尝试设置字符数组的字符时程序崩溃 [英] Program crashes when trying to set a character of a char array

问题描述

我有这个我的程序的奇怪的行为，我不能弄清楚。我的教授给我一个在我的程序中的缺陷，我只是复制一个字符指针，当我构造一个对象，而不是一个新的副本的整个数组，所以你可以愚弄它。他用类似的代码演示了这个。

I got this weird behavior of my programm, that i cant figure out. My professor showed me a flaw in my programm, where i just copy a char pointer when i construct an object instead of making a new copy of the whole array, so you can fool around with it. He demonstrated this with similar code like this.

对于代码：

char sweat[] ="Sweater";
warenkorb = new WareImKorb(new Textil (205366,4.2,sweat,40),2,warenkorb);
sweat[0] = '\0';

现在如果我改为：

char* sweat ="Sweater";

程序运行正常，直到我试试sweat [0] ='\0';

the program runs fine till i try sweat[0] = '\0'; It simply crahes then.

但是，这是有效的：
char cc [] =Sweater;
char * sweat = cc;

However this works: char cc[] ="Sweater"; char* sweat = cc;

这真的让我很烦，我不明白，为什么版本1不工作。

It is really bugging me, that i dont understand, why version 1 does not work. Hope you guys can help me out, or else i will go crazy wondering about this.

推荐答案

char* sweat ="Sweater";
sweat[0] = '\0'

汗水指向一个CONSTANT数据。 Sweater是const文本数据，驻留在只读存储器中的某处，并且 sweat 指向此数据。 它不会复制它。因此，当你做 sweat [0] ='\0'，它试图改变CONSTANT数据的字符。因此错误。顺便说一句，一个好的编译器应该给出警告，如果你不写 const 在你的声明，如 const char * sweater =Sweater。请参阅此处的警告： http://www.ideone.com/P47vv

Here sweat points to a CONSTANT data. "Sweater" is const literal data, residing somewhere in read-only memory, and sweat points to this data as such. It doesn't make a copy of it. So when you do sweat[0]='\0', it tries to change first character of the CONSTANT data. Hence the error. By the way, a good compiler should give warning if you don't write const in your declaration, as const char* sweater = "Sweater". See the warning here : http://www.ideone.com/P47vv

但是当你写 char sweat [] =Sweater时，会创建一个char数组，从CONSTANT数据复制数据毛衣;该数组的元素本身是可修改的！

But when you write char sweat[] = "Sweater", an array of char is created, copying the data from the CONSTANT data which is "Sweater"; that array's element itself is modifiable!

让我们看一个有趣的事情：因为在第一种情况下，它不会复制const数据，所以无论你声明多少个变量（都指向同一个数据），所有变量的地址都是相同的。请参阅：

Lets see an interesting thing: since in the first case, it doesn't make a copy of the const data, so no matter how many variables you declare (all pointing to the same data), the address would be same for all variables. See this:

#include<cstdio>
int main() {
        char* sweat  ="Sweater";        //notice the warning
        const char* another ="Sweater"; //no warning!
        std::printf("%p\n", sweat);     //print the address
        std::printf("%p\n", another);   //print the address
        return 0;
}

输出：

0x8048610
0x8048610

相同的地址！

在这里查看自己： http：// www.ideone.com/VcyM6

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