返回一个新字符串，其中包含来自其他两个字符串的字符

Question

Write a function common_char that takes two strings as arguments and returns a new string that contains a single copy of all characters that appear in either of the two strings.

For example, string1: hello ; string2: world ; the new string is : hellowrd ( o and l were already in array from hello).

May use string function here.In other words, all characters in string1 are copied into the new string, but characters in string 2 are copied only characters that are not in string1. That is past exam question and the university did not provide answer. Here is my code.

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

char *common_char(char *string1, char *string2) {
    int str_length1 = strlen(string1);
    int str_length2 = strlen(string2);
    char *new_string = malloc(str_length1+str_length2+1);

    for (int index_1 = 0; index_1 < str_length1; index_1++) {
        for (int index_2 = 0; index_2 < str_length2; index_2++) {
            if (string1[index_1] == string2[index_2]) {
            
            }
        }
    } 
}

int main(void) {


    return 0;
}

My idea is to find duplicate characters in string 2 and string 1 according to the nested loop, but there is a problem with the conditional statement, there is red line, also how to copy the character of the non-duplicate string? I know strcopy(), but how to remove the repeated characters?

Answer 1

I've come up with a solution that uses dynamic memory and resizes the result char* each time a new char must be added. There are two loops, the first iterates the b string and the second loop checks that non of char of the b string is repeated in the a string, if it is not repeated, then adds it. Hope you understand the realloc to resize dynamically the char* each time it must be added an element.

Firstly I initialize the result string to the size of string a so it can be all copied inside. The ordering method I think it is called bubble method.

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

char* common_char(char* a, char* b)  {
    
    char* result = (char*)malloc(sizeof(char)*strlen(a)+1);
    int i = 0;
    int j = 0;
    int repeated = 0;
    
    strcpy(result,a);
    
    for(i=0; i<strlen(b); i++) {
        for(j=0; j<strlen(result); j++) {
            if(b[i] == a[j]) {
                repeated = 1;
            }
        }
        if(!repeated) {
            result = (char*)realloc(result,strlen(result)+sizeof(char));
            result[strlen(result)] = b[i];
            result[strlen(result)+1] = '\0';
        }
        repeated = 0;
    }
    
    return result;
}

int main()
{
    char a[] = "hello";
    char b[] = "world";
    
    char* result = common_char(a,b);
    
    printf("%s", result);
    
    return 0;
}

EDIT: I've modified the code to make it function. About the comment of memory allocation, I've modified the declaration of result to give it space for the '\0'. When doing the realloc, I've already considered that the realloc does not increment the strlen() because strlen() is a counter till the '\0' not of the size of the variable.