I am trying to remove duplicated letters in each word from string.(I haven't specified it for upper and lower case letters yet)
Input:
Ii feel good todday!!
thhis iss fixed
Output:
I fel god today!
this is fixed
I am calling this function in the main and i have to use the result in another function. That's why I call it by reference.
int main(){
char string[100];
printf("Enter a string:");
gets(string);
dup_letters_rule(&string);
return 0;
}
void dup_letters_rule(char *str_[]){
char new_str_[100];
int i=0, j=0;
printf("Fixed duplicates:\n");
while(*str_[i]!='\0'){
if(*str_[i]== *str_[i+1] && *str_[i+1]!='\0'){
while(*str_[i]==*str_[i+1] && *str_[i+1]!='\0'){
i++;
}
*str_[i]=new_str_[j];
j++;
i++;
}
else{
*str_[i]=new_str_[j];
j++;
i++;
}
}
new_str_[j]='\0';
puts(new_str_);
}
It works like:
void dup_letters_rule(char *str_[]){
char *new_str_=*str_, *temp=*str_;
temp++;
printf("Fixed duplicates:\n");
while(*new_str_!='\0'){
if(*new_str_== *temp && *temp!='\0'){
while(*new_str_==*temp && *temp!='\0'){
new_str_++;
temp++;
}
putchar(*new_str_);
new_str_++;
temp++;
}
else{
putchar(*new_str_);
new_str_++;
temp++;
}
}
}
But then, I can't use *str_ string in another function.
The code can be simplified.
We can keep an int value that is the previous char seen and compare it against the current char and only "copy it out" if they are different. (i.e. we only need two pointers).
We also have to use tolower because Ii goes to I.
Although a second/output buffer could be used, the function can do the cleanup "in-place". Then, the caller can use the cleaned up buffer. This is what we'd normally want to do.
If the caller needs to keep the original string, it can save the original to a temp buffer and call the function with the temp
I had to refactor your code. I tested it against your sample input. It is annotated:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
void
dup_letters_rule(char *src)
{
char *dst = src;
int prev = -1;
// rchr -- the "raw" char
// lchr -- the result of tolower(rchr)
// prev -- the previous value of lchr (starts with -1 to force output of
// first char)
for (int rchr = *src++; rchr != 0; rchr = *src++) {
// get lowercase char
int lchr = tolower((unsigned char) rchr);
// output if _not_ a dup
if (lchr != prev)
*dst++ = rchr;
// remember this char for the next iteration
prev = lchr;
}
*dst = 0;
}
int
main(void)
{
char *cp;
char buf[1000];
while (1) {
cp = fgets(buf,sizeof(buf),stdin);
if (cp == NULL)
break;
// get rid of newline
buf[strcspn(buf,"\n")] = 0;
// eliminate dups
dup_letters_rule(buf);
// output the clean string
printf("%s\n",buf);
}
return 0;
}
UPDATE:
can i print the clean string in the dup_letters_rule function? – hamster
Sure, of course. We're the programmers, so we can do whatever we want ;-)
There is a maxim for functions: Do one thing well
In many actual (re)use cases, we don't want the simple/low level function to do printing. That is the usual.
But, we could certainly add printing to the function. We'd move the printf from main into the function itself.
To get the best of both worlds, we can use two functions. One that just does the transformation. And, a second that calls the simple function and then prints the result.
Here's a slight change that illustrates that. I renamed my function and created dup_letters_rule with the printf embedded in it:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
void
dup_letters_rule_basic(char *src)
{
char *dst = src;
int prev = -1;
// rchr -- the "raw" char
// lchr -- the result of tolower(rchr)
// prev -- the previous value of lchr (starts with -1 to force output of
// first char)
for (int rchr = *src++; rchr != 0; rchr = *src++) {
// get lowercase char
int lchr = tolower((unsigned char) rchr);
// output if _not_ a dup
if (lchr != prev)
*dst++ = rchr;
// remember this char for the next iteration
prev = lchr;
}
*dst = 0;
}
void
dup_letters_rule(char *buf)
{
dup_letters_rule_basic(buf);
// output the clean string
printf("%s\n",buf);
}
int
main(void)
{
char *cp;
char buf[1000];
while (1) {
cp = fgets(buf,sizeof(buf),stdin);
if (cp == NULL)
break;
// get rid of newline
buf[strcspn(buf,"\n")] = 0;
dup_letters_rule(buf);
}
return 0;
}
UPDATE #2:
and why it's not char *dst = *src; but char *dst = src; – hamster
This is basic C. We want dst to have the same value/contents that src does. Just as if we did:
int x = 23;
int y = x;
If we do what you're suggesting, the compiler flags the statement:
bad.c: In function ‘dup_letters_rule_basic’:
bad.c:8:14: warning: initialization of ‘char *’ from ‘char’ makes pointer from integer without a cast [-Wint-conversion]
char *dst = *src;
^
Doing char *dst = *src [as you suggest] is using * in two different ways.
Doing char *dst says that dst is defined as a pointer to a char.
Doing *src here [which is the initializer for dst and is an expression], the * is the dereference operator. It says "fetch the value (a char) pointed to by src". Not what we want.
Perhaps this would be more clear if we didn't use an initializer. We use a definition (without an initializer) and set the initial value of dst with an assignment statement:
char *dst; // define a char pointer (has _no_ initial value)
dst = src; // assign the value of dst from the value of src
The assignment [statement] can occur anywhere after the definition and before the for loop/statement. Here's the first few lines of the function body:
char *dst;
int prev = -1;
dst = src;
To remove the duplicate consecutive characters from a string in-place, keep track of position in string where the next character, which is not same as its previous character, to be write and check current processing character with previous character (ignore the difference in their case) except when the character is the first character of string because the first character does not have any character previous to it. If current processing character is same as previous character then move to next character in the string and if they are not same then overwrite the character at tracked position with current processing character and increment tracked position pointer by 1.
Its implementation:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
void remove_consecutive_dup_chars (char * pstr) {
if (pstr == NULL) {
printf ("Invalid input..\n");
return;
}
/* Pointer to keep track of position where next character
* to be write in order to remove consecutive duplicate character.
*/
char * p = pstr;
for (unsigned int i = 0; pstr[i] ; ++i) {
if ((i) && (tolower (pstr[i]) == tolower (pstr[i - 1]))) {
continue;
}
*p++ = pstr[i];
}
/* Add the null terminating character.
*/
*p = '\0';
}
int main (void) {
char buf[256] = {'\0'};
strcpy (buf, "Ii feel good todday!!");
remove_consecutive_dup_chars (buf);
printf ("%s\n", buf);
strcpy (buf, "thhis iss fixed");
remove_consecutive_dup_chars (buf);
printf ("%s\n", buf);
strcpy (buf, "");
remove_consecutive_dup_chars (buf);
printf ("%s\n", buf);
strcpy (buf, "aaaaaa zzzzzz");
remove_consecutive_dup_chars (buf);
printf ("%s\n", buf);
return 0;
}
Output:
I fel god today!
this is fixed
a z
Related
So what i have is a string(str) that i get from fgets(str, x, stdin);.
If i write for example "Hello World" i want to be able to add a character infront of each word in the string.
To get this "Hello? World?" as an example. I think i've made it alot harder for myself by trying to solve it this way:
add(char *s, char o, char c){
int i, j = 0;
for (i = 0; s[i] != '\0'; i++) {
if (s[i] != o) {
s[j] = s[i];
}
else {
s[j] = c;
}
j++;
}
}
add(str, ' ','?');
printf("\n%s", str);
This will read out "Hello?World" without the spaces. Now the only way i see this working is if i move everything after the first "?" one to the right while also making the positon of the "W" to a space and a "?" at the end. But for much longer strings i can't see myself doing that.
You can't safely extend a string with more characters without insuring the buffer that holds the string is big enough. So let's devise a solution that counts how many additional characters are needed, allocate a buffer big enough to hold a string of that length, then do the copy loop. Then return the new string back to the caller.
char* add(const char* s, char o, char c)
{
size_t len = strlen(s);
const char* str = s;
char* result = NULL;
char* newstring = NULL;
// count how many characters are needed for the new string
while (*str)
{
len += (*str== o) ? 2 : 1;
str++;
}
// allocate a result buffer big enough to hold the new string
result = malloc(len + 1); // +1 for null char
// now copy the string and insert the "c" parameter whenever "o" is seen
newstring = result;
str = s;
while (*str)
{
*newstring++ = *str;
if (*str == o)
{
*newstring++ = c;
}
str++;
}
*newString = '\0';
return result;
}
Then your code to invoke is as follows:
char* newstring g= add(str, ' ','?');
printf("\n%s", newstring);
free(newstring);
#include <stdio.h>
#include <string.h>
int main(void) {
char text[] = "Hello World";
for(char* word = strtok(text, " .,?!"); word; word = strtok(NULL, " .,?!"))
printf("%s? ", word);
return 0;
}
Example Output
Success #stdin #stdout 0s 4228KB
Hello? World?
IDEOne Link
Knowing the amount of storage available when you reach a position where the new character will be inserted, you can check whether the new character will fit in the available storage, move from the current character through end-of-string to the right by one and insert the new character, e.g.
#include <stdio.h>
#include <string.h>
#define MAXC 1024
char *add (char *s, const char find, const char replace)
{
char *p = s; /* pointer to string */
while (*p) { /* for each char */
if (*p == find) {
size_t remain = strlen (p); /* get remaining length */
if ((p - s + remain < MAXC - 1)) { /* if space remains for char */
memmove (p + 1, p, remain + 1); /* move chars to right by 1 */
*p++ = replace; /* replace char, advance ptr */
}
else { /* warn if string full */
fputs ("error: replacement will exceed storage.\n", stderr);
break;
}
}
p++; /* advance to next char */
}
return s; /* return pointer to beginning of string */
}
...
(note: the string must be mutable, not a string-literal, and have additional storage for the inserted character. If you need to pass a string-literal or you have no additional storage in the current string, make a copy as shown by #Selbie in his answer)
Putting together a short example with a 1024-char buffer for storage, you can do something like:
#include <stdio.h>
#include <string.h>
#define MAXC 1024
char *add (char *s, const char find, const char replace)
{
char *p = s; /* pointer to string */
while (*p) { /* for each char */
if (*p == find) {
size_t remain = strlen (p); /* get remaining length */
if ((p - s + remain < MAXC - 1)) { /* if space remains for char */
memmove (p + 1, p, remain + 1); /* move chars to right by 1 */
*p++ = replace; /* replace char, advance ptr */
}
else { /* warn if string full */
fputs ("error: replacement will exceed storage.\n", stderr);
break;
}
}
p++; /* advance to next char */
}
return s; /* return pointer to beginning of string */
}
int main (void) {
char buf[MAXC];
if (!fgets (buf, MAXC, stdin))
return 1;
buf[strcspn(buf, "\n")] = 0;
puts (add (buf, ' ', '?'));
}
Example Use/Output
$ ./bin/str_replace_c
Hello World?
Hello? World?
Look things over and let me know if you have questions.
Just for fun, here's my implementation. It modifies the string in-place and in O(n) time. It assumes that the char-buffer is large enough to hold the additional characters, so it's up to the calling code to ensure that.
#include <stdio.h>
void add(char *s, char o, char c)
{
int num_words = 0;
char * p = s;
while(*p) if (*p++ == o) num_words++;
char * readFrom = p;
char * writeTo = p+num_words;
char * nulByte = writeTo;
// Insert c-chars, iterating backwards to avoid overwriting chars we have yet to read
while(readFrom >= s)
{
*writeTo = *readFrom;
if (*writeTo == o)
{
--writeTo;
*writeTo = c;
}
writeTo--;
readFrom--;
}
// If our string doesn't end in a 'c' char, append one
if ((nulByte > s)&&(*(nulByte-1) != c))
{
*nulByte++ = c;
*nulByte = '\0';
}
}
int main(int argc, char ** argv)
{
char test_string[1000] = "Hello World";
add(test_string, ' ','?');
printf("%s\n", test_string);
return 0;
}
The program's output is:
$ ./a.out
Hello? World?
I'm having looping issues with my code. I have a method that takes in two char arrays (phrase, characters). The characters array holds characters that must be read individually and compared to the phrase. If it matches, every occurrence of the character will be removed from the phrase.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
//This method has two parameters: (str, c)
//It will remove all occurences of var 'c'
//inside of 'str'
char * rmstr(char * c, char * str) {
//Declare counters and pointers
int stemp = 0;
int ctemp = 0;
char *p = str;
char *d = c;
//Retrieve str count
while(str[stemp] != '\0') {
stemp++;
}
//Retrieve c count
while(c[ctemp] != '\0') {
ctemp++;
}
//Output information
printf("String Count: %d\n",stemp);
printf("Character Count: %d\n",ctemp);
//Iterate through arrays
for (int i = 0; i != stemp; i++) {
for (int j = 0; j != ctemp; j++) {
if (c[j] != str[i]){
*p++ = str[i];
}
break;
}
printf("%s\n",str);
}
*p = 0;
return str;
}
int main()
{
char c[256] = "ema";
char input[256] = "Great message!";
char *result = rmstr(c, input);
printf("%s", result);
return 0;
}
In this case, the input would be "Great Message!" and the character I'd like to remove all occurrences of the characters: e, m, a (As specified in main).
Using the code as it is above, the output is as follows:
Grat mssag!
It is only looping through 1 iteration and removing 'e'. I would like it to loop through 'm' and 'a' as well.
After you fix your break; that was causing your inner loop to exit, it may make sense to reorder your loops and loop over the chars to remove while checking against the characters in str. This is more of a convenience allowing you to shuffle each character down by one in str if it matches a character is c. If you are using the functions in string.h like memmove to move characters down, it doesn't really matter.
A simple implementation using only pointers to manually work through str removing all chars in c could look something like the following:
#include <stdio.h>
char *rmstr (char *str, const char *chars)
{
const char *c = chars; /* set pointer to beginning of chars */
while (*c) { /* loop over all chars with c */
char *p = str; /* set pointer to str */
while (*p) { /* loop over each char in str */
if (*p == *c) { /* if char in str should be removed */
char *sp = p, /* set start pointer at p */
*ep = p + 1; /* set end pointer at p + 1 */
do
*sp++ = *ep; /* copy end to start to end of str */
while (*ep++); /* (nul-char copied on last iteration) */
}
p++; /* advance to next char in str */
}
c++; /* advance to next char in chars */
}
return str; /* return modified str */
}
int main (void) {
char c[] = "ema";
char input[] = "Great message!";
printf ("original: %s\n", input);
printf ("modified: %s\n", rmstr (input, c));
return 0;
}
(there are many ways to do this -- how is largely up to you. whether you use pointers as above, or get the lengths and use string-indexes is also a matter of choice)
Example Use/Output
$ ./bin/rmcharsinstr
original: Great message!
modified: Grt ssg!
If you did want to use memmove (to address the overlapping nature of the source and destination) to move the remaining characters in str down by one each time the character in str matches a character in c, you could leave the loops in your original order, e.g.
#include <string.h>
char *rmstr (char *str, const char *chars)
{
char *p = str; /* set pointer to str */
while (*p) { /* loop over each char in str */
const char *c = chars; /* set pointer to beginning of chars */
while (*c) { /* loop over all chars with c */
while (*c == *p) { /* while the character matches */
memmove (p, p + 1, strlen (p)); /* shuffle down by 1 */
c = chars; /* reset c = chars to check next */
}
c++; /* advance to next char in chars */
}
p++; /* advance to next char in str */
}
return str; /* return modified str */
}
(make sure you understand why you must reset c = chars; in this case)
Finally, if you really wanted the shorthand way of doing it, you could use strpbrk and memmove and reduce your function to:
#include <string.h>
char *rmstr (char *str, const char *chars)
{
/* simply loop using strpbrk removing the character found */
for (char *p = strpbrk (str, chars); p; p = strpbrk (str, chars))
memmove (p, p+1, strlen(p));
return str; /* return modified str */
}
(there is always more than one way to skin-the-cat in C)
The output is the same. Look things over here and let me know if you have further questions.
I want to write a program in C that displays each word of a whole sentence (taken as input) at a seperate line. This is what I have done so far:
void manipulate(char *buffer);
int get_words(char *buffer);
int main(){
char buff[100];
printf("sizeof %d\nstrlen %d\n", sizeof(buff), strlen(buff)); // Debugging reasons
bzero(buff, sizeof(buff));
printf("Give me the text:\n");
fgets(buff, sizeof(buff), stdin);
manipulate(buff);
return 0;
}
int get_words(char *buffer){ // Function that gets the word count, by counting the spaces.
int count;
int wordcount = 0;
char ch;
for (count = 0; count < strlen(buffer); count ++){
ch = buffer[count];
if((isblank(ch)) || (buffer[count] == '\0')){ // if the character is blank, or null byte add 1 to the wordcounter
wordcount += 1;
}
}
printf("%d\n\n", wordcount);
return wordcount;
}
void manipulate(char *buffer){
int words = get_words(buffer);
char *newbuff[words];
char *ptr;
int count = 0;
int count2 = 0;
char ch = '\n';
ptr = buffer;
bzero(newbuff, sizeof(newbuff));
for (count = 0; count < 100; count ++){
ch = buffer[count];
if (isblank(ch) || buffer[count] == '\0'){
buffer[count] = '\0';
if((newbuff[count2] = (char *)malloc(strlen(buffer))) == NULL) {
printf("MALLOC ERROR!\n");
exit(-1);
}
strcpy(newbuff[count2], ptr);
printf("\n%s\n",newbuff[count2]);
ptr = &buffer[count + 1];
count2 ++;
}
}
}
Although the output is what I want, I have really many black spaces after the final word displayed, and the malloc() returns NULL so the MALLOC ERROR! is displayed in the end.
I can understand that there is a mistake at my malloc() implementation, but I do not know what it is.
Is there another more elegant or generally better way to do it?
http://www.cplusplus.com/reference/clibrary/cstring/strtok/
Take a look at this, and use whitespace characters as the delimiter. If you need more hints let me know.
From the website:
char * strtok ( char * str, const char * delimiters );
On a first call, the function expects a C string as argument for str, whose first character is used as the starting location to scan for tokens. In subsequent calls, the function expects a null pointer and uses the position right after the end of last token as the new starting location for scanning.
Once the terminating null character of str is found in a call to strtok, all subsequent calls to this function (with a null pointer as the first argument) return a null pointer.
Parameters
str
C string to truncate.
Notice that this string is modified by being broken into smaller strings (tokens).
Alternativelly [sic], a null pointer may be specified, in which case the function continues scanning where a previous successful call to the function ended.
delimiters
C string containing the delimiter characters.
These may vary from one call to another.
Return Value
A pointer to the last token found in string.
A null pointer is returned if there are no tokens left to retrieve.
Example
/* strtok example */
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] ="- This, a sample string.";
char * pch;
printf ("Splitting string \"%s\" into tokens:\n",str);
pch = strtok (str," ,.-");
while (pch != NULL)
{
printf ("%s\n",pch);
pch = strtok (NULL, " ,.-");
}
return 0;
}
For the fun of it here's an implementation based on the callback approach:
const char* find(const char* s,
const char* e,
int (*pred)(char))
{
while( s != e && !pred(*s) ) ++s;
return s;
}
void split_on_ws(const char* s,
const char* e,
void (*callback)(const char*, const char*))
{
const char* p = s;
while( s != e ) {
s = find(s, e, isspace);
callback(p, s);
p = s = find(s, e, isnotspace);
}
}
void handle_word(const char* s, const char* e)
{
// handle the word that starts at s and ends at e
}
int main()
{
split_on_ws(some_str, some_str + strlen(some_str), handle_word);
}
malloc(0) may (optionally) return NULL, depending on the implementation. Do you realize why you may be calling malloc(0)? Or more precisely, do you see where you are reading and writing beyond the size of your arrays?
Consider using strtok_r, as others have suggested, or something like:
void printWords(const char *string) {
// Make a local copy of the string that we can manipulate.
char * const copy = strdup(string);
char *space = copy;
// Find the next space in the string, and replace it with a newline.
while (space = strchr(space,' ')) *space = '\n';
// There are no more spaces in the string; print out our modified copy.
printf("%s\n", copy);
// Free our local copy
free(copy);
}
Something going wrong is get_words() always returning one less than the actual word count, so eventually you attempt to:
char *newbuff[words]; /* Words is one less than the actual number,
so this is declared to be too small. */
newbuff[count2] = (char *)malloc(strlen(buffer))
count2, eventually, is always one more than the number of elements you've declared for newbuff[]. Why malloc() isn't returning a valid ptr, though, I don't know.
You should be malloc'ing strlen(ptr), not strlen(buf). Also, your count2 should be limited to the number of words. When you get to the end of your string, you continue going over the zeros in your buffer and adding zero size strings to your array.
Just as an idea of a different style of string manipulation in C, here's an example which does not modify the source string, and does not use malloc. To find spaces I use the libc function strpbrk.
int print_words(const char *string, FILE *f)
{
static const char space_characters[] = " \t";
const char *next_space;
// Find the next space in the string
//
while ((next_space = strpbrk(string, space_characters)))
{
const char *p;
// If there are non-space characters between what we found
// and what we started from, print them.
//
if (next_space != string)
{
for (p=string; p<next_space; p++)
{
if(fputc(*p, f) == EOF)
{
return -1;
}
}
// Print a newline
//
if (fputc('\n', f) == EOF)
{
return -1;
}
}
// Advance next_space until we hit a non-space character
//
while (*next_space && strchr(space_characters, *next_space))
{
next_space++;
}
// Advance the string
//
string = next_space;
}
// Handle the case where there are no spaces left in the string
//
if (*string)
{
if (fprintf(f, "%s\n", string) < 0)
{
return -1;
}
}
return 0;
}
you can scan the char array looking for the token if you found it just print new line else print the char.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *s;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
int len = strlen(s);
char delim =' ';
for(int i = 0; i < len; i++) {
if(s[i] == delim) {
printf("\n");
}
else {
printf("%c", s[i]);
}
}
free(s);
return 0;
}
char arr[50];
gets(arr);
int c=0,i,l;
l=strlen(arr);
for(i=0;i<l;i++){
if(arr[i]==32){
printf("\n");
}
else
printf("%c",arr[i]);
}
char * removeChar(char * str, char c){
int len = strlen(str);
int i = 0;
int j = 0;
char * copy = malloc(sizeof(char) * (len + 1));
while(i < len){
if(str[i] != c){
copy[j] = str[i];
j++;
i++;
}else{
i++;
}
}
if(strcmp(copy, str) != 0){
strcpy(str,copy);
}else{
printf("Error");
}
return copy;
}
int main(int argc, char * argv[]){
char str[] = "Input string";
char * input;
input = removeChar(str,'g');
printf("%s\n", input);
free(input);
return 0;
}
I don't know why every time I try to run it ,it always says uninitialized variable and sticks in the strcpy line and printf line.
Basically this function is to take a string, and a character and removes the that character from the string (because I am learning malloc so that's why I wrote the function like this).
After the while loop do:
copy[j] = '\0';
to NULL-terminate your string; that way it can work with methods coming from <string.h>, which assume that the string is nul-terminated.
PS: One warning you should see is about not returning copy in your function in any case, because now if the condition of the if statement is wrong, your function won't return something valid, so add this:
return copy;
at the end of your function (which is now corrected with your edit).
Other than that, the only warning you should still get are for the unused arguments of main(), nothing else:
prog.c: In function 'main':
prog.c:32:14: warning: unused parameter 'argc' [-Wunused-parameter]
int main(int argc, char * argv[]){
^~~~
prog.c:32:27: warning: unused parameter 'argv' [-Wunused-parameter]
int main(int argc, char * argv[]){
^~~~
While you copy over bytes from str to copy, you don't add a terminating null byte at the end. As a result, strcmp reads past the copied characters into unitialized memory, possibly past the end of the allocated memory block. This invokes undefined behavior.
After your while loop, add a terminating null byte to copy.
Also, you never return a value if the if block at the end is false. You need to return something for that, probably the copied string.
char * removeChar(char * str, char c){
int len = strlen(str);
int i = 0;
int j = 0;
char * copy = malloc(sizeof(char) * (len + 1));
while(i < len){
if(str[i] != c){
copy[j] = str[i];
j++;
i++;
}else{
i++;
}
}
// add terminating null byte
copy[j] = '\0';
if(strcmp(copy, str) != 0){
strcpy(str,copy);
}
// always return copy
return copy;
}
You never initialised input and the some compilers don't notice,
that the the value is never used before the line
input = removeChar(str, 'g');
in your code. So they emit the diagnostic just to be sure.
strcpy(str, copy)
gets stuck in your code, as copy never got a closing 0 byte and
so depends on the nondeterministic content of your memory at the
moment of the allocation of the memory backing copy, how long strcpy
will run and if you get eventually a SIGSEGV (or similar).
strcpy will loop until it finds a 0 byte in your memory.
For starters to remove a character from a string there is no need to create dynamically a character array and then copy this array into the original string.
Either you should write a function that indeed removes the specified character from a string or a function that creates a new string based on the source string excluding the specified character.
It is just a bad design that only confuses users. That is the function is too complicated and uses redundant functions like malloc, strlen, strcmp and strcpy. And in fact it has a side effect that is not obvious. Moreover there is used incorrect type int for the length of a string instead of the type size_t.
As for your function implementation then you forgot to append the terminating zero '\0' to the string built in the dynamically allocated array.
If you indeed want to remove a character from a string then the function can look as it is shown in the demonstrative program.
#include <stdio.h>
char * remove_char(char *s, char c)
{
char *p = s;
while (*p && *p != c) ++p;
for ( char *q = p; *p++; )
{
if (*p != c) *q++ = *p;
}
return s;
}
int main( void )
{
char str[] = "Input string";
puts(str);
puts(remove_char(str, 'g'));
return 0;
}
The program output is
Input string
Input strin
If you are learning the function malloc and want to use it you in any case should try to implement a correct design.
To use malloc you could write a function that creates a new string based on the source string excluding the specified character. For example
#include <stdio.h>
#include <stdlib.h>
char * remove_copy_char(const char *s, char c)
{
size_t n = 0;
for (const char *p = s; *p; ++p)
{
if (*p != c) ++n;
}
char *result = malloc(n + 1);
if (result)
{
char *q = result;
for (; *s; ++s)
{
if (*s != c) *q++ = *s;
}
*q = '\0';
}
return result;
}
int main( void )
{
char *str = "Input string";
puts(str);
char *p = remove_copy_char(str, 'g');
if ( p ) puts(p );
free(p);
return 0;
}
The program output will be the same as above.
Input string
Input strin
Pay attention to the function declaration
char * remove_copy_char(const char *s, char c);
^^^^^^
In this case the source string can be a string literal.
char *str = "Input string";
I want to write a program in C that displays each word of a whole sentence (taken as input) at a seperate line. This is what I have done so far:
void manipulate(char *buffer);
int get_words(char *buffer);
int main(){
char buff[100];
printf("sizeof %d\nstrlen %d\n", sizeof(buff), strlen(buff)); // Debugging reasons
bzero(buff, sizeof(buff));
printf("Give me the text:\n");
fgets(buff, sizeof(buff), stdin);
manipulate(buff);
return 0;
}
int get_words(char *buffer){ // Function that gets the word count, by counting the spaces.
int count;
int wordcount = 0;
char ch;
for (count = 0; count < strlen(buffer); count ++){
ch = buffer[count];
if((isblank(ch)) || (buffer[count] == '\0')){ // if the character is blank, or null byte add 1 to the wordcounter
wordcount += 1;
}
}
printf("%d\n\n", wordcount);
return wordcount;
}
void manipulate(char *buffer){
int words = get_words(buffer);
char *newbuff[words];
char *ptr;
int count = 0;
int count2 = 0;
char ch = '\n';
ptr = buffer;
bzero(newbuff, sizeof(newbuff));
for (count = 0; count < 100; count ++){
ch = buffer[count];
if (isblank(ch) || buffer[count] == '\0'){
buffer[count] = '\0';
if((newbuff[count2] = (char *)malloc(strlen(buffer))) == NULL) {
printf("MALLOC ERROR!\n");
exit(-1);
}
strcpy(newbuff[count2], ptr);
printf("\n%s\n",newbuff[count2]);
ptr = &buffer[count + 1];
count2 ++;
}
}
}
Although the output is what I want, I have really many black spaces after the final word displayed, and the malloc() returns NULL so the MALLOC ERROR! is displayed in the end.
I can understand that there is a mistake at my malloc() implementation, but I do not know what it is.
Is there another more elegant or generally better way to do it?
http://www.cplusplus.com/reference/clibrary/cstring/strtok/
Take a look at this, and use whitespace characters as the delimiter. If you need more hints let me know.
From the website:
char * strtok ( char * str, const char * delimiters );
On a first call, the function expects a C string as argument for str, whose first character is used as the starting location to scan for tokens. In subsequent calls, the function expects a null pointer and uses the position right after the end of last token as the new starting location for scanning.
Once the terminating null character of str is found in a call to strtok, all subsequent calls to this function (with a null pointer as the first argument) return a null pointer.
Parameters
str
C string to truncate.
Notice that this string is modified by being broken into smaller strings (tokens).
Alternativelly [sic], a null pointer may be specified, in which case the function continues scanning where a previous successful call to the function ended.
delimiters
C string containing the delimiter characters.
These may vary from one call to another.
Return Value
A pointer to the last token found in string.
A null pointer is returned if there are no tokens left to retrieve.
Example
/* strtok example */
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] ="- This, a sample string.";
char * pch;
printf ("Splitting string \"%s\" into tokens:\n",str);
pch = strtok (str," ,.-");
while (pch != NULL)
{
printf ("%s\n",pch);
pch = strtok (NULL, " ,.-");
}
return 0;
}
For the fun of it here's an implementation based on the callback approach:
const char* find(const char* s,
const char* e,
int (*pred)(char))
{
while( s != e && !pred(*s) ) ++s;
return s;
}
void split_on_ws(const char* s,
const char* e,
void (*callback)(const char*, const char*))
{
const char* p = s;
while( s != e ) {
s = find(s, e, isspace);
callback(p, s);
p = s = find(s, e, isnotspace);
}
}
void handle_word(const char* s, const char* e)
{
// handle the word that starts at s and ends at e
}
int main()
{
split_on_ws(some_str, some_str + strlen(some_str), handle_word);
}
malloc(0) may (optionally) return NULL, depending on the implementation. Do you realize why you may be calling malloc(0)? Or more precisely, do you see where you are reading and writing beyond the size of your arrays?
Consider using strtok_r, as others have suggested, or something like:
void printWords(const char *string) {
// Make a local copy of the string that we can manipulate.
char * const copy = strdup(string);
char *space = copy;
// Find the next space in the string, and replace it with a newline.
while (space = strchr(space,' ')) *space = '\n';
// There are no more spaces in the string; print out our modified copy.
printf("%s\n", copy);
// Free our local copy
free(copy);
}
Something going wrong is get_words() always returning one less than the actual word count, so eventually you attempt to:
char *newbuff[words]; /* Words is one less than the actual number,
so this is declared to be too small. */
newbuff[count2] = (char *)malloc(strlen(buffer))
count2, eventually, is always one more than the number of elements you've declared for newbuff[]. Why malloc() isn't returning a valid ptr, though, I don't know.
You should be malloc'ing strlen(ptr), not strlen(buf). Also, your count2 should be limited to the number of words. When you get to the end of your string, you continue going over the zeros in your buffer and adding zero size strings to your array.
Just as an idea of a different style of string manipulation in C, here's an example which does not modify the source string, and does not use malloc. To find spaces I use the libc function strpbrk.
int print_words(const char *string, FILE *f)
{
static const char space_characters[] = " \t";
const char *next_space;
// Find the next space in the string
//
while ((next_space = strpbrk(string, space_characters)))
{
const char *p;
// If there are non-space characters between what we found
// and what we started from, print them.
//
if (next_space != string)
{
for (p=string; p<next_space; p++)
{
if(fputc(*p, f) == EOF)
{
return -1;
}
}
// Print a newline
//
if (fputc('\n', f) == EOF)
{
return -1;
}
}
// Advance next_space until we hit a non-space character
//
while (*next_space && strchr(space_characters, *next_space))
{
next_space++;
}
// Advance the string
//
string = next_space;
}
// Handle the case where there are no spaces left in the string
//
if (*string)
{
if (fprintf(f, "%s\n", string) < 0)
{
return -1;
}
}
return 0;
}
you can scan the char array looking for the token if you found it just print new line else print the char.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *s;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
int len = strlen(s);
char delim =' ';
for(int i = 0; i < len; i++) {
if(s[i] == delim) {
printf("\n");
}
else {
printf("%c", s[i]);
}
}
free(s);
return 0;
}
char arr[50];
gets(arr);
int c=0,i,l;
l=strlen(arr);
for(i=0;i<l;i++){
if(arr[i]==32){
printf("\n");
}
else
printf("%c",arr[i]);
}