为什么的malloc（sizeof的（指针））工作？

问题描述

这下code正常工作：

This following code works fine:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    struct node{
        int a, b, c, d, e;
    };
    struct node *ptr = NULL;
    printf("Size of pointer ptr is %lu bytes\n",sizeof (ptr));
    printf("Size of struct node is %lu bytes\n",sizeof (struct node));
    ptr = (struct node*)malloc(sizeof (ptr));               //Line 1
//    ptr = (struct node*)malloc(sizeof (struct node));    //Line 2

    ptr->a = 1; ptr->b = 2; ptr->c = 3; ptr->d = 4; ptr->e = 5;
    printf("a: %d, b: %d, c: %d, d: %d, e: %d\n",
            ptr->a,ptr->b,ptr->c,ptr->d,ptr->e);
    return 0;
}

在遵从为：

gcc -Wall file.c

我的问题是：为什么是这样的好

My question is: why is this fine?

的malloc 分配被指定在它的参数的字节数。在这里， sizeof的PTR 是我的64位Linux机器上的8个字节。我以为的malloc 将提供8个字节但它是如何访问所有的变量A，B，C，D，E？它是用gcc只能还是我失去了标准C的东西吗？

malloc allocates the number of bytes which are specified in it's argument. Here sizeof ptr is 8 bytes on my 64-bit linux machine. I thought malloc will provide 8 bytes but then how is it accessing all the variables a,b,c,d,e? Is it with gcc only or am I missing something with standard C?

据我所知，2号线应该在那里，而不是1号线，但无论是线路的正常工作。为什么呢？

As far as I know "Line 2" should be there instead of "Line 1" but either of the line works fine. Why?

推荐答案

您不确定这里的行为。

的malloc 将分配8个字节（如你所说），但这种转换是坏：

malloc will allocate 8 bytes (as you say), but this cast is "bad":

ptr = (struct node*)malloc(sizeof (ptr));

这行之后， PTR 将指向一个内存块，它只有8个字节分配，剩下的都是一些随机的字节。因此，使得

After this line, ptr will point to a memory block, which has only 8 allocated bytes, the rest are some "random" bytes. So, making

ptr->a = 1; ptr->b = 2; ptr->c = 3; ptr->d = 4; ptr->e = 5;

您真正改变一些内存，不仅的malloc分配。

you actually change some memory, not only the allocated by malloc.

在换句话说，你重写内存，你不应该去碰。

In other words, you are rewriting memory, you're not supposed to touch.

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