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I have an array of strings read from a file. I'd like to take each string in the existing array and copy it to a new array unit the first instance of a tab character, and then move to the next element in the array.
What would be the best way to do this?
Thanks
You can use standard C function strchr. For example if you have two character arrays like
char s1[12] = "Hello\tWorld";
char s2[12];
then you can write
char *p = strchr( s1, '\t' );
if ( p != NULL )
{
memcpy( s2, s1, p - s1 );
s2[p - s1] = '\0';
}
else
{
strcpy( s2, s1 );
}
For two dimensional arrays you can do the same in a loop.
You could create a function until_tabs that takes an array of strings and the array's length. Then we allocate a same sized array of char pointers and iterate over the original array.
For each string input we can use strchr to look for a tab. If it's absent, just duplicate the string with strdup. Otherwise allocate an adequately sized buffer for the new string and copy everything before '\t' into it with strncpy.
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
char **until_tabs(char **strings, size_t n) {
char **result = malloc(sizeof(char *) * n);
for (size_t i = 0; i < n; i++) {
char *tab = strchr(strings[i], '\t');
if (!tab) {
result[i] = strdup(strings[i]);
continue;
}
size_t size = tab-strings[i];
result[i] = malloc(size+1);
strncpy(result[i], strings[i], size);
result[i][size] = '\0';
}
return result;
}
int main(void) {
char *arr[] = {"hello", "world", "foo\tbar"};
char **arr2 = until_tabs(arr, 3);
for (size_t i = 0; i < 3; i++) {
printf("%s\n", arr2[i]);
}
return 0;
}
Output:
hello
world
foo
For each string in the array of strings:
Find the length of the string, but only up to a potential '\t'
size_t length == strcspn(source_string, "\t");
Allocated needed memory
char *destination_string = malloc(length + 1); // +1 for the null character.
if (destination_string == NULL) Handle_Allocation_Error();
Copy it
memcpy(destination_string, source_string, length);
destination_string[length] = '\0';
// or
sprintf(destination_string, ".*s", (int) length, source_string);
// or
strncpy(destination_string, source_string, length);
destination_string[length] = '\0';
I really want to change all spaces ' ' in my char array for NULL -
#include <string.h>
void ReplaceCharactersInString(char *pcString, char *cOldChar, char *cNewChar) {
char *p = strtok(pcString, cOldChar);
strcpy(pcString, p);
while (p != NULL) {
strcat(pcString, p);
p = strtok(cNewChar, cOldChar);
}
}
int main() {
char pcString[] = "I am testing";
ReplaceCharactersInString(pcString, " ", NULL);
printf(pcString);
}
OUTPUT: Iamtesting
If I simply put the printf(p) function before:
p = strtok(cNewChar, cOldChar);
In the result I have what I need - but the problem is how to store it in pcString (directly)?
Or there is maybe a better solution to simply do it?
While some functions expect a [single] string to be pre-parsed to: I\0am\0testing, that is rare.
And, if you have multiple spaces/delimiters, you'll get (e.g.) foo\0\0bar, which you probably don't want.
And, your printf in main will only print the first token in the string because it will stop on the first EOS (i.e. '\0').
(i.e.) You probably don't want strcpy/strcat.
More likely, you want to fill an array of char * pointers to the tokens you parse.
So, you'd want to pass down char **argv, then do: argv[argc++] = strtok(...); and then do: return argc
Here's how I would refactor your code:
#include <stdio.h>
#include <string.h>
#define ARGMAX 100
int
ReplaceCharactersInString(int argmax,char **argv,char *pcString,
const char *delim)
{
char *p;
int argc;
// allow space for NULL termination
--argmax;
for (argc = 0; argc < argmax; ++argc, ++argv) {
// get next token
p = strtok(pcString,delim);
if (p == NULL)
break;
// zap the buffer pointer
pcString = NULL;
// store the token in the [returned] array
*argv = p;
}
*argv = NULL;
return argc;
}
int
main(void)
{
char pcString[] = "I am testing";
int argc;
char **av;
char *argv[ARGMAX];
argc = ReplaceCharactersInString(ARGMAX,argv,pcString," ");
printf("argc: %d\n",argc);
for (av = argv; *av != NULL; ++av)
printf("'%s'\n",*av);
return 0;
}
Here's the output:
argc: 3
'I'
'am'
'testing'
strcat strcpy should not be used when the source and destination overlap in memory.
Iterate through the array and replace the matching character with the desired character.
Since zeros are part of the string, printf will stop at the first zero and strlen can't be used for the length to print. sizeof can be used as pcString is defined in the same scope.
Note that ReplaceCharactersInString would not work a second time as it would stop at the first zero. The function could be written to accept a length parameter and loop using the length.
#include <stdio.h>
#include <stdlib.h>
void ReplaceCharactersInString(char *pcString, char cOldChar,char cNewChar){
while ( pcString && *pcString) {//not NULL and not zero
if ( *pcString == cOldChar) {//match
*pcString = cNewChar;//replace
}
++pcString;//advance to next character
}
}
int main ( void) {
char pcString[] = "I am testing";
ReplaceCharactersInString ( pcString, ' ', '\0');
for ( int each = 0; each < sizeof pcString; ++each) {
printf ( "pcString[%02d] = int:%-4d char:%c\n", each, pcString[each], pcString[each]);
}
return 0;
}
You want to split the string into individual tokens separated by spaces such as "I\0am\0testing\0". You can use strtok() for this but this function is error prone. I suggest you allocate an array of pointers and make them point to the words. Note that splitting the source string is sloppy and does not allow for tokens to be adjacent such as in 1+1. You could allocate the strings instead.
Here is an example:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char **split_string(const char *str, char *delim) {
size_t i, len, count;
const char *p;
/* count tokens */
p = str;
p += strspn(p, delim); // skip initial delimiters
count = 0;
while (*p) {
count++;
p += strcspn(p, delim); // skip token
p += strspn(p, delim); // skip delimiters
}
/* allocate token array */
char **array = calloc(sizeof(*array, count + 1);
p = str;
p += strspn(p, delim); // skip initial delimiters
for (i = 0; i < count; i++) {
len = strcspn(p, delim); // token length
array[i] = strndup(p, len); // allocate a copy of the token
p += len; // skip token
p += strspn(p, delim); // skip delimiters
}
/* array ends with a null pointer */
array[count] = NULL;
return array;
}
int main() {
const char *pcString = "I am testing";
char **array = split_string(pcString, " \t\r\n");
for (size_t i = 0; array[i] != NULL; i++) {
printf("%zu: %s\n", i, array[i]);
}
return 0;
}
The strtok function pretty much does exactly what you want. It basically replaces the next delimiter with a '\0' character and returns the pointer to the current token. The next time you call strtok, you should pass a NULL argument (see the documentation for strtok) and it will point to the next token, which will again be delimited by '\0'. Read some more examples of correct strtok usage.
My str_split function returns (or at least I think it does) a char** - so a list of strings essentially. It takes a string parameter, a char delimiter to split the string on, and a pointer to an int to place the number of strings detected.
The way I did it, which may be highly inefficient, is to make a buffer of x length (x = length of string), then copy element of string until we reach delimiter, or '\0' character. Then it copies the buffer to the char**, which is what we are returning (and has been malloced earlier, and can be freed from main()), then clears the buffer and repeats.
Although the algorithm may be iffy, the logic is definitely sound as my debug code (the _D) shows it's being copied correctly. The part I'm stuck on is when I make a char** in main, set it equal to my function. It doesn't return null, crash the program, or throw any errors, but it doesn't quite seem to work either. I'm assuming this is what is meant be the term Undefined Behavior.
Anyhow, after a lot of thinking (I'm new to all this) I tried something else, which you will see in the code, currently commented out. When I use malloc to copy the buffer to a new string, and pass that copy to aforementioned char**, it seems to work perfectly. HOWEVER, this creates an obvious memory leak as I can't free it later... so I'm lost.
When I did some research I found this post, which follows the idea of my code almost exactly and works, meaning there isn't an inherent problem with the format (return value, parameters, etc) of my str_split function. YET his only has 1 malloc, for the char**, and works just fine.
Below is my code. I've been trying to figure this out and it's scrambling my brain, so I'd really appreciate help!! Sorry in advance for the 'i', 'b', 'c' it's a bit convoluted I know.
Edit: should mention that with the following code,
ret[c] = buffer;
printf("Content of ret[%i] = \"%s\" \n", c, ret[c]);
it does indeed print correctly. It's only when I call the function from main that it gets weird. I'm guessing it's because it's out of scope ?
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define DEBUG
#ifdef DEBUG
#define _D if (1)
#else
#define _D if (0)
#endif
char **str_split(char[], char, int*);
int count_char(char[], char);
int main(void) {
int num_strings = 0;
char **result = str_split("Helo_World_poopy_pants", '_', &num_strings);
if (result == NULL) {
printf("result is NULL\n");
return 0;
}
if (num_strings > 0) {
for (int i = 0; i < num_strings; i++) {
printf("\"%s\" \n", result[i]);
}
}
free(result);
return 0;
}
char **str_split(char string[], char delim, int *num_strings) {
int num_delim = count_char(string, delim);
*num_strings = num_delim + 1;
if (*num_strings < 2) {
return NULL;
}
//return value
char **ret = malloc((*num_strings) * sizeof(char*));
if (ret == NULL) {
_D printf("ret is null.\n");
return NULL;
}
int slen = strlen(string);
char buffer[slen];
/* b is the buffer index, c is the index for **ret */
int b = 0, c = 0;
for (int i = 0; i < slen + 1; i++) {
char cur = string[i];
if (cur == delim || cur == '\0') {
_D printf("Copying content of buffer to ret[%i]\n", c);
//char *tmp = malloc(sizeof(char) * slen + 1);
//strcpy(tmp, buffer);
//ret[c] = tmp;
ret[c] = buffer;
_D printf("Content of ret[%i] = \"%s\" \n", c, ret[c]);
//free(tmp);
c++;
b = 0;
continue;
}
//otherwise
_D printf("{%i} Copying char[%c] to index [%i] of buffer\n", c, cur, b);
buffer[b] = cur;
buffer[b+1] = '\0'; /* extend the null char */
b++;
_D printf("Buffer is now equal to: \"%s\"\n", buffer);
}
return ret;
}
int count_char(char base[], char c) {
int count = 0;
int i = 0;
while (base[i] != '\0') {
if (base[i++] == c) {
count++;
}
}
_D printf("Found %i occurence(s) of '%c'\n", count, c);
return count;
}
You are storing pointers to a buffer that exists on the stack. Using those pointers after returning from the function results in undefined behavior.
To get around this requires one of the following:
Allow the function to modify the input string (i.e. replace delimiters with null-terminator characters) and return pointers into it. The caller must be aware that this can happen. Note that supplying a string literal as you are doing here is illegal in C, so you would instead need to do:
char my_string[] = "Helo_World_poopy_pants";
char **result = str_split(my_string, '_', &num_strings);
In this case, the function should also make it clear that a string literal is not acceptable input, and define its first parameter as const char* string (instead of char string[]).
Allow the function to make a copy of the string and then modify the copy. You have expressed concerns about leaking this memory, but that concern is mostly to do with your program's design rather than a necessity.
It's perfectly valid to duplicate each string individually and then clean them all up later. The main issue is that it's inconvenient, and also slightly pointless.
Let's address the second point. You have several options, but if you insist that the result be easily cleaned-up with a call to free, then try this strategy:
When you allocate the pointer array, also make it large enough to hold a copy of the string:
// Allocate storage for `num_strings` pointers, plus a copy of the original string,
// then copy the string into memory immediately following the pointer storage.
char **ret = malloc((*num_strings) * sizeof(char*) + strlen(string) + 1);
char *buffer = (char*)&ret[*num_strings];
strcpy(buffer, string);
Now, do all your string operations on buffer. For example:
// Extract all delimited substrings. Here, buffer will always point at the
// current substring, and p will search for the delimiter. Once found,
// the substring is terminated, its pointer appended to the substring array,
// and then buffer is pointed at the next substring, if any.
int c = 0;
for(char *p = buffer; *buffer; ++p)
{
if (*p == delim || !*p) {
char *next = p;
if (*p) {
*p = '\0';
++next;
}
ret[c++] = buffer;
buffer = next;
}
}
When you need to clean up, it's just a single call to free, because everything was stored together.
The string pointers you store into the res with ret[c] = buffer; array point to an automatic array that goes out of scope when the function returns. The code subsequently has undefined behavior. You should allocate these strings with strdup().
Note also that it might not be appropriate to return NULL when the string does not contain a separator. Why not return an array with a single string?
Here is a simpler implementation:
#include <stdlib.h>
char **str_split(const char *string, char delim, int *num_strings) {
int i, n, from, to;
char **res;
for (n = 1, i = 0; string[i]; i++)
n += (string[i] == delim);
*num_strings = 0;
res = malloc(sizeof(*res) * n);
if (res == NULL)
return NULL;
for (i = from = to = 0;; from = to + 1) {
for (to = from; string[to] != delim && string[to] != '\0'; to++)
continue;
res[i] = malloc(to - from + 1);
if (res[i] == NULL) {
/* allocation failure: free memory allocated so far */
while (i > 0)
free(res[--i]);
free(res);
return NULL;
}
memcpy(res[i], string + from, to - from);
res[i][to - from] = '\0';
i++;
if (string[to] == '\0')
break;
}
*num_strings = n;
return res;
}
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]);
}
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]);
}