查找字符串中的常见字符

Question

Hi i have two different string and i need to find the common character in the string. I managed to get the common string but i need to return "null string" for input that does not have same character.

Current issues:

input1: abc
input2: def
output: ' // it shld be "null string";

heres my code:

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

void strInterset(char * str1, char * str2, char * str3);

int main() {
    char str1[50], str2[50], str3[50];

    printf("Enter str1: \n");
    scanf("%s", str1);
    printf("Enter str2: \n");
    scanf("%s", str2);
    strInterset(str1, str2, str3);

    if (*str3 == '\0')
        printf("strIntersect(): null string\n");
    else
        printf("strIntersect(): %s\n", str3);

    return 0;
}

void strInterset(char * str1, char * str2, char * str3) {
    int i = 0, j;
    for (i; *(str1 + i) != '\0'; i++) {
        for (j = 0; *(str2 + j) != '\0'; j++) {
            if ( *(str2 + j) == *(str1 + i)) {
                strcpy(str3, str1 + i);
                str3++;
            }
        }
    }
}

Answer 1

The reason is that strInterset() only calls strcpy() if a match is found, and doesn't modify str3 or the data it points to otherwise. The fix is simple - before the loop in strInterset() add the statement

 *str3 = '\0';

If a match is found, strcpy() will still be called. If not, the test being done in main() will succeed.

Initialising the array in main() to zero will also work for the FIRST call of strInterset() . It may not work for subsequent calls though (unless main() reinitialises str3 before every call). Hence it is better to do the initialisation in strInterset() .

Answer 2

Initialise your str3 explictly to NULL by declaring it like

char str3[50] = { NULL }; 

If you do not do that its an unitialised array with junk/indeterminate values.

Answer 3

You should initialize str3 to NULL , like this:

char str3[50] = {0};

since if you don't do that, it will remain an unitialized array, which means that it will invoke Undefined Behavior when accessed, since its values are garbage.

Moreover, even if common character(s) exist, str3 will not be NULL-terminated.

I would personally change your function to this:

void strInterset(char * str1, char * str2, char * str3) {
    int i = 0, j;
    *str3 = '\0'; // NULL terminate
    for (i; *(str1 + i) != '\0'; i++) {
        ...
}

Output:

Enter str1: abc
Enter str2: dfg
strIntersect(): null string

---

PS: Compile with warnings enabled and you will get:

prog.c: In function 'strInterset':
prog.c:26:5: warning: statement with no effect [-Wunused-value]
     for (i; *(str1 + i) != '\0'; i++) {
     ^~~

Just change it to: for (; *(str1 + i) != '\\0'; i++) { , or even better for (int i = 0; *(str1 + i) != '\\0'; i++) { . This is not your problem, but is good to fix warnings.

Answer 4

The quick fix should be str3[0] = '\\0' as second line in main .

But there are more ways you can make your program better:

• Why you need a full char str3[50] ? You could instead use the return value of the function like this: char strInterset(char * str1, char * str2); And then add return in the appropriate places. (By the way: should that be spellect intersect ?)

• Your program uses two nested for -loops (which is slow on large inputs). Instead, you could make an array where each entry corresponds to one character value (look at an ascii-table). The array then can contain a true/false whether the character appeared. You initialize all entries to 0. Then you run through the first string and for each character you set its entry in the array to 1. Then you run through the second string and for each character you check whether the entry in the array is 1. If such a case is found then you found a character that appears in both strings.

Answer 5

Here is your function which checks for the common characters and adds them to the result string without the repetition. If you do not care about the repetitions just remove the second if (and only the line where the if is, but leave its body in the brackets )

char *strcomm(const char *s1, const char *s2, char *s3)
{
    const char *tmp;
    char *tmps3 = s3;
    *s3 = 0;
    while(*s1)
    {
        tmp = s2;
        while(*tmp)
        {
            if(*s1 == *tmp)
                if(strchr(s3,*s1) == NULL)
                {
                    *s3++ = *s1;
                    *s3 = 0;
                }
            tmp++;
        }
        s1++;
    }
    return tmps3;
}