下面的代码在第2行接收到seg错误:
char *str = "string";
str[0] = 'z'; // could be also written as *str = 'z'
printf("%s\n", str);
虽然这个方法非常有效:
char str[] = "string";
str[0] = 'z';
printf("%s\n", str);
用MSVC和GCC测试。
当前回答
char *str = "string";
分配一个指向字符串字面量的指针，编译器将其放入可执行文件中不可修改的部分;
char str[] = "string";
分配并初始化一个可修改的本地数组
其他回答
The C FAQ that @matli linked to mentions it, but no one else here has yet, so for clarification: if a string literal (double-quoted string in your source) is used anywhere other than to initialize a character array (ie: @Mark's second example, which works correctly), that string is stored by the compiler in a special static string table, which is akin to creating a global static variable (read-only, of course) that is essentially anonymous (has no variable "name"). The read-only part is the important part, and is why the @Mark's first code example segfaults.
当您试图访问不可访问的内存时，会导致分割错误。
Char *str是一个指向不可修改的字符串的指针(这是导致segfault的原因)。
而char str[]是一个数组，可以修改。
参见C常见问题，问题1.32
Q: What is the difference between these initializations? char a[] = "string literal"; char *p = "string literal"; My program crashes if I try to assign a new value to p[i]. A: A string literal (the formal term for a double-quoted string in C source) can be used in two slightly different ways: As the initializer for an array of char, as in the declaration of char a[] , it specifies the initial values of the characters in that array (and, if necessary, its size). Anywhere else, it turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. In an expression context, the array is converted at once to a pointer, as usual (see section 6), so the second declaration initializes p to point to the unnamed array's first element. Some compilers have a switch controlling whether string literals are writable or not (for compiling old code), and some may have options to cause string literals to be formally treated as arrays of const char (for better error catching).
这些答案大部分都是正确的，但为了更清楚一点……
人们所说的“只读内存”是ASM术语中的文本段。它是内存中加载指令的同一个地方。出于安全等明显的原因，这是只读的。当创建一个初始化为字符串的char*时，字符串数据被编译到文本段中，程序初始化指向文本段的指针。所以如果你想改变它，就死定了。段错误。
当作为数组编写时，编译器将初始化的字符串数据放在数据段中，这与全局变量等存在的位置相同。这个内存是可变的，因为数据段中没有指令。这一次，当编译器初始化字符数组(仍然只是一个char*)时，它指向的是数据段而不是文本段，您可以在运行时安全地更改文本段。
The
char *str = "string";
Line定义了一个指针，并将其指向一个字面值字符串。字面值字符串是不可写的，所以当你这样做:
str[0] = 'z';
你会得到一个隔离失误。在某些平台上，字面值可能位于可写内存中，因此您不会看到段错误，但无论如何它都是无效代码(导致未定义的行为)。
线:
char str[] = "string";
分配一个字符数组并将字面值字符串复制到该数组中，该数组是完全可写的，因此后续更新没有问题。
