C++指针解释
char* pstr[] = { "Robert Redford", // Initializing a pointer array
"Hopalong Cassidy",
"Lassie",
"Slim Pickens",
"Boris Karloff",
"Oliver Hardy"
};
如果我像下面这样写：
*pstr[0] = 'X';
程序可以编译，但执行此语句时崩溃。为什么？我认为 *pstr[0] 是“R”，这样我就可以从“R”更改为“X”。 谢谢！
char* pstr[] = { "Robert Redford", // Initializing a pointer array
"Hopalong Cassidy",
"Lassie",
"Slim Pickens",
"Boris Karloff",
"Oliver Hardy"
};
If I write like below:
*pstr[0] = 'X';
The program can compile but crashes when this statement is executed. Why? I thought *pstr[0] is 'R' so that I can change from 'R' to 'X'.
Thanks!
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编译器应该警告您：
您在这里所做的是将一些常量 char 数组分配给可变 char 指针，很像：
结果是，在运行时，一个指向您的二进制文件的指针，并且您正在编写到。但那是只读内存，所以......崩溃。
The compiler should have warned you:
What you're doing here is assigning some constant char arrays to a mutable char pointer, much like:
The result is, at run time, a pointer that points into your binary, and that you're writing to. But that's readonly memory, so.. crash.
您正在创建一个指向
const char*数组的指针。所以 pstr[0] 指向一个 const char* 并且你不能改变它的值。以这种方式编写此代码是一个很好的编程习惯，因此如果您尝试更改该值而不是运行时错误，您将收到编译器错误：
You are making a pointer to an array of
const char*s. So pstr[0] is pointing to aconst char*and you can't change it's value.It's a good programming practice to write this code in this way so you'll get a compiler error if you try to change the value instead of runtime error:
字符串文字是常量，因此您无法更改它们。
允许从
const char*到char*的转换，只是为了允许 C 获得const之前的旧 C 代码。The string literals are constants, so you cannot change them.
Conversion from
const char*tochar*is allowed just to allow old C code from the time before C gotconst.pstr是一个包含const *char元素的数组。修改只读存储器的结果是未定义的。如果您知道每个字符串的最大大小，您可以像这样声明pstrpstris an array withconst *charelements. The result of modifying read-only memory is undefined. If you know the max size of each string you can declarepstrlike so