c - C 中的字符串反转 : What is it I am doing wrong?

Question

我正在阅读 K&R C，主要是为了刷我的 C 技能，并且在尝试编写一个程序来反转给定的字符串时，我遇到了一个困扰我的错误，最糟糕的是，我无法调试 - 也不知道可能是什么原因。

我的代码如下:

#include <stdio.h>
#include <string.h>

char * reverse(char *string);

int main(int argc, char *argv[])
{
    printf("Please input a string: \t");

    char string[256];

    scanf("%s", string);

    char *reversed = reverse(string);

    printf("The reversed string is %s\n", reversed);

    return 0;
}

char * reverse(char string[])
{
    int size = strlen(string);
    printf("DEBUG: The size of the string that we got as input was: %d\n", size);
    int counter;
    char reversed[size + 1];

    for(counter = size - 1; counter >= 0; counter--) {
        reversed[size - counter] = string[counter];
        printf("DEBUG: The character copied now was %c and was at index %d\n", string[counter], counter);
    }

    reversed[size + 1] = '\0';

    printf("DEBUG: The reversed string is %s\n", reversed);

    return reversed;
}

(请原谅乱七八糟的调试语句。除此之外，随时纠正您可能看到的任何错误，也随时提出改进建议)

现在，我的代码可以正常工作(大部分)，但错误是它复制了我没有输入的字符。以下是两次测试运行的(有趣的)结果:

第一个:

nlightnfotis@crunchbang:~/SoftwareExperiments$ ./reverse
Please input a string:  fotis
DEBUG: The size of the string that we got as input was: 5
DEBUG: The character copied now was s and was at index 4
DEBUG: The character copied now was i and was at index 3
DEBUG: The character copied now was t and was at index 2
DEBUG: The character copied now was o and was at index 1
DEBUG: The character copied now was f and was at index 0
DEBUG: The reversed string is $sitof
The reversed string is $sitof

(注意 $)

第二个:

nlightnfotis@crunchbang:~/SoftwareExperiments$ ./reverse
Please input a string:  lol
DEBUG: The size of the string that we got as input was: 3
DEBUG: The character copied now was l and was at index 2
DEBUG: The character copied now was o and was at index 1
DEBUG: The character copied now was l and was at index 0
DEBUG: The reversed string is lol
The reversed string is lol

此处描述更准确:

比我更有知识和经验的人可以向我解释我的代码有什么问题，或者给我一些提示，说明为什么我会遇到这个令人沮丧的错误吗？

Answer 1

你正在返回一个局部变量:

char * reverse(char string[]) {    
  char reversed[size + 1];
  ....
  return reversed;
}

分配在堆栈上的局部变量 reversed 在函数 reverse 返回后不复存在。因此 main 对它的任何引用都会导致未定义的行为。

要解决此问题，您可以执行以下任一操作:

1. 使函数 void 并修改输入数组。

2. 将数组reversed 声明为静态的，以便其生命周期更改为程序的生命周期。

3. 动态分配(然后取消分配)反向