为什么这不给分段违例的错吗？ [英] Why does this not give a segmentation violation fault?

问题描述

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

int noOfIntegers = 2;

struct stack {
  int *s;
  int top;
} st;

void push(int item) {
  st.top++;
  st.s[top] = item;
}

int main() {
  st.s = malloc (2 * (sizeof(int)));
  st.top = -1;
  push(1);
  push(2);
  push(3);
  return 0;
}

我不明白这是为什么不给分段错误，因为我已经malloc分配空间2 INT 的，而且我插入3到数组由指针指向。

I don't understand why this is not giving a segmentation fault, as I have malloc'd space for 2 int's, and am inserting 3 into the array pointed to by the pointer.

推荐答案

的malloc 分配从堆内存空间;你已经在堆中分配2 整数，并写出3 整数。结果

malloc allocates space from the heap memory; you have allocated 2 ints on the Heap, and write 3 ints.

你的程序存储器通常在中等大小的块分配给对内存管理更有效。

Memory for your program is typically allocated in moderately sized chunks to be more efficient for the memory manager.

请求1字节？你可能有8个字节。结果
请求7个字节？你可能有8个字节。结果
要求14个字节？你可能有16个字节。结果
的（或者至少其16字节...）的

Request 1 byte? You probably got 8 bytes.
Request 7 bytes? You probably got 8 bytes.
Request 14 bytes? You probably got 16 bytes.
(or maybe its 16-bytes at a minimum...)

您只的保证的访问您实际要求的内存，但你有时可以编写超出了几个字节，而不会引起问题。只是有没有答应，过多的写操作是安全的。您的可能的获得赛格故障，则可能不会！

You are only guaranteed access to the memory you actually requested, but you can sometimes write beyond that for a few bytes without causing a problem. There's just no promise that excessive writes will be safe. You might get a seg-fault, you might not!

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