c - 为什么我的removeString函数没有删除字符？

Question

我正在完成一本书中的练习，除了 printf 和 scanf 之外，我不允许使用任何库函数。

我编写了一个接受字符串的程序。然后，您可以输入要从字符串中删除的字符数。然后选择在哪个索引处执行此操作。

我的尝试是循环遍历我想要保留在函数末尾的字符，然后用我想要丢失的函数替换它们。

出于某种原因，它的行为恰恰相反。

如果我输入字符串:“错误的儿子”然后选择删除 6 个字符并从数组的第 4 个索引开始，

我应该得到:“儿子”

相反，我得到了“the wro”。

我已经为此苦苦挣扎了 4 个小时。

你能告诉我哪里出错了吗？

#include <stdio.h>

void removeString(char source[], int start, int count);
int stringLength(const char string[]);

    int main(void)
    {
        int start, count;
        char source[20]; 

        printf("What's your string?\n");
        scanf("%[^\n]s", source);

        printf("How many characters do you want to remove?\n");
        scanf("%i", &count);

        // Choose which index to start deletion from
        printf("Where are we starting from?\n");
        scanf("%i", &start);

        // Call function to remove characters from string
        removeString(source, start, count);

        // print out result, stored in source
        printf("Here's the result: %s\n", source);

        return 0;
    }

    void removeString(char source[], int start, int count)
    {
        int i, j;

        // Find length of string
        int length = stringLength(source);


        //loop through the last few characters that we DO want to keep until we hit the end of the string 
        for(i = (start+count); i <= length; i++)
        {
            //loop backwards through the string, from the first of the ones we want to keep, stop at the 'start' of the index
            for(j = (start+count)-1; j <= start; j--)
            {
                // assign 'i' which is the last few characters we want to keep and put them in the place of the characters we want to delete
                source[j] = source[i];
            }
        }

        // assign a null character to end the string once we've finished modifying
        source[count + 1] = '\0';
    }

    int stringLength(const char string[])
    {
        int count =0;

        while(string[count] != '\0')
            count++;

        return count;
    }

Answer 1

您只需要一个 for 循环即可实现这一点。假设我们可以成功避免超出字符串末尾，您可以编写如下内容:

for(i = start; i < start + count && i < length; i++) {
   source[i] = source[i+count];
}

那么你只需要在正确的位置设置一个空字符即可。但是，这不算数。也许是 length-count-1 (或考虑 start+count)，因为您要删除“count”个字符。