sizeof typedef指针 [英] sizeof typedef pointer

问题描述

我有一个这样定义的结构:

I have a struct that defined like that:

typedef struct my_struct
{
    int numbers[10];
}
*my_struct;

有没有办法找出它的大小?

Is there a way to find out its size?

sizeof(my_struct);// return size of a pointer

推荐答案

结构类型本身使用struct进行拼写，因此您可以说:

The struct type itself is spelled with struct, so you can say:

sizeof (struct my_struct)

如果您还没有给结构命名，那将是不可能的:

This would not work if you hadn't also given your struct a name, which would have been possible:

typedef struct { int numbers[10]; } * foo;  /* struct type has no name */
foo p = malloc(1000);
p->numbers[3] = 81;

我会说所有这些都是可怜的代码，它们毫无理由地简洁.我只是保持所有名称唯一，并为所有名称命名，而不是为别名指针命名.例如:

I'd say all of this is poor code that is needlessly terse for no reason. I would just keep all the names unique, and name everything, and not alias pointers, for that matter. For example:

typedef struct my_struct_s my_struct;

my_struct * create_my_struct(void);
void destroy_my_struct(my_struct * p);

struct my_struct_s
{
    int numbers[10];
};

所有内容都有唯一的名称，typedef与struct定义分开，并且指针是显式的.

Everything has a unique name, the typedef is separate from the struct definition, and pointers are explicit.

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