C: Same chars from two strings - c

Question

I have two strings:
char *str1 = "this is a test";
char *str2 = "ts bd a";
I'm trying to write a function that returns a new string with the same chars from the two string without duplicates (also ' ' is duplicate). eg.:
char *retStr = GetSameChars(str1, str2); //returns "ts a";
How can I do that?
What I'm tried:
char *GetSameChars(char str1[], char str2[]) {
int found = -1, i , j = 0, biggest, index = 0;
char *retArr, *star = '*';
int str1Len, str2Len, count = 0;
str1Len = strlen(str1);
str2Len = strlen(str2);
biggest = str1Len > str2Len ? str1Len : str2Len;
retArr = (char *)malloc(sizeof(char) * count);
for (i = 0; i < str1Len; i++) {
for (j = 0; j < str2Len; j++) {
if (str1[i] == str2[j] && found == -1) {
count++;
found = j;
} else
if (str2[j] == str2[found])
str2[j] = star; //Throw an exception
}
found = -1;
}
retArr = (char *)malloc(sizeof(char) * count);
j = 0;
for (i = 0; i < str2Len; i++)
if (str2[i] != '*')
retArr[j++] = str2[i];
for (i = 0; i < str2Len; i++)
printf("%c", retArr[i]);
}
When I tried the line str2[j] = star; I got an exception.
What is my mistake?

Answer 1

My recommendations would be: keep it simple; get to know the C standard library; write less, test more.
Some specific problems with your code: you pass the wrong variable to malloc(); you estimate the answer to fit in the size of the larger of the two strings but it will actually fit into the smaller of the two; you modify an argument string str2[j] = star -- you should be treating the arguments as readonly; you malloc() retArr twice unnecessarily, leaking the first one when you allocate the second; your algorithm simply doesn't work.
Although a lookup table, as others have suggested, would be more efficient, let's use the standard library routine strchr() to solve this problem:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *getSameChars(const char *string1, const char *string2) {
size_t string1_length = strlen(string1);
size_t string2_length = strlen(string2);
size_t shortest = string1_length < string2_length ? string1_length : string2_length;
char *common_pointer, *common = malloc(shortest + 1);
*(common_pointer = common) = '\0';
for (size_t i = 0; i < string1_length; i++) {
// character found in both input strings, but not yet in common string
if (strchr(string2, string1[i]) != NULL && strchr(common, string1[i]) == NULL) {
*common_pointer++ = string1[i];
*common_pointer = '\0';
}
}
return common;
}
int main() {
char *stringA = "this is a test";
char *stringB = "ts bd a";
char *result = getSameChars(stringA, stringB);
printf("%s\n", result);
free(result);
return(0);
}

Answer 2

Your code complains because you are trying to assign a pointer to a char, to get the value inside a pointer you need to use the * operator like so:
*star;
a good way to check if a letter have already appeared(if you want to use it on all of the ascii table then 128) is to use a lookup table. first you will need to declare an array the length of all letters in the alphabet like so:
char lut[26];
If it is a global variable then it will be set to 0, then all you need to do is go to the index of the char you got and mark it as 1, a simple if will later be able to determine if a letter has already appeard.
example:
lut[toupper(somechar) - 'A'] = 1;
In this example you set the char in the lookup table that is equivalent to the somechar variable as 1, marking it has already appeared.
hope this helps.