Removing consecutive repeated characters from string using C - c

Question

I'm trying to remove consecutive repeated characters from a given string.
Example:
bssdffFdcrrrtttii ***#
output is supposed to be:
bsdfFdcrti *#
This code doesn't work and only prints the first char (b), I want to learn about my mistake.
when I'm doing a printf test, it works but not for spaces.
I think the problem might be with the new char array.
void Ex6() {
char* string[80];
scanf("%s", &string);
puts(removeDup(string));
}
char* removeDup(char *string) {
int i, c = 0;
char* newString[80];
for (i = 0; i < strlen(string); i++) {
if (string[i] != string[i + 1]) {
newString[c++] = string[i];
}
}
return newString;
}

Answer 1

There are several problems with your program:
The declaration of newString should be char newString[80], i.e., an array of characters and not an array of pointers-to-characters, and likewise for the declaration in Ex6.
The call to scanf should then be scanf("%s", string), since string is already the address of an array of characters, but...
Use fgets to read a string from the user to ensure that you read whitespace, if it's important, and that the buffer is not exceeded.
newString is allocated on the stack and so should not be returned to the caller. It is better to do a char *newString = strdup(string), or, slightly less sloppy, char *newString = malloc(strlen(string)+1), which will call malloc for a block of memory sufficient to hold the original string, and thus the version without duplicates -- the comments rightly point out that this could be optimized. In principle, the caller, i.e., Ex6, must free the returned pointer to avoid a memory leak but it hardly matters in such a short program.
The result needs a null terminator: newString[c] = '\0'.
Otherwise, the removeDup function seems to work correctly.
So, putting all of that together:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* removeDup(const char *string)
{
size_t i, c = 0;
size_t string_len = strlen(string);
char *newString = malloc(string_len + 1);
for (i = 0; i < string_len; i++) {
if (string[i] != string[i + 1]) {
newString[c++] = string[i];
}
}
newString[c] = '\0';
return newString;
}
#define MAX_STRING_LEN 80
void Ex6() {
char string[MAX_STRING_LEN];
char* result;
if (fgets(string, MAX_STRING_LEN, stdin) != NULL) {
result = removeDup(string);
printf("%s", result);
free(result);
}
}
Finally, I agree with #tadman's comment. Since the input string must anyway be traversed to calculate the length, we may as well optimize the size of the result string:
char* removeDup(const char *string)
{
size_t i, c = 0;
char *newString;
for (i = 0; string[i] != '\0'; i++)
c += (string[i] != string[i + 1]);
newString = malloc(c + 1);
for (i = c = 0; string[i] != '\0'; i++) {
if (string[i] != string[i + 1]) {
newString[c++] = string[i];
}
}
newString[c] = '\0';
return newString;
}

Answer 2

There are quite a few issues in your program. It wouldn't even compile let alone run. Also, the most problematic issue is that you are returning a pointer to a local variable from a function that ceases its scope upon completion. A simplified version of your program is as follows:
void Ex6()
{
char string[80];
scanf("%s", string);
int i, c = 0;
char newString[80];
for (i = 0; i < strlen(string); i++) {
if (string[i] != string[i + 1]) {
newString[c++] = string[i];
}
}
newString[c] = '\0';
puts(newString);
}

Answer 3

You can do it with O(n) time and O(1) space, by modifying existing string:
#include <stdio.h>
char* removeDup(char* input) {
char* newTail = input, *oldTail = input;
while (*oldTail) {
if (*newTail == *oldTail) {
++oldTail;
} else {
*++newTail = *oldTail++;
}
}
return newTail;
}
int main() {
char string[] = "bssdffFdcrrrtttii ***#";
char* newEnd = removeDup(string);
char* tmp = string;
while (tmp != newEnd) {
printf("%c", *tmp++);
}
//Print the last char if string had any duplicates
if(*tmp) {
printf("%c", *tmp++);
}
return 0;
}