C中的字符串反转：我做错了什么？

Question

我正在读K＆amp; RC，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，我有一个困扰的错误，其中，最糟糕的是，我是无法调试 - 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;
}

（请原谅那些乱丢代码逻辑的调试语句。除此之外，请随时纠正您可能看到的任何错误，并随时提出建议以改进它）

现在，我的代码正在运行（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）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

（请注意$）

和第二个：

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

这里描述得更准确：

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，或者可能会给我一个暗示，为什么我面对这个令人沮丧的错误？

Answer 1

您正在返回一个本地变量：

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

在函数reversed返回后，在堆栈上分配的局部变量reverse不再存在。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. 创建函数void并修改输入数组。

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

3. 动态分配（以及稍后取消分配）reversed

Answer 2

总是一样的错误，一遍又一遍......

我

char reversed[size + 1];
// ...
return reversed;

你返回一个自动数组，一旦函数返回就超出范围 - 未定义的行为。将reversed变为static变量以避免这种情况（然后在您开始相信它的魔力之前阅读static关键字。）

II。

char string[256];
scanf("%s", string);

当输入带有空格的字符串时，潜在的缓冲区溢出和错误。将其更改为

fgets(string, sizeof(string), stdin);

III。

char reversed[size + 1];
// ...
reversed[size + 1] = '\0';

另一个缓冲区溢出。在C中，数组从0开始索引。

---

是时候读一本好的C书了。

Answer 3

除了codaddict的帖子和H2CO3的好解释之外，还有一个错误：

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

这将导致出界索引。说size = 10，然后size +1 =11。 char数组reversed的索引值是 0,1,2,3,...,10。因此，reversed[11]会让您遇到麻烦。

Answer 4

好的，注释上面发现的错误加上我自己的答案：

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;

    /* BUGBUG: You are using a stack variable to store the return string.
     * char *reversed = malloc(sizeof(char) * (size + 1))
     * to allocate a string that can be returned safely.
     */
    char reversed[size + 1];

    for(counter = size - 1; counter >= 0; counter--) {
        /* BUGBUG: You are setting the wrong char in "reversed", it should be
         * reversed[size - 1 - counter].  You want the "size - 1" char from the original
         * to be copied to the 0 position of the reversed
         */
        reversed[size - counter] = string[counter];
        printf("DEBUG: The character copied now was %c and was at index %d\n", string[counter], counter);
    }

    /* BUGBUG: You are setting the null character one past the end of the string.
     * Here you want reversed[size], which with 0-indexing is the size+1'st
     * character.
     */
    reversed[size + 1] = '\0';

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

    /* BUGBUG: Just to stress this -- you cannot expect this to work; that it
     * does so is accidental because the call stack is not cleaned.  If the calling
     * function called another function (say printf) then the printf is likely to 
     * overwrite the contents of reversed.
     */
    return reversed;
}