写入 c 字符串 [英] Writing into c-string

问题描述

我的代码段错误，我不知道为什么.

my code segfaults and I don't know why.

 1  #include <stdio.h>
 2
 3  void overwrite(char str[], char x) {
 4    int i;
 5    for (i = 0; str[i] != ''; i++)
 6      str[i] = x;
 7  }
 8
 9  int main(void) {
10    char *s = "abcde";
11    char x = 'X';
12    overwrite(s, x);
13    printf("%s
", s);
14    return 0;
15  }

gdb 调试器告诉我，问题出在第 6 行，我想将一个 char 存储到 c-string 中(如果我使用左值指针解引用，这是同样的问题.)这是他所说的:

The gdb debugger tells me, that problem is on the line 6, where I want to store a char, into c-string (if I use lvalue pointer dereferencing, it's the same problem.) This is what he says:

(gdb) run
Starting program: /tmp/x/x 

Breakpoint 1, overwrite (str=0x8048500 "abcde", x=88 'X') at x.c:5
5         for (i = 0; str[i] != ''; i++)
(gdb) s
6           str[i] = x;
(gdb) 

Program received signal SIGSEGV, Segmentation fault.
0x080483e3 in overwrite (str=0x8048500 "abcde", x=88 'X') at x.c:6
6           str[i] = x;
(gdb) q

我正在从 K&R-C 书中学习，这是第 2.8 章中的简化示例(删除功能).我不知道问题出在哪里.

I am learning from K&R-C book and this is simplified example from chapter 2.8 (the remove function). I have no idea where is the problem.

推荐答案

因为 char*s = "abcde";在只读内存中创建字符串.试试

because char*s = "abcde"; creates string in readonly memory. try

char s[] = "abcde";

解释:char* 是指针，abcde"是在只读内存中创建的 -> 不可变.

explanation: char* is pointer, and "abcde" is created in readonly memory -> immutable.

char[]是数组，完全存放在栈上，从内存中初始化，所以是可变的

char[] is array, which is wholly stored on stack and initialized from the memory, so is mutable

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