使用sizeof分配是否会为结构指针产生错误的大小? [英] Does allocating using sizeof yield wrong size for structure pointers?

问题描述

使用valgrind读取此消息:大小为4的无效读/写

Using valgrind to read this I get: Invalid write/read of size 4

 struct Person{
        char* name;
        int age;
    };

    struct Person* create_person(char *name, int age)
    {
        struct Person* me = (struct Person*)malloc(sizeof(struct Person*));
        assert(me!= NULL); //make sure that the statement is not null
        me->name = name;
        me->age = age;

        return me;
    }

使用带有valgrind的干净日志

Using this got clean log with valgrind

struct Person{
    char* name;
    int age;
};

struct Person* create_person(char *name, int age)
{
    struct Person* me = (struct Person*)malloc(sizeof(struct Person*)+4);
    assert(me!= NULL); //make sure that the statement is not null
    me->name = name;
    me->age = age;

    return me;
}

为什么我应该明确放置sizeof(struct+intSize)以避免此错误? sizeof不能获得结构的整个大小吗?

Why should I explicitly put sizeof(struct+intSize) to avoid this error? sizeof don't get the whole size of a struct?

推荐答案

您在调用malloc时使用了错误的大小.

You are using the wrong size in the call to malloc.

struct Person* me = (struct Person*)malloc(sizeof(struct Person*));
                                                  ^^^^^^^^^^^^^^^

那是指针的大小，而不是对象的大小.您需要使用:

That is a size of a pointer, not the size of an object. You need to use:

struct Person* me = (struct Person*)malloc(sizeof(struct Person));

为避免此类错误，请使用以下模式，并且不要强制转换malloc的返回值(请参见

To avoid errors like this, use the following pattern and don't cast the return value of malloc (See Do I cast the result of malloc?):

struct Person* me = malloc(sizeof(*me));

malloc(sizeof(struct Person*)+4)起作用是一个巧合.您的struct有一个指针和一个int.在您的平台上显示sizeof(int)为4.因此，sizeof(struct Person*)+4恰好与struct Person的大小匹配.

It's a coincidence that malloc(sizeof(struct Person*)+4) works. Your struct has a pointer and an int. It appears sizeof(int) on your platform is 4. Hence, sizeof(struct Person*)+4 happen to match the size of struct Person.

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