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c - 为什么在写入使用字符串文字初始化的 "char *s"而不是 "char s[]"时出现段错误?

转载 作者:太空狗 更新时间:2023-10-29 16:13:39 26 4
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以下代码在第 2 行收到段错误:

char *str = "string";
str[0] = 'z'; // could be also written as *str = 'z'
printf("%s\n", str);

虽然这非常有效:

char str[] = "string";
str[0] = 'z';
printf("%s\n", str);

使用 MSVC 和 GCC 测试。

最佳答案

请参阅 C 常见问题解答,Question 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:

  1. 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).
  2. 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).

关于c - 为什么在写入使用字符串文字初始化的 "char *s"而不是 "char s[]"时出现段错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/164194/

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