Question

我不知道我的代码中有什么问题，但是当我编译时，我得到：

warning: passing arg 2 of `strcspn' makes pointer from integer without a cast

以下是代码：

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

#define STR_LEN 50

int main(void) {
    int i = 0, j = 0, length = 0, count1 = 0, count2 = 0, count3 = 0;
    char letter3 = 'a', letter2 = 'a', string[STR_LEN] = { 0 };

    length = strlen(string);
    printf("Enter a sentence: ");
    fgets(string, STR_LEN, stdin);

    for (i = 0; i < length; i++) {
        for (j = 0; j < length; j++) {
            if (string[i] == string[j]) {
                count1++;
            } else {
                count1 = 0;
            }
        }
        if (count1 > count3) {
            count2 = count3;
            count3 = count1;
            letter2 = letter3;
            letter3 = string[i];
        } else
        if (count1 > count2) {
            count2 = count1;
            letter2 = string[i];
        }
    }

    string[strcspn(string, letter2)] = letter3;
    string[strcspn(string, letter3)] = letter2;

    printf("\n %s", string);

    system("pause");

    return 0;
}

该代码应该从用户那里得到一个句子并用第二个公用字母切换句子中最常见的字母。

Answer 1

当前问题

strcspn()函数接受两个字符串作为参数，但是您传递的是字符串和字符。你需要以某种方式将字符转换为字符串。一种方法是：

int sep[2] = "";
sep[0] = letter2;
string[strcspn(string, sep)] = letter3;
sep[0] = letter3;
string[strcspn(string, sep)] = letter2;

但是，第一次调用会将第一次出现letter2更改为letter3;第二个调用会更改letter3的第一次出现（可能是前一次调用中刚刚替换的那个）letter2。这不是转换字符串的完整工作 - 您需要扫描整个字符串进行更改。

实施解决方案

一种可能性是：

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

#define NULL_VALUE '\0'

static inline void map(char *str, int len, int c_old, int c_new)
{
    for (int i = 0; i < len; i++)
    {
        if (str[i] == c_old)
            str[i] = c_new;
    }
}

int main(void)
{
    char buffer[4096];

    printf("Enter a sentence: ");
    if (fgets(buffer, sizeof(buffer), stdin) == 0)
        return 0;
    int length = strlen(buffer);
    if (length > 0)
        buffer[--length] = '\0';

    putchar('\n');
    printf("Original [%s]\n", buffer);

    int count[256] = { 0 };
    for (int i = 0; i < length; i++)
    {
        if (isalpha((unsigned char)buffer[i]))
            count[(unsigned char)buffer[i]]++;
    }

    int max1_count = 0;
    int max2_count = 0;
    char max1_value = '\0';
    char max2_value = '\0';
    for (int i = 0; i < 256; i++)
    {
        if (count[i] > max1_count)
        {
            max2_count = max1_count;
            max2_value = max1_value;
            max1_count = count[i];
            max1_value = i;
        }
        else if (count[i] > max2_count)
        {
            max2_count = count[i];
            max2_value = i;
        }
    }

    /*
    ** Since a string is a sequence of non-null character codes followed
    ** by a null byte, it is safe to use '\0' as the temporary value in
    ** the three-step swap operation
    */
    if (max2_count > 0)
    {
        map(buffer, length, max1_value, NULL_VALUE);
        map(buffer, length, max2_value, max1_value);
        map(buffer, length, NULL_VALUE, max2_value);
    }

    printf("Revised  [%s]\n", buffer);

    return 0;
}

使用宏NULL_VALUE的唯一原因是三条map()行的对称性是不言而喻的。

示例运行

我调用了程序ccswap19，我使用Bash'here strings'来提供数据 - putchar('\n');表示输出显示在提示的单独行上。如果您以交互方式运行程序，则在“原始”打印之前会有一个空行。

$ ccswap19 <<< "The hidden costs of the exodus are now revealed for all to see."
Enter a sentence: 
Original [The hidden costs of the exodus are now revealed for all to see.]
Revised  [Tho hiddon cests ef tho oxedus aro new rovoalod fer all te soo.]
$ ccswap19 <<< "aaaaaaaaaaaa"
Enter a sentence: 
Original [aaaaaaaaaaaa]
Revised  [aaaaaaaaaaaa]
$ ccswap19 <<< "aaaabaaaaaaa"
Enter a sentence: 
Original [aaaabaaaaaaa]
Revised  [bbbbabbbbbbb]
$

Answer 2

strcspn（const char * str1，const char * str2）查找第一个字符串中第二个字符串中任何字符的第一个实例。您传递的是字符而不是字符串作为第二个参数。您需要函数strchr（const char * string，int character），它查找单个字符。

交换字符串中两个最常见字母的出现次数

2 个答案:

当前问题

实施解决方案

示例运行