结构的尺寸如何与不同的数据类型而变化 [英] how size of a structure varies with different data types

问题描述

我使用Linux 32位操作系统，
和GCC编译器。

I am using Linux 32 bit os, and GCC compiler.

我尝试了三种不同类型的结构。
在第一个结构我已经定义只有一个字符变量。这个结构的大小是1这是正确的。

I tried with three different type of structure. in the first structure i have defined only one char variable. size of this structure is 1 that is correct.

在第二结构中我已经定义只有一个 INT 变量。这里的结构尺寸显示4，这也是正确的。

in the second structure i have defined only one int variable. here size of the structure is showing 4 that is also correct.

但在第三结构时，我定义了一个字符和一个 INT 这意味着总规模应该是5，却是露出8输出谁能请解释结构是如何分配的？

but in the third structure when i defined one char and one int that means total size should be 5, but the output it is showing 8. Can anyone please explain how a structure is assigned?

typedef struct struct_size_tag
{
    char c;
    //int i;
}struct_size;

int main()
{
        printf("Size of structure:%d\n",sizeof(struct_size));
        return 0;
}

输出：结构的大小：1

typedef struct struct_size_tag
{
    //char c;
    int i;
}struct_size;

int main()
{
        printf("Size of structure:%d\n",sizeof(struct_size));
        return 0;
}

输出：结构尺寸：4

typedef struct struct_size_tag
{
    char c;
    int i;
}struct_size;

int main()
{
        printf("Size of structure:%d\n",sizeof(struct_size));
        return 0;
}

输出：

推荐答案

在大小的差别是由于对齐。编译器可以自由选择填充字节，使一个结构的总大小不一定其各元素的总和。

The difference in size is due to alignment. The compiler is free to choose padding bytes, which make the total size of a structure not necessarily the sum of its individual elements.

如果一个结构的填充是不希望的，因为它可能具有与某些硬件要求（或其他原因）接口，编译器通常支持包装结构，因此在填充被禁用。

If the padding of a structure is undesired, because it might have to interface with some hardware requirement (or other reasons), compilers usually support packing structures, so the padding is disabled.

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