为什么数组多个接收值比它宣布举行 [英] Why can an array receive values more than it is declared to hold

问题描述

int main(void)
{
    char name1[5];
    int count;
    printf("Please enter names\n");
    count = scanf("%s",name1);
    printf("You entered name1 %s\n",name1);
    return 0;
}

当我进入了超过5个字符，它打印的字符作为我进入，这是超过5个，但字符数组被声明为：

When I entered more than 5 characters, it printed the characters as I entered, it was more than 5, but the char array is declared as:

char name1[5];

为什么发生这种情况。

Why did this happened

推荐答案

由于该字符的存储空间之后存储的地址。这是非常危险的，可能导致死机。

Because the characters are stored on the addresses after the 'storage space'. This is very dangerous and can lead to crashes.

例如。假设您输入的名字：迈克尔和变量名1开始在位于0x1000

E.g. suppose you enter name: Michael and the name1 variable starts at 0x1000.

name1: M       i     c      h      a      e      l     \0
      0x1000 0x1001 0x1002 0x1003 0x1004 0x1005 0x1006 0x1007
      [................................]

分配的空间被示出具有[...]
这意味着从0x1005内存将被覆盖。

The allocated space is shown with [...] This means from 0x1005 memory is overwritten.

解决方案：

仅复制5个字符（包括\\ 0结尾），或检查输入的字符串的长度复制之前。

Copy only 5 characters (including the \0 at the end) or check the length of the entered string before you copy it.

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