用指针引用字符串
可能的重复：
为什么简单的 C 代码会收到分段错误？
为什么代码片段2 的行为与片段 1 不同吗？
//Code snippet 1
char pstr[] = "helloworld";
char *p = pstr;
p[2] = 'd';
//Code snippet 2
char *p = "helloworld";
p[2] = 'd'; //error: access violation
PS请原谅我的无知。
Possible Duplicate:
Why does simple C code receive segmentation fault?
Why code snippet 2 doesn't behave like snippet 1?
//Code snippet 1
char pstr[] = "helloworld";
char *p = pstr;
p[2] = 'd';
//Code snippet 2
char *p = "helloworld";
p[2] = 'd'; //error: access violation
P.S Forgive my ignorance.
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评论（4）
第一个片段创建一个 char 数组并将其内容初始化为“helloworld”。然后你改变它的第三个元素。
第二个只是创建一个指向 char 的指针，该指针指向字符串文字。然后您尝试更改该文字的第三个字符。在许多现代编译器生成的代码中，字符串文字是不可写的。
编辑：
GCC 曾经有一个
-fwritable-strings选项，使字符串文字可写，因为周围有依赖于此行为的遗留代码。该选项已在 GCC 4.0 发行版系列中删除。The first snippet creates an array of char and initializes its contents to "helloworld". Then you change its third element.
The second one simply creates a pointer to char that points to a string literal. Then you attempt to change the third character of that literal. String literals are not writable in code produced by many modern compilers.
EDIT:
GCC used to have an
-fwritable-stringsoption that enabled string literals to be writable, since there is legacy code around that depends on this behaviour. That option was removed in the GCC 4.0 release series."helloworld"是一个const char数组。类型系统中存在一个漏洞，允许您使用char*指向它，因为存在大量使用char *指向只读数据的代码，并且这是安全的。但是 const_cast 规则适用，即使您创建了一个指向 const 的指针，您实际上也无法写入 const 数据。
"helloworld"is an array ofconst char. There's a hole in the type system which allows you to point to it with achar*, because a lot of code exists which uses achar *to point to readonly data and this is safe.But
const_castrules apply, you can't actually write to theconstdata even if you make a non-const pointer to it.如果您能告诉我们他们的行为有何不同，将会有所帮助。
但作为猜测，我认为你的问题是第二种形式的“p”指向只读内存中的字符串。尝试通过指针“p”写入将导致程序失败。
我可以告诉你，Gnu c++ 编译器会警告你这一点。
It would help if you could tell us in what way they're behaving differently.
But as a guess, I think your problem is that the second form has 'p' pointing to a string in read-only memory. Attempts to write through the pointer 'p' would result in a program failure.
I can tell you that the Gnu c++ compiler will warn you about this.
我猜当你说“行为不像”时，你的意思是一个抛出非法访问异常（或类似的东西），而另一个给出编译时警告或错误？
答案是，在第一种情况下，您将创建一个指向您自己的内存的指针，并将 c9ontents 复制到其中。此时编译器会忘记它曾经是一个指向静态内存的指针；然而，运行时系统不会忘记。
在另一种情况下，编译器“知道”p 是指向静态内存的指针，因此有机会说“哇，伙计，不能这样做”。
但这只是一个猜测，不知道它到底有什么不同。它还将依赖于编译器和实现。
I'm guessing when you say "doesn't behave like" you mean one throws an illegal access exception (or something similar) while the other gives a compile time warning or error?
The answer is that in the first case, you're creating a pointer to your own memory, and copying the c9ontents into it. At that point the compiler forgets it used to be a pointer to static memory; the run time system, however, doesn't forget.
In the other case, the compiler "knows" p is a pointer to static memory, and so has the chance to say "whoa, dude, can't do that".
But this is a bit of a guess without knowing exactly what it does differently. it's also going to be compiler and implementation dependent.