计算结构大小而不填充字节的函数 [英] function that calculate Size of Structure without padding bytes

问题描述

我想制作一个无需填充字节即可计算结构大小的函数.

Hi i want to make a function which calculate the size of structure without padding bytes.

示例:

struct test{
    int   x;
    char  y;
    int   z;
};

如果我计算上述结构的大小，我必须得到 9 个字节(即没有填充字节)

If i calculate the size of above structure i must get 9 byte (i.e. without padding bytes)

并考虑我可能不知道 struct 中存在的变量的情况.

And consider the case where i might not know the variable present in struct.

示例:

struct test{
    int   x;
    ....
    ....
    ....
    int   z;
};

所以如果我计算尺寸，它应该给出正确的尺寸.

so if i calculate the size it should give correct size.

是否可以编写这样的函数?我尝试阅读结构，但我找不到任何解决方案.我看到有一些编译器选项可以从中获得，但我不想要任何构建编译器选项.

Is it possible do write such function?I tried reading on structure's but i dint find any solution.I saw there is some compiler option are present from that i can get but i dont want any in build compiler option.

推荐答案

不，C 没有足够的内省来实现这一点.在运行时，程序没有任何信息可以用来了解"结构中的哪些字段，或者它们的类型.

No, C doesn't have enough introspection to make this possible. At run-time, there is no information left that the program can use to "know" which fields are in a structure, or what their types are.

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