C string to array -uppercase character bug [closed] - c

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Given this string "red, blue, green" create an array that contains these colors as its elements. The code I've written below works but when I change the first letter of the colors to uppercase I get the output- Red, Blu\301-!Wree\316. How can I make this code more dynamic to work with words that start with uppercase as well? Thank you.
#include <stdio.h>
#include <stdlib.h>
int findLength(char string[]){
int l =0;
for(l = 0; string[l]!='\0'; l++){
}
return l;
};
char *stringToArray(char string[]){
int i = 0;
int j = 0;
char c = ',';
int n = 0;
int l = findLength(string);
char *str = (char *)malloc(l * sizeof(char));
while(string[i] != l){
if(string[i] == c || string[i] != '\0'){
for(n = j; n < i; n++){
str[j++] += string[n];
}
}
i++;
}
printf("%s\n", str);
str = '\0';
return str;
}
int main(int argc, const char * argv[]) {
char *string = "red, blue, green";
//char *string = "Red, Blue, Green";
char *str = stringToArray(string);
free(str);
return 0;
}

The strange behaviour doesn't have anything to do with whether your strings have upparcase letters or not. Your termination condition for the loop in stringToArray is wrong:
int l = findLength(string);
while (string[i] != l) ...
The condition should be
while (i < l) ...
or, as you have already used in findLength:
while (string[i] != '\0') ...
Because the condition is wrong – l is 16 in your case and none of the letters have an ASCII value of 16 – you go beyond the valid bounds of the string, which leads to undefined behaviour.
At the moment, you just copy the old string to the new one, albeit in a very strange fashion. Your inner loop makes use of three variables, of which it increments two. That's very confusing. It probably also doesn't do what you think, because the condition:
if (string[i] == c || string[i] != '\0') ..
is true for all letters of the string, provided that the opuer loop should consider only valid characters up to, but not including the end of the string.
Finally, if you want to copy the string, you should allocate süpace for the terminating character:
char *str = malloc(l + 1);
When you want to append the final null character:
str = '\0';
You actually set the while allocated string to null, which leads to a memory leak. (The free in main doesn't produce an error, because free can legally take ´NULL` as argument.) Instead, use:
str[l] = '\0';
With these fixes, you now have a program that copies the original string. The (POSIX) library function strdup does this more effectively. If you want to return an array of strings, you must reflect your function to return a pointer to pointers of chars.
Below is a possible implementation of that behaviour. (It uses the approach to allocate memory for everything on the heap. If you always expect three short strings that may not be the best solution.)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char **stringToArray(const char *str)
{
char **res;
const char *p = str;
int n = 0;
while (p) {
p = strchr(p, ',');
if (p) p++;
n++;
}
res = malloc((n + 1) * sizeof(*res));
p = str;
n = 0;
while (p) {
const char *begin;
size_t len;
while (*p == ' ') p++;
begin = p;
p = strchr(p, ',');
if (p) {
len = p - begin;
p++;
} else {
len = strlen(begin);
}
res[n] = malloc(len + 1);
memcpy(res[n], begin, len);
res[n][len] = '\0';
n++;
}
res[n] = NULL;
return res;
}
int main(int argc, const char * argv[])
{
char *str = "Vermilion, Ultramarine, Chartreuse";
char **res = stringToArray(str);
int i = 0;
for (i = 0; res[i]; i++) {
puts(res[i]);
free(res[i]);
}
free(res);
return 0;
}

you have a few mistakes...
I've corrected them for you:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int findLength(char string[]){
int l = 1;
for (int i = 0; string[i] != '\0'; i++){
if (string[i] == ',')// to check the end of a color
l++;
}
return l;
};
char **stringToArray(char string[]){//added a * for array of satrings
int i = 0;
int j = 0;
char c = ',';
int n = 0;
int l = findLength(string);
char **str = (char **)malloc(l * sizeof(char*)+l);
char *pos = string;
for (int i = 0; i < l-1; i++) //getting each color to the array
{
char *c =strchr(string, ',');
int index = c - pos;
string[index] = 0;
str[i] = _strdup(pos); //copying the color to the array
pos = c + 1;
string = string +1 +index; // next color
}
str[l - 1] = _strdup(pos); //copying last color
for (int i = 0; i < l; i++) //printing the results
{
printf("%s\n",str[i]);
}
return str;
}
int main(int argc, const char * argv[]) {
char string[] = "red,blue,green"; //deleted spaces
char **str = stringToArray(string);
getchar();
free(str);
return 0;
}
also added comments for you to understand.

Related

Is there a way if string repeats to return only repeated letters once?

I made code which will for string "aabbcc" return "abc" but in cases when there is more letters like "aaa" it will return "aa" instead of just one.
Here is the code I made.
void Ponavljanje(char *s, char *p) {
int i, j = 0, k = 0, br = 0, m = 0;
for (i = 0; i < strlen(s) - 1; i++) {
for (j = i + 1; j < strlen(s); j++) {
if (s[i] == s[j]) {
br++;
if (br == 1) {
p[k++] = s[i];
}
}
}
br = 0;
}
p[k] = '\0';
puts(p);
}
For "112233" output should be "123" or for "11122333" it should be also "123".
Avoid repeated calls to strlen(s). A weak compiler may not see that s is unchanged and call strlen(s) many times, each call insuring a cost of n operations - quite inefficient. #arkku.1 Instead simply stop iterating when the null character detected.
Initialize a boolean list of flags for all char to false. When a character occurs, set the flag to prevent subsequent usage. Be careful when indexing that list as char can be negative.
Using a const char *s allows for wider allocation and helps a compiler optimization.
Example:
#include <stdbool.h>
#include <limits.h>
void Ponavljanje(const char *s, char *p) {
const char *p_original = p;
bool occurred[CHAR_MAX - CHAR_MIN + 1] = { 0 }; // all values set to 0 (false)
while (*s) {
if (!occurred[*s - CHAR_MIN]) {
occurred[*s - CHAR_MIN] = true;
*p++ = *s;
}
s++;
}
*p = '\0';
puts(p_original);
}
1 #wrongway4you comments that many compilers may assume the string did not change and optimize out the repeated strlen() call. A compliant compiler cannot do that though without restrict unless it is known that in all calls, s and p do not overlap. A compiler otherwise needs to assume p may affect s and warrant a repeated strlen() call.
does the work with a complexity O(n)
I suppose programming can give rmg
void Ponavljanje(char *s,char *p)
{
char n[256] = {0};
int i = 0;
while (*s) {
switch (n[(unsigned char) *s]) {
case 0:
n[(unsigned char) *s] = 1;
break;
case 1:
p[i++] = *s;
n[(unsigned char) *s] = 2;
}
s += 1;
}
p[i] = 0;
puts(p);
}
While the inner loop checks br to only copy the output on the first repetition, the outer loop still passes over each repetition in s on future iterations. Hence each further occurrence of the same character will run a separate inner loop after br has already been reset.
With aaa as the input, both the first and the second a cause the inner loop to find a repetition, giving you aa. In fact, you always get one occurrence fewer of each character in the output than there is in the input, which means it only works for 1 or 2 occurrences in the input (resulting in 0 and 1 occurrences, respectively, in the output).
If you only want to remove the successive double letters, then this function would be sufficient, and the examples given in the question would fit:
#include <stdio.h>
void Ponavljanje(char *s,char *p)
{
char dd = '\0';
char *r;
if(s == NULL || p == NULL)
return;
r = p;
while(*s){
if(*s != dd){
*r = *s;
dd = *s;
r++;
}
s++;
}
*r = '\0';
puts(p);
}
int main(void)
{
char s[20] = "1111332222";
char p[20];
Ponavljanje(s,p);
}
Here is something that works regardless of order:
#include <stdio.h>
#include <string.h>
void
repeat(char *s, char *p)
{
int slen;
int sidx;
int pidx;
int plen;
int schr;
slen = strlen(s);
plen = 0;
for (sidx = 0; sidx < slen; ++sidx) {
schr = s[sidx];
// look for duplicate char
int dupflg = 0;
for (pidx = 0; pidx < plen; ++pidx) {
if (p[pidx] == schr) {
dupflg = 1;
break;
}
}
// skip duplicate chars
if (dupflg)
continue;
p[plen++] = schr;
}
p[plen] = 0;
puts(p);
}
int
main(void)
{
char p[100];
repeat("112233",p);
repeat("123123",p);
return 0;
}
Note: As others have mentioned, strlen should not be placed in the loop condition clause of the for [because the length of s is invariant]. Save strlen(s) to a separate variable and loop to that limit
Here is a different/faster version that uses a histogram so that only a single loop is required:
#include <stdio.h>
#include <string.h>
void
repeat(char *s, char *p)
{
char dups[256] = { 0 };
int slen;
int sidx;
int pidx;
int plen;
int schr;
slen = strlen(s);
sidx = 0;
plen = 0;
for (sidx = 0; sidx < slen; ++sidx) {
schr = s[sidx] & 0xFF;
// look for duplicate char
if (dups[schr])
continue;
dups[schr] = 1;
p[plen++] = schr;
}
p[plen] = 0;
puts(p);
}
int
main(void)
{
char p[100];
repeat("112233",p);
repeat("123123",p);
return 0;
}
UPDATE #2:
I would suggest iterating until the terminating NUL byte
Okay, here's a full pointer version that is as fast as I know how to make it:
#include <stdio.h>
#include <string.h>
void
repeat(char *s, char *p)
{
char dups[256] = { 0 };
char *pp;
int schr;
pp = p;
for (schr = *s++; schr != 0; schr = *s++) {
schr &= 0xFF;
// look for duplicate char
if (dups[schr])
continue;
dups[schr] = 1;
*pp++ = schr;
}
*pp = 0;
puts(p);
}
int
main(void)
{
char p[100];
repeat("112233",p);
repeat("123123",p);
return 0;
}

My program can't seem to separate words at certain cases [closed]

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 4 years ago.
Improve this question
I am mostly concerned with the find and separation function. When I reused them in another program, I found that they don't always work and sometimes show segmentation fault. I tried every possible thing I could do, but nothing worked.
#include <stdio.h>
#include<stdlib.h>
#include <string.h>
int find(char *argument)
{
int count = 0;
int i = 0;
int state = 0;
while (argument[i] != '\0')
{
if (argument[i] == ' ' || argument[i + 1] == '\0')
{
state = 0;
}
else if (state == 0)
{
state = 1;
count++;
}
i++;
}
return count;
}
char **sepration(int args, char str[])
{
int i = 0;
char **argv = (char **)malloc(args);
int k = 0;
int len = strlen(str);
while (i < len)
{
if (str[i] != ' ')
{
char *st = &str[i];
argv[k] = st;
k++;
while (str[i] != ' ' && str[i] != '\0' && i < len)
{
i++;
}
str[i] = '\0';
}
i++;
}
return argv;
}
int main()
{
char argument[100];
scanf("%[^\n]s", &argument);
//finds the no of words in the given argument
int args = find(argument);
char **argv = (char **)malloc(args + 1);
//stores the word separately in **char pointer array
argv = sepration(args, argument);
//adds the arguments whichs are numbers
add(args, argv);
}
The problem with your solution is that you are not allocating the memory properly.
Your double pointer argv should work as an array of strings. Hence argv should have enough memory allocated for the same. Then next question is how to allocate sufficient memory for the same. As argv should hold argc number of strings hence proper memory allocation will be char **argv = malloc(args * sizeof(char *)); or,
char *argv[argc];
Modified code will look like
char **sepration(int args, char str[])
{
int i = 0;
char **argv = malloc(args * sizeof(char *));
int k = 0;
int len = strlen(str);
while (i < len)
{
if (str[i] != ' ')
{
char *st = &str[i];
argv[k] = st;
k++;
while (str[i] != ' ' && str[i] != '\0' && i < len)
{
i++;
}
str[i] = '\0';
}
i++;
}
return argv;
}
int main()
{
char argument[100];
scanf("%[^\n]", argument);
//finds the no of words in the given argument
int args = find(argument);
char **argv = malloc(args * sizeof(char *));
//stores the word separately in **char pointer array
argv = sepration(args, argument);
for(int i = 0; i < args; i++)
printf("%s\n", argv[i]);
}
Instead of char **argv = malloc(args * sizeof(char *)); you can use char *argv[argc]; as well. This will also give the same result.

Efficiently replace a substring in a string

I have made two functions that find a substring index and substitute that substring in the string. I'm glad I jury rigged this at all, given that similar questions previously asked were never answered/marked as closed without any help. Is there a cleaner method?
void destroy_substr(int index, int len)
{
int i;
for (i = index; i < len; i++)
{
string[i] = '~';
}
}
void find_substr_index(char* substr)
{
int i;
int j;
int k;
int count;
int len = strlen(substr);
for (i = 0; i < strlen(string); i++)
{
if (string[i] == substr[0])
{
for(j = i, k = 0; k < len; j++, k++)
{
if (string[j] == substr[k])
{
count++;
}
if (count == len)
destroy_substr((j - len + 1), len);
}
j = 0;
k = 0;
count = 0;
}
}
}
Your code seems like you're trying to re-inventing your own wheel.
By using standard C functions, which is strstr() and memset(), you can achieve the same result as you expected.
#include <stdio.h>
#include <string.h>
char string[] = "foobar foobar foobar";
char substr[] = "foo";
char replace = '~';
int main() {
int substr_size = strlen(substr);
// Make a copy of your `string` pointer.
// This is to ensure we can safely modify this pointer value, without 'touching' the original one.
char *ptr = string;
// while true (infinite loop)
while(1) {
// Find pointer to next substring
ptr = strstr(ptr, substr);
// If no substring found, then break from the loop
if(ptr == NULL) { break; }
// If found, then replace it with your character
memset(ptr, replace, substr_size);
// iIncrement our string pointer, pass replaced substring
ptr += substr_size;
}
printf("%s\n", string);
return 0;
}
How about this:
#include <stdio.h>
#include <string.h>
int main(int argc, char **argv)
{
char string[] = "HELLO hello WORLD world HELLO hello ell";
char substring[] = "ell";
int stringLength = strlen(string);
int substringLength = strlen(substring);
printf("Before: %s\n", string);
if(substringLength <= stringLength)
{
int i;
int j;
for(i = 0, j = stringLength - substringLength + 1; i < j; )
{
if(memcmp(&string[i], substring, substringLength) == 0)
{
memset(&string[i], '~', substringLength);
i += substringLength;
}
else
{
i++;
}
}
}
printf("After: %s\n", string);
return 0;
}
Key ideas are:
You only need to scan the string (stringLength - substringLength) times
You can use functions from string.h to do the comparison and to replace the substring
You can copy the new string in place. If you want to support insertion of longer strings you will need to manage memory with malloc()/realloc(). If you want to support insertion of smaller strings you'll need to advance the pointer to the beginning by the length of the replacement string, copy the rest of the string to that new location, then zero the new end of the string.
#include <stdio.h>
#include <string.h>
#include <err.h>
int main(int argc, char **argv)
{
char *str = strdup("The fox jumps the dog\n");
char *search = "fox";
char *replace = "cat";
size_t replace_len = strlen(replace);
char *begin = strstr(str, search);
if (begin == NULL)
errx(1, "substring not found");
if (strlen(begin) < replace_len)
errx(1, "replacement too long");
printf("%s", str);
memcpy(begin, replace, replace_len);
printf("%s", str);
return 0;
}

How to remove the character at a given index from a string in C?

How do I remove a character from a string?
If I have the string "abcdef" and I want to remove "b" how do I do that?
Removing the first character is easy with this code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char word[] = "abcdef";
char word2[10];
strcpy(word2, &word[1]);
printf("%s\n", word2);
return 0;
}
and
strncpy(word2, word, strlen(word) - 1);
will give me the string without the last character, but I still didn't figure out how to remove a char in the middle of a string.
memmove can handle overlapping areas, I would try something like that (not tested, maybe +-1 issue)
char word[] = "abcdef";
int idxToDel = 2;
memmove(&word[idxToDel], &word[idxToDel + 1], strlen(word) - idxToDel);
Before: "abcdef"
After: "abdef"
Try this :
void removeChar(char *str, char garbage) {
char *src, *dst;
for (src = dst = str; *src != '\0'; src++) {
*dst = *src;
if (*dst != garbage) dst++;
}
*dst = '\0';
}
Test program:
int main(void) {
char* str = malloc(strlen("abcdef")+1);
strcpy(str, "abcdef");
removeChar(str, 'b');
printf("%s", str);
free(str);
return 0;
}
Result:
>>acdef
My way to remove all specified chars:
void RemoveChars(char *s, char c)
{
int writer = 0, reader = 0;
while (s[reader])
{
if (s[reader]!=c)
{
s[writer++] = s[reader];
}
reader++;
}
s[writer]=0;
}
char a[]="string";
int toBeRemoved=2;
memmove(&a[toBeRemoved],&a[toBeRemoved+1],strlen(a)-toBeRemoved);
puts(a);
Try this . memmove will overlap it.
Tested.
Really surprised this hasn't been posted before.
strcpy(&str[idx_to_delete], &str[idx_to_delete + 1]);
Pretty efficient and simple. strcpy uses memmove on most implementations.
int chartoremove = 1;
strncpy(word2, word, chartoremove);
strncpy(((char*)word2)+chartoremove, ((char*)word)+chartoremove+1,
strlen(word)-1-chartoremove);
Ugly as hell
The following will extends the problem a bit by removing from the first string argument any character that occurs in the second string argument.
/*
* delete one character from a string
*/
static void
_strdelchr( char *s, size_t i, size_t *a, size_t *b)
{
size_t j;
if( *a == *b)
*a = i - 1;
else
for( j = *b + 1; j < i; j++)
s[++(*a)] = s[j];
*b = i;
}
/*
* delete all occurrences of characters in search from s
* returns nr. of deleted characters
*/
size_t
strdelstr( char *s, const char *search)
{
size_t l = strlen(s);
size_t n = strlen(search);
size_t i;
size_t a = 0;
size_t b = 0;
for( i = 0; i < l; i++)
if( memchr( search, s[i], n))
_strdelchr( s, i, &a, &b);
_strdelchr( s, l, &a, &b);
s[++a] = '\0';
return l - a;
}
This is an example of removing vowels from a string
#include <stdio.h>
#include <string.h>
void lower_str_and_remove_vowel(int sz, char str[])
{
for(int i = 0; i < sz; i++)
{
str[i] = tolower(str[i]);
if(str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u')
{
for(int j = i; j < sz; j++)
{
str[j] = str[j + 1];
}
sz--;
i--;
}
}
}
int main(void)
{
char str[101];
gets(str);
int sz = strlen(str);// size of string
lower_str_and_remove_vowel(sz, str);
puts(str);
}
Input:
tour
Output:
tr
Use strcat() to concatenate strings.
But strcat() doesn't allow overlapping so you'd need to create a new string to hold the output.
I tried with strncpy() and snprintf().
int ridx = 1;
strncpy(word2,word,ridx);
snprintf(word2+ridx,10-ridx,"%s",&word[ridx+1]);
Another solution, using memmove() along with index() and sizeof():
char buf[100] = "abcdef";
char remove = 'b';
char* c;
if ((c = index(buf, remove)) != NULL) {
size_t len_left = sizeof(buf) - (c+1-buf);
memmove(c, c+1, len_left);
}
buf[] now contains "acdef"
This might be one of the fastest ones, if you pass the index:
void removeChar(char *str, unsigned int index) {
char *src;
for (src = str+index; *src != '\0'; *src = *(src+1),++src) ;
*src = '\0';
}
This code will delete all characters that you enter from string
#include <stdio.h>
#include <string.h>
#define SIZE 1000
char *erase_c(char *p, int ch)
{
char *ptr;
while (ptr = strchr(p, ch))
strcpy(ptr, ptr + 1);
return p;
}
int main()
{
char str[SIZE];
int ch;
printf("Enter a string\n");
gets(str);
printf("Enter the character to delete\n");
ch = getchar();
erase_c(str, ch);
puts(str);
return 0;
}
input
a man, a plan, a canal Panama
output
A mn, pln, cnl, Pnm!
Edit : Updated the code zstring_remove_chr() according to the latest version of the library.
From a BSD licensed string processing library for C, called zString
https://github.com/fnoyanisi/zString
Function to remove a character
int zstring_search_chr(char *token,char s){
if (!token || s=='\0')
return 0;
for (;*token; token++)
if (*token == s)
return 1;
return 0;
}
char *zstring_remove_chr(char *str,const char *bad) {
char *src = str , *dst = str;
/* validate input */
if (!(str && bad))
return NULL;
while(*src)
if(zstring_search_chr(bad,*src))
src++;
else
*dst++ = *src++; /* assign first, then incement */
*dst='\0';
return str;
}
Exmaple Usage
char s[]="this is a trial string to test the function.";
char *d=" .";
printf("%s\n",zstring_remove_chr(s,d));
Example Output
thisisatrialstringtotestthefunction
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 50
void dele_char(char s[],char ch)
{
int i,j;
for(i=0;s[i]!='\0';i++)
{
if(s[i]==ch)
{
for(j=i;s[j]!='\0';j++)
s[j]=s[j+1];
i--;
}
}
}
int main()
{
char s[MAX],ch;
printf("Enter the string\n");
gets(s);
printf("Enter The char to be deleted\n");
scanf("%c",&ch);
dele_char(s,ch);
printf("After Deletion:= %s\n",s);
return 0;
}
#include <stdio.h>
#include <string.h>
int main(){
char ch[15],ch1[15];
int i;
gets(ch); // the original string
for (i=0;i<strlen(ch);i++){
while (ch[i]==ch[i+1]){
strncpy(ch1,ch,i+1); //ch1 contains all the characters up to and including x
ch1[i]='\0'; //removing x from ch1
strcpy(ch,&ch[i+1]); //(shrinking ch) removing all the characters up to and including x from ch
strcat(ch1,ch); //rejoining both parts
strcpy(ch,ch1); //just wanna stay classy
}
}
puts(ch);
}
Let's suppose that x is the "symbol" of the character you want to remove
,my idea was to divide the string into 2 parts:
1st part will countain all the characters from the index 0 till (and including) the target character x.
2nd part countains all the characters after x (not including x)
Now all you have to do is to rejoin both parts.
This is what you may be looking for while counter is the index.
#include <stdio.h>
int main(){
char str[20];
int i,counter;
gets(str);
scanf("%d", &counter);
for (i= counter+1; str[i]!='\0'; i++){
str[i-1]=str[i];
}
str[i-1]=0;
puts(str);
return 0;
}
I know that the question is very old, but I will leave my implementation here:
char *ft_strdelchr(const char *str,char c)
{
int i;
int j;
char *s;
char *newstr;
i = 0;
j = 0;
// cast to char* to be able to modify, bc the param is const
// you guys can remove this and change the param too
s = (char*)str;
// malloc the new string with the necessary length.
// obs: strcountchr returns int number of c(haracters) inside s(tring)
if (!(newstr = malloc(ft_strlen(s) - ft_strcountchr(s, c) + 1 * sizeof(char))))
return (NULL);
while (s[i])
{
if (s[i] != c)
{
newstr[j] = s[i];
j++;
}
i++;
}
return (newstr);
}
just throw to a new string the characters that are not equal to the character you want to remove.
Following should do it :
#include <stdio.h>
#include <string.h>
int main (int argc, char const* argv[])
{
char word[] = "abcde";
int i;
int len = strlen(word);
int rem = 1;
/* remove rem'th char from word */
for (i = rem; i < len - 1; i++) word[i] = word[i + 1];
if (i < len) word[i] = '\0';
printf("%s\n", word);
return 0;
}
This is a pretty basic way to do it:
void remove_character(char *string, int index) {
for (index; *(string + index) != '\0'; index++) {
*(string + index) = *(string + index + 1);
}
}
I am amazed none of the answers posted in more than 10 years mention this:
copying the string without the last byte with strncpy(word2, word, strlen(word)-1); is incorrect: the null terminator will not be set at word2[strlen(word) - 1]. Furthermore, this code would cause a crash if word is an empty string (which does not have a last character).
The function strncpy is not a good candidate for this problem. As a matter of fact, it is not recommended for any problem because it does not set a null terminator in the destination array if the n argument is less of equal to the source string length.
Here is a simple generic solution to copy a string while removing the character at offset pos, that does not assume pos to be a valid offset inside the string:
#include <stddef.h>
char *removeat_copy(char *dest, const char *src, size_t pos) {
size_t i;
for (i = 0; i < pos && src[i] != '\0'; i++) {
dest[i] = src[i];
}
for (; src[i] != '\0'; i++) {
dest[i] = src[i + 1];
}
dest[i] = '\0';
return dest;
}
This function also works if dest == src, but for removing the character in place in a modifiable string, use this more efficient version:
#include <stddef.h>
char *removeat_in_place(char *str, size_t pos) {
size_t i;
for (i = 0; i < pos && str[i] != '\0'; i++)
continue;
for (; str[i] != '\0'; i++)
str[i] = str[i + 1];
return str;
}
Finally, here are solutions using library functions:
#include <string.h>
char *removeat_copy(char *dest, const char *src, size_t pos) {
size_t len = strlen(src);
if (pos < len) {
memmove(dest, src, pos);
memmove(dest + pos, src + pos + 1, len - pos);
} else {
memmove(dest, src, len + 1);
}
return dest;
}
char *removeat_in_place(char *str, size_t pos) {
size_t len = strlen(str);
if (pos < len) {
memmove(str + pos, str + pos + 1, len - pos);
}
return str;
}
A convenient, simple and fast way to get rid of \0 is to copy the string without the last char (\0) with the help of strncpy instead of strcpy:
strncpy(newStrg,oldStrg,(strlen(oldStrg)-1));

C program to remove repeated char from a string

I came across a interview question that asked to remove the repeated char from a given string, in-place.
So if the input was "hi there" the output expected was "hi ter". It was also told to consider only alphabetic repititions and all the
alphabets were lower case. I came up with the following program. I have comments to make my logic clear. But the program does not work as expectd for some inputs. If the input is "hii" it works, but if its "hi there" it fails. Please help.
#include <stdio.h>
int main()
{
char str[] = "programming is really cool"; // original string.
char hash[26] = {0}; // hash table.
int i,j; // loop counter.
// iterate through the input string char by char.
for(i=0,j=0;str[i];)
{
// if the char is not hashed.
if(!hash[str[i] - 'a'])
{
// hash it.
hash[str[i] - 'a'] = 1;
// copy the char at index i to index j.
str[j++] = str[i++];
}
else
{
// move to next char of the original string.
// do not increment j, so that later we can over-write the repeated char.
i++;
}
}
// add a null char.
str[j] = 0;
// print it.
printf("%s\n",str); // "progamin s ely c" expected.
return 0;
}
when str[i] is a non-alphabet, say a space and when you do:
hash[str[i] - 'a']
your program can blow.
ASCII value of space is 32 and that of a is 97 so you are effectively accessing array hash with a negative index.
To solve this you can ignore non-alphabets by doing :
if(! isalpha(str[i]) {
str[j++] = str[i++]; // copy the char.
continue; // ignore rest of the loop.
}
This is going to break on any space characters (or anything else outside the range 'a'..'z') because you are accessing beyond the bounds of your hash array.
void striprepeatedchars(char *str)
{
int seen[UCHAR_MAX + 1];
char *c, *n;
memset(seen, 0, sizeof(seen));
c = n = str;
while (*n != '\0') {
if (!isalpha(*n) || !seen[(unsigned char) *n]) {
*c = *n;
seen[(unsigned char) *n]++;
c++;
}
n++;
}
*c = '\0';
}
This is code golf, right?
d(s){char*i=s,*o=s;for(;*i;++i)!memchr(s,*i,o-s)?*o++=*i:0;*o=0;}
...
// iterate through the input string char by char.
for(i=0,j=0;str[i];)
{
if (str[i] == ' ')
{
str[j++] = str[i++];
continue;
}
// if the char is not hashed.
if(!hash[str[i] - 'a'])
{
...
#include <stdio.h>
#include <string.h>
int hash[26] = {0};
static int in_valid_range (char c);
static int get_hash_code (char c);
static char *
remove_repeated_char (char *s)
{
size_t len = strlen (s);
size_t i, j = 0;
for (i = 0; i < len; ++i)
{
if (in_valid_range (s[i]))
{
int h = get_hash_code (s[i]);
if (!hash[h])
{
s[j++] = s[i];
hash[h] = 1;
}
}
else
{
s[j++] = s[i];
}
}
s[j] = 0;
return s;
}
int
main (int argc, char **argv)
{
printf ("%s\n", remove_repeated_char (argv[1]));
return 0;
}
static int
in_valid_range (char c)
{
return (c >= 'a' && c <= 'z');
}
static int
get_hash_code (char c)
{
return (int) (c - 'a');
}
char *s;
int i = 0;
for (i = 0; s[i]; i++)
{
int j;
int gap = 0;
for (j = i + 1; s[j]; j++)
{
if (gap > 0)
s[j] = s[j + gap];
if (!s[j])
break;
while (s[i] == s[j])
{
s[j] = s[j + gap + 1];
gap++;
}
}
}

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