为什么我不能写一个字符串字面量，而我*可以*写一个字符串对象？ [英] Why can't I write to a string literal while I *can* write to a string object?

问题描述

如果我定义如下，

  char * s1 =Hello 
  

为什么我不能做下面的事情，

  * s1 ='w'; //给分段错误...为什么？ 
  

如果我执行下面的操作该怎么办，

  string s1 =hello; 
  

我可以这样做，

  * s1 ='w'; 
   
 
解决方案
因为Hello / code>创建一个常量char []。这会衰减到 const char * 而不是 char * 。在C ++中，字符串文字是只读的。 
 
 
 
但是当你这样做
  string s1 =hello; 
  
您将const char *hello复制到 s1 。区别在于第一个示例 s1  指向只读hello，第二个示例只读hello复制到非常量 s1 中，允许您访问复制的字符串中的元素，以做他们想要的。
 
 
 
如果你想对char *做同样的事情，你需要为char数据分配空间并将hello复制到它中。
  char hello [] =hello; //创建一个足够大的char数组来保存hello
 hello [0] ='w'; //写入数组中的第0个字符
  
 
If i define something like below, 
char  *s1 = "Hello";
why I can't do something like below,
*s1 = 'w'; // gives segmentation fault ...why???
What if I do something like below,
string s1 = "hello";
Can I do something like below,
*s1 = 'w'; 

解决方案
Because "Hello" creates a const char[]. This decays to a const char* not a char*. In C++ string literals are read-only. You've created a pointer to such a literal and are trying to write to it.

But when you do
string s1 = "hello";
You copy the const char* "hello" into s1. The difference being in the first example s1 points to read-only "hello" and in the second example read-only "hello" is copied into non-const s1, allowing you to access the elements in the copied string to do what you wish with them.

If you want to do the same with a char* you need to allocate space for char data and copy hello into it
char hello[] = "hello"; // creates a char array big enough to hold "hello"
hello[0] = 'w';           //  writes to the 0th char in the array


                        
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