.htm

          Byte 0 Byte 1 Byte 2  Byte 3
0x1000
0x1004  X0     X1     X2      X3
0x1008
0x100C         Y0     Y1      Y2

If it had stored them in 4-byte, 1-byte and 4-byte form, it would have taken 2 cpu cycles to retrieve int c and some bit-shifting to get the actual value of c aligned properly for use as an int.

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