C中的程序返回两个字符串之间的类似字符的数量，而无需多次计数字母

Question

我目前正在尝试在C中制作一个程序，以返回两个字符串之间的类似字符的数量，但没有多次计算字母。

检查两个字符串是否完全一样。

例如，使用“周”和“弱”，它应该返回3，因为字母'w'，'e'和'k'在两个词中都是。

两个字符串的大小仅为4。

我制作了“包含”功能，以检查我们是否尚未比较一个角色，例如使用“测试”和“ tast” Got 2“ T”，因此它的返回超出了预期。

目前，我的代码似乎可以正常工作，但我不明白为什么...

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

#define STRLEN 4

bool contains(char tab[], char c){
    for(int i=0; i<STRLEN; i++){
        if(tab[i] == c){
            return 1;
        }
    }
    return 0;
}

int likeness(char* password, char* wordchosen) {

    if(strlen(password)!=STRLEN && strlen(wordchosen)!=STRLEN) {
        return -1;
    }

    if(!strcmp(password, wordchosen)) {
        return 4;
    }

    int like = 0;
    char found[STRLEN];

    for(int i=0; i<STRLEN; i++) {
        for(int j=0; j<STRLEN; j++) {
            /*printf("%c : %c\n", password[i], wordchosen[j]);
            printf("\n");*/
            if(password[i] == wordchosen[j] && !(contains(found, wordchosen[j]))) {
                like++;
            }
        }
        found[i] = password[i];
    }

    return like;
}

int main() {
    printf("week and week = %d\n", likeness("week", "week")); // should be 4 (same word even if only 3 (w, e, k))
    printf("test and tast = %d\n", likeness("test", "tast")); // should be 2 (t, s)
    printf("week and weak = %d\n", likeness("week", "weak")); // should be 3 (w, e, k)
    printf("snet and sent = %d\n", likeness("snet", "sent")); // should be 4 (s, n, e, t)
    printf("gree and green = %d\n", likeness("gree", "gren")); // should be 3 (g, r, e)
    printf("mail and main = %d\n", likeness("mail", "main")); // should be 3 (m, a, i)
    printf("same = %d\n", likeness("same", "same")); // should be 4
    // opposite side in the likeness func. return different result...
    return 0;
}

Answer 1

数组<代码>发现未初始化，

char found[STRLEN];

因此函数包含使用非专业化阵列调用不确定的行为。

另一个问题是，在

    for(int j=0; j<STRLEN; j++) {
        /*printf("%c : %c\n", password[i], wordchosen[j]);
        printf("\n");*/
        if(password[i] == wordchosen[j] && !(contains(found, wordchosen[j]))) {
            like++;
        }
    }

字母密码[i]中存在字符串Word Chosen 多次一次，尚未在数组中写入。

也似乎是，如果语句

if(strlen(password)!=STRLEN && strlen(wordchosen)!=STRLEN) {
    return -1;
}

您应该写这件事，

if(strlen(password)!=STRLEN || strlen(wordchosen)!=STRLEN) {
    return -1;
}

而不是为此注意，而不是您的手动书面函数包含您可以使用标准C String函数strchr或memchr。

这是一个演示程序。

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

#define STRLEN 4

int likeness( const char *password, const char *wordchosen ) 
{
    if ( strlen( password ) != STRLEN || strlen( wordchosen ) != STRLEN ) 
    {
        return -1;
    }

    if ( strcmp( password, wordchosen ) == 0 ) 
    {
        return STRLEN;
    }

    int like = 0;
    char found[STRLEN + 1] = "";

    for ( ; *password; ++password )
    {
        if ( strchr( wordchosen, *password ) != NULL && 
             strchr( found, *password ) == NULL )
        {
            found[like++] = *password;
        }            
    }

    return like;
}

int main( void )
{
    printf("week and week = %d\n", likeness("week", "week")); // should be 4 (same word even if only 3 (w, e, k))
    printf("test and tast = %d\n", likeness("test", "tast")); // should be 2 (t, s)
    printf("week and weak = %d\n", likeness("week", "weak")); // should be 3 (w, e, k)
    printf("snet and sent = %d\n", likeness("snet", "sent")); // should be 4 (s, n, e, t)
    printf("gree and gren = %d\n", likeness("gree", "gren")); // should be 3 (g, r, e)
    printf("mail and main = %d\n", likeness("mail", "main")); // should be 3 (m, a, i)
    printf("same = %d\n", likeness("same", "same")); // should be 4
}

程序输出是

week and week = 4
test and tast = 2
week and weak = 3
snet and sent = 4
gree and gren = 3
mail and main = 3
same = 4

因为您可以看到函数定义不取决于strlen的值。

也可以在不使用辅助阵列的情况下定义该函数。

给你。

int likeness( const char *password, const char *wordchosen ) 
{
    if ( strlen( password ) != STRLEN || strlen( wordchosen ) != STRLEN ) 
    {
        return -1;
    }

    if ( strcmp( password, wordchosen ) == 0 ) 
    {
        return STRLEN;
    }

    int like = 0;

    for ( const char *p = password; *p; ++p )
    {
        if ( memchr( password, *p, p - password ) == NULL && 
             strchr( wordchosen, *p ) != NULL )
        {
            ++like;
        }            
    }

    return like;
}