How to replace characters by strtok function - C? - c

Question

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?

Answer 1

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'

Answer 2

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;
}

Answer 3

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;
}

Answer 4

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.