C中结构的大小 [英] size of struct in C

问题描述

可能的重复:
为什么不是’t sizeof for struct 等于每个成员的 sizeof 之和?

考虑以下 C 代码:

#include <stdio.h>    

struct employee
{
  int id;
  char name[30];  
};

int main()
{
  struct employee e1;      
  printf("%d %d %d", sizeof(e1.id), sizeof(e1.name), sizeof(e1));
  return(0);
}

输出为:

4 30 36

为什么结构的大小不等于其各个组件变量的大小之和?

Why is the size of the structure not equal to the sum of the sizes of its individual component variables?

推荐答案

编译器可能会为对齐要求添加填充.请注意，这不仅适用于结构体字段之间的填充，还适用于结构体的末尾(以便结构体类型的数组将每个元素正确对齐).

The compiler may add padding for alignment requirements. Note that this applies not only to padding between the fields of a struct, but also may apply to the end of the struct (so that arrays of the structure type will have each element properly aligned).

例如:

struct foo_t {
    int x;
    char c;
};

即使 c 字段不需要填充，结构体通常也会有一个 sizeof(struct foo_t) == 8(在 32 位系统上 -而是具有 32 位 int 类型的系统)，因为在 c 字段之后需要 3 个字节的填充.

Even though the c field doesn't need padding, the struct will generally have a sizeof(struct foo_t) == 8 (on a 32-bit system - rather a system with a 32-bit int type) because there will need to be 3 bytes of padding after the c field.

请注意，系统(如 x86 或 Cortex M3)可能不需要填充，但出于性能原因，编译器可能仍会添加它.

Note that the padding might not be required by the system (like x86 or Cortex M3) but compilers might still add it for performance reasons.

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