为什么不能为指针赋值？ [英] Why can't I assign values to pointers?

问题描述

阅读常见问题和其他一切之后，我仍然感到困惑。如果我有一个以这种方式初始化的字符指针：

After reading the faq's and everything else I can find, I'm still confused. If I have a char pointer that is initialised in this fashion:

char * s =Hello world！

字符串在只读内存中，我不能这样更改：

The string is in read-only memory and I cannot change it like this:

*s = 'W';

以使Wello world！这我明白，但我不能，对我的生活，了解如何使它不是只读。我必须使用数组而不是指针吗？像这里？

to make "Wello world!". This I understand, but I can't, for the life of me, understand how to make it NOT read-only. Do I have to use an array instead of a pointer? Like here?

这是我的代码：

char *s = str;
char *e = s;
while (*e != '\0')
e++;
e--;
char *temp;
//Swop the string around
while (s <= e) {
    *temp = *s;
    *s = *e;
    *e = *temp;
    e--;
    s++;
}

错误消息只是一个分段错误。如果这是一个真的愚蠢的问题，提前道歉。

The error message is just a segmentation fault. Apologies in advance if this is a really stupid question.

非常感谢所有的帮助。完成所有的建议后，我得到这个：

Thanks so much for all the help. After taking all your advice, I get this:

void something(char * str) {
    char *store = str;
    char *s = new char[strlen(str) + 1]; //Allocate memory. Nice one.
    strcpy(s, str);
    char *e = new char[strlen(str) + 1];
    strcpy(e, str);
    while (*e != '\0')
    	e++;
    e--;
    char temp; //no longer a pointer
    while (s <= e) {
    	cout << *e;
    	temp = *s;
    	*s = *e;
    	*e = temp;
    	e--;
    	s++;

    }
    delete [] e;
    delete [] s;        
}

但是，函数结尾处的删除似乎是导致自己的分割错误。为什么？

however, the deletes at the end of the function seem to be causing their own segmentation faults. Why?

为了感兴趣，错误是由于在e和s指针递增之后访问它们。一个更简单的解决方案是：

For interest's sake: The faults were due to accessing the e and s pointers after they were incremented. A much simpler solution followed from that:

void something(char * str) {
    char *s = new char[strlen(str) + 1];
    strcpy(s, str);
    char temp;
    int j = strlen(str) - 1;
    for (int i = 0; i <= strlen(str)/2; i++) {
    	cout << s << endl;
    	temp = s[i];
    	s[i] = s[j];
    	s[j] = temp;
    	j--;
    }
    delete [] s;
}

推荐答案

例如：

char buf[128];
const char *src = "Hello World";
strncpy(buf, src, 127); // one less - we always 0-terminate
buf[127] = '\0';

// you can now modify buf
buf[0] = 'W';

您的代码不起作用的原因是您尚未为副本分配任何内存字符串 - 你刚刚做了第二个指针到同一只读存储器。 （然后试图复制它？我不知道其余的代码是做什么）。你需要在某个地方获得一些非只读内存，使用标准库将它复制到更容易

The reason your code doesn't work is that you haven't allocated any memory for the copy of the string - you've just made a second pointer to the same read-only memory. (And then tried to copy it? I'm not quite sure what the rest of the code is doing.) You need to get some non-read-only memory somewhere, and it's much easier to use the standard library to copy it into that new memory, rather than writing the loop yourself.

如果你不知道字符串的长度，你也可以使用malloc（或者，更好的做什么drschnz的回答说，并使用 new char [] ）：

In the case when you don't know the length of the string beforehand, you can also use malloc (or, even better, do what drschnz's answer says and use new char[]):

const char *src = "Hello world";
char *buf = malloc(strlen(src) + 1);   // or = new char[strlen(src) + 1];
strcpy(buf, src);
// you can now modify buf
// later, you need to free it
free(buf);                             // or delete [] buf;

另外，如果你使用C ++，你可以使用std :: string： p>

Also, if you're using C++, you can just use a std::string:

std::string myString("Hello world");
myString[0] = "W";

希望有帮助。

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