尝试设置 char 数组的字符时程序崩溃
我的程序出现了这种奇怪的行为，我无法弄清楚.我的教授向我展示了我的程序中的一个缺陷，我只是在构造一个对象时复制一个 char 指针，而不是制作整个数组的新副本，所以你可以玩弄它.他用类似的代码证明了这一点.
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.
代码:
char sweat[] ="Sweater";
warenkorb = new WareImKorb(new Textil (205366,4.2,sweat,40),2,warenkorb);
sweat[0] = '';
现在如果我成功了:
char* sweat ="Sweater";
程序运行良好，直到我尝试汗水[0] = '';然后它就会崩溃.
the program runs fine till i try sweat[0] = ''; It simply crahes then.
但是这有效:char cc[] ="毛衣";char* 汗水 = 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 不起作用.希望你们能帮帮我，否则我会疯了.
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] = ''
这里的sweat 指向一个CONSTANT 数据.毛衣"是常量字面量数据，驻留在只读内存中的某处，而 sweat 就这样指向该数据.它不会复制它.因此，当您执行 sweat[0]='' 时，它会尝试更改 CONSTANT 数据的第一个字符.因此错误.顺便说一句，如果你没有在声明中写 const，一个好的编译器应该会给出警告，例如 const char* shirt = "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]='', 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
但是当你编写char shirt[] = "Sweater" 时，会创建一个char 数组，从"Sweater" 的CONSTANT 数据中复制数据；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(都指向相同的数据)，地址都是相同的所有变量.看到这个:
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
", sweat); //print the address
std::printf("%p
", another); //print the address
return 0;
}
输出:
0x8048610
0x8048610
意味着，两个 printfs 都打印相同的地址！
Means, both printfs print the same address!
在这里看到自己:http://www.ideone.com/VcyM6
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