问题描述
如果我定义如下，
char *s1 = "Hello";
为什么我不能做下面这样的事情，
why I can't do something like below,
*s1 = 'w'; // gives segmentation fault ...why???
如果我执行以下操作会怎样，
What if I do something like below,
string s1 = "hello";
我可以做类似下面的事情吗，
Can I do something like below,
*s1 = 'w';
推荐答案
因为 "Hello" 创建了一个 const char[].这会衰减为 const char* 而不是 char*.在 C++ 中，字符串文字是只读的.您已经创建了一个指向此类文字的指针，并且正在尝试写入.
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.
但是当你这样做时
string s1 = "hello";
您将 const char* "hello" 复制到 s1 中.不同之处在于在第一个示例中 s1 指向 只读hello"，而在第二个示例中只读hello"被复制到 非常量 s1，允许您访问复制的字符串中的元素以对它们执行您想要的操作.
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.
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 复制到其中
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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