#include <stdio.h>
#include <string.h>
int main(){
char name[] = "eseumdesconhecidolheoferecerflores.issoeimpulse.cities";
char *str;
printf("%s\n", name)
str = strtok(name, ".cities");
printf("%s\n", str);
return 0;
}
This is the output:
eseumdesconhecidolheoferecerflores.issoeimpulse.cities
umd
I have no idea what is happening at all. What I want is for the output of strtok to be a pointer to "eseumdesconhecidolheoferecerflores.issoeimpulse"
The delimiter argument to strtok is a string containing individual characters used to separate the string.
You specified delimiters ., c, i, t, e, and s.
So it's no surprise the output is umd for the first token, since it is surrounded by characters in your delimiter string.
If you want to find a whole string, you should use strstr instead.
For example:
char name[] = "eseumdesconhecidolheoferecerflores.issoeimpulse.cities";
char *pos;
pos = strstr(name, ".cities");
if (pos)
{
*pos = '\0';
printf("%s\n", name);
}
char line[] = "COPY\tSTART\t0\tCOPY";
char *tmp;
tmp = strtok(line, "\t");
printf("%s", tmp);
This code's output is COPY. And when
char line[] = "\tSTART\t0\tCOPY";
Output is START.
But! I want to check there is nothing in front of string START.
That is I think \t is first delimiter so output of strtok(line, "\t") is NULL.
But real output is START.
Is there any misunderstanding? What can I do?
As per the man page of strtok() (emphasis mine)
A sequence of two or more contiguous delimiter bytes in the parsed string is considered to be a single delimiter. Delimiter bytes at the start or end of the string are ignored. Put another way: the tokens returned by strtok() are always nonempty strings.
So, what you're experiencing is the right behaviour of strtok().
OTOH, strtok() will return NULL if there is no more tokens, so as per you have expected, returning NULL for the initial delimiter will convey wrong message and it will be confusing. So, the bottom line is,
if a token is present
the tokens returned by strtok() are always nonempty strings.
if a token is not present
strtok() will return NULL.
Note: it is useful to mention that before using the retured token, always check for NULL.
What can I do?
Build your own function, not exactly how strtok works but you can get some idea:
#include <stdio.h>
#include <string.h>
char *scan(char **pp, char c)
{
char *s, *p;
p = strchr(*pp, c);
if (p) *p++ = '\0';
s = *pp;
*pp = p;
return s;
}
int main(void)
{
char line1[] = "COPY\tSTART\t0\tCOPY";
char line2[] = "\tSTART\t0\tCOPY";
char *p;
puts("Line 1");
p = line1;
while (p) {
printf("%s\n", scan(&p, '\t'));
}
puts("Line 2");
p = line2;
while (p) {
printf("%s\n", scan(&p, '\t'));
}
return 0;
}
Output:
Line 1
COPY
START
0
COPY
Line 2
START
0
COPY
I was trying to write a calculator program and so part of this I need to evaluate an expression.
So I need to perform the operation based on the operator given. I am taking the whole expression into a string.
For example it might be 5+6 or 5*6.
So I have written it in this way:
char input1[20] = "";
char input2[20] = "";
char output[20] = "";
char *arg1= NULL, *arg2 = NULL;
int value;
getinput ( input1); //Function for getting the expression
strcpy (input2, input1);
if ( arg1 = strtok (input1, "*"))
{
arg2 = strtok (NULL, "");
value = atoi(arg1) * atoi(arg2);
}
else
{
char* arg1, *arg2;
arg1 = strtok ( input2, "+");
arg2 = strtok ( NULL, "");
value = atoi (arg1) + atoi(arg2);
}
sprintf (output,"%d", value);
printf ("The output value is %s", output);
This code works only if I give expression having multiplication. For example it works only if I give 5*6. This is not working if I give 5+6.
The problem is in the else part. It is not able to tokenize the string input2.
Can't I tokenize two different strings in a single program.
Where am I wrong? Can someone explain me this concept of why strtok is not working for secong string?
the first strtok will not return NULL for "3+5", but rather a pointer to the token "3+5"
(so the else statement won't get executed).
now the problem is that the second call to strtok (around line #12 in your code) will return NULL, and the subsequent call atoi(NULL) will segfault.
The first strtok call will not return NULL (unless your input string is either empty or only contains '*' characters), so the else statement will not be executed for a string like "5+6".
You probably want to use strchr (or similar) to figure out what operation is to be performed, and then get the operands.
case of input "3+5" result of strtok(input1, "*") is "3+5", That else clause is not executed because it should not be NULL.
Second parameter for strtok is an array use like this strtok (input1, "+*"). That means that it will tokenize if they get '*' or '+'.
I code this like below:
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
int main()
{
char ss[100] = "123+321";
int a = atoi(ss);
int aLength = (int)floor(log10((double)a))+1;
int b = atoi(ss+aLength);
if(ss[aLength] == '+')
printf("%d + %d = %d\n", a, b, (a+b));
else
printf("%d * %d = %d\n", a, b, (a*b));
return 0;
}
/* strchr example */
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] = "This is a sample string";
char * pch;
printf ("Looking for the 's' character in \"%s\"...\n",str);
pch=strchr(str,'s');
while (pch!=NULL)
{
printf ("found at %d\n",pch-str+1);
pch=strchr(pch+1,'s');
}
return 0;
}
How would I index the str so that I would replace every 's' with 'r'.
Thanks.
You don't need to index the string. You have a pointer to the character you want to change, so assign via the pointer:
*pch = 'r';
In general, though, you index using []:
ptrdiff_t idx = pch - str;
assert(str[idx] == 's');
You can use the following function:
char *chngChar (char *str, char oldChar, char newChar) {
char *strPtr = str;
while ((strPtr = strchr (strPtr, oldChar)) != NULL)
*strPtr++ = newChar;
return str;
}
It simply runs through the string looking for the specific character and replaces it with the new character. Each time through (as with yours), it starts with the address one beyond the previous character so as to not recheck characters that have already been checked.
It also returns the address of the string, a trick often used so that you can use the return value as well, such as with:
printf ("%s\n", chngChar (myName, 'p', 'P'));
void reeplachar(char *buff, char old, char neo){
char *ptr;
for(;;){
ptr = strchr(buff, old);
if(ptr==NULL) break;
buff[(int)(ptr-buff)]=neo;
}
return;
}
Usage:
reeplachar(str,'s','r');
Provided that your program does really search the positions without fault (I didn't check), your question would be how do I change the contents of an object to which my pointer pch is already pointing?
Please explain to me the working of strtok() function. The manual says it breaks the string into tokens. I am unable to understand from the manual what it actually does.
I added watches on str and *pch to check its working when the first while loop occurred, the contents of str were only "this". How did the output shown below printed on the screen?
/* 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;
}
Output:
Splitting string "- This, a sample string." into tokens:
This
a
sample
string
the strtok runtime function works like this
the first time you call strtok you provide a string that you want to tokenize
char s[] = "this is a string";
in the above string space seems to be a good delimiter between words so lets use that:
char* p = strtok(s, " ");
what happens now is that 's' is searched until the space character is found, the first token is returned ('this') and p points to that token (string)
in order to get next token and to continue with the same string NULL is passed as first
argument since strtok maintains a static pointer to your previous passed string:
p = strtok(NULL," ");
p now points to 'is'
and so on until no more spaces can be found, then the last string is returned as the last token 'string'.
more conveniently you could write it like this instead to print out all tokens:
for (char *p = strtok(s," "); p != NULL; p = strtok(NULL, " "))
{
puts(p);
}
EDIT:
If you want to store the returned values from strtok you need to copy the token to another buffer e.g. strdup(p); since the original string (pointed to by the static pointer inside strtok) is modified between iterations in order to return the token.
strtok() divides the string into tokens. i.e. starting from any one of the delimiter to next one would be your one token. In your case, the starting token will be from "-" and end with next space " ". Then next token will start from " " and end with ",". Here you get "This" as output. Similarly the rest of the string gets split into tokens from space to space and finally ending the last token on "."
strtok maintains a static, internal reference pointing to the next available token in the string; if you pass it a NULL pointer, it will work from that internal reference.
This is the reason strtok isn't re-entrant; as soon as you pass it a new pointer, that old internal reference gets clobbered.
strtok doesn't change the parameter itself (str). It stores that pointer (in a local static variable). It can then change what that parameter points to in subsequent calls without having the parameter passed back. (And it can advance that pointer it has kept however it needs to perform its operations.)
From the POSIX strtok page:
This function uses static storage to keep track of the current string position between calls.
There is a thread-safe variant (strtok_r) that doesn't do this type of magic.
strtok will tokenize a string i.e. convert it into a series of substrings.
It does that by searching for delimiters that separate these tokens (or substrings). And you specify the delimiters. In your case, you want ' ' or ',' or '.' or '-' to be the delimiter.
The programming model to extract these tokens is that you hand strtok your main string and the set of delimiters. Then you call it repeatedly, and each time strtok will return the next token it finds. Till it reaches the end of the main string, when it returns a null. Another rule is that you pass the string in only the first time, and NULL for the subsequent times. This is a way to tell strtok if you are starting a new session of tokenizing with a new string, or you are retrieving tokens from a previous tokenizing session. Note that strtok remembers its state for the tokenizing session. And for this reason it is not reentrant or thread safe (you should be using strtok_r instead). Another thing to know is that it actually modifies the original string. It writes '\0' for teh delimiters that it finds.
One way to invoke strtok, succintly, is as follows:
char str[] = "this, is the string - I want to parse";
char delim[] = " ,-";
char* token;
for (token = strtok(str, delim); token; token = strtok(NULL, delim))
{
printf("token=%s\n", token);
}
Result:
this
is
the
string
I
want
to
parse
The first time you call it, you provide the string to tokenize to strtok. And then, to get the following tokens, you just give NULL to that function, as long as it returns a non NULL pointer.
The strtok function records the string you first provided when you call it. (Which is really dangerous for multi-thread applications)
strtok modifies its input string. It places null characters ('\0') in it so that it will return bits of the original string as tokens. In fact strtok does not allocate memory. You may understand it better if you draw the string as a sequence of boxes.
To understand how strtok() works, one first need to know what a static variable is. This link explains it quite well....
The key to the operation of strtok() is preserving the location of the last seperator between seccessive calls (that's why strtok() continues to parse the very original string that is passed to it when it is invoked with a null pointer in successive calls)..
Have a look at my own strtok() implementation, called zStrtok(), which has a sligtly different functionality than the one provided by strtok()
char *zStrtok(char *str, const char *delim) {
static char *static_str=0; /* var to store last address */
int index=0, strlength=0; /* integers for indexes */
int found = 0; /* check if delim is found */
/* delimiter cannot be NULL
* if no more char left, return NULL as well
*/
if (delim==0 || (str == 0 && static_str == 0))
return 0;
if (str == 0)
str = static_str;
/* get length of string */
while(str[strlength])
strlength++;
/* find the first occurance of delim */
for (index=0;index<strlength;index++)
if (str[index]==delim[0]) {
found=1;
break;
}
/* if delim is not contained in str, return str */
if (!found) {
static_str = 0;
return str;
}
/* check for consecutive delimiters
*if first char is delim, return delim
*/
if (str[0]==delim[0]) {
static_str = (str + 1);
return (char *)delim;
}
/* terminate the string
* this assignmetn requires char[], so str has to
* be char[] rather than *char
*/
str[index] = '\0';
/* save the rest of the string */
if ((str + index + 1)!=0)
static_str = (str + index + 1);
else
static_str = 0;
return str;
}
And here is an example usage
Example Usage
char str[] = "A,B,,,C";
printf("1 %s\n",zStrtok(s,","));
printf("2 %s\n",zStrtok(NULL,","));
printf("3 %s\n",zStrtok(NULL,","));
printf("4 %s\n",zStrtok(NULL,","));
printf("5 %s\n",zStrtok(NULL,","));
printf("6 %s\n",zStrtok(NULL,","));
Example Output
1 A
2 B
3 ,
4 ,
5 C
6 (null)
The code is from a string processing library I maintain on Github, called zString. Have a look at the code, or even contribute :)
https://github.com/fnoyanisi/zString
This is how i implemented strtok, Not that great but after working 2 hr on it finally got it worked. It does support multiple delimiters.
#include "stdafx.h"
#include <iostream>
using namespace std;
char* mystrtok(char str[],char filter[])
{
if(filter == NULL) {
return str;
}
static char *ptr = str;
static int flag = 0;
if(flag == 1) {
return NULL;
}
char* ptrReturn = ptr;
for(int j = 0; ptr != '\0'; j++) {
for(int i=0 ; filter[i] != '\0' ; i++) {
if(ptr[j] == '\0') {
flag = 1;
return ptrReturn;
}
if( ptr[j] == filter[i]) {
ptr[j] = '\0';
ptr+=j+1;
return ptrReturn;
}
}
}
return NULL;
}
int _tmain(int argc, _TCHAR* argv[])
{
char str[200] = "This,is my,string.test";
char *ppt = mystrtok(str,", .");
while(ppt != NULL ) {
cout<< ppt << endl;
ppt = mystrtok(NULL,", .");
}
return 0;
}
For those who are still having hard time understanding this strtok() function, take a look at this pythontutor example, it is a great tool to visualize your C (or C++, Python ...) code.
In case the link got broken, paste in:
#include <stdio.h>
#include <string.h>
int main()
{
char s[] = "Hello, my name is? Matthew! Hey.";
char* p;
for (char *p = strtok(s," ,?!."); p != NULL; p = strtok(NULL, " ,?!.")) {
puts(p);
}
return 0;
}
Credits go to Anders K.
Here is my implementation which uses hash table for the delimiter, which means it O(n) instead of O(n^2) (here is a link to the code):
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define DICT_LEN 256
int *create_delim_dict(char *delim)
{
int *d = (int*)malloc(sizeof(int)*DICT_LEN);
memset((void*)d, 0, sizeof(int)*DICT_LEN);
int i;
for(i=0; i< strlen(delim); i++) {
d[delim[i]] = 1;
}
return d;
}
char *my_strtok(char *str, char *delim)
{
static char *last, *to_free;
int *deli_dict = create_delim_dict(delim);
if(!deli_dict) {
/*this check if we allocate and fail the second time with entering this function */
if(to_free) {
free(to_free);
}
return NULL;
}
if(str) {
last = (char*)malloc(strlen(str)+1);
if(!last) {
free(deli_dict);
return NULL;
}
to_free = last;
strcpy(last, str);
}
while(deli_dict[*last] && *last != '\0') {
last++;
}
str = last;
if(*last == '\0') {
free(deli_dict);
free(to_free);
deli_dict = NULL;
to_free = NULL;
return NULL;
}
while (*last != '\0' && !deli_dict[*last]) {
last++;
}
*last = '\0';
last++;
free(deli_dict);
return str;
}
int main()
{
char * str = "- This, a sample string.";
char *del = " ,.-";
char *s = my_strtok(str, del);
while(s) {
printf("%s\n", s);
s = my_strtok(NULL, del);
}
return 0;
}
strtok() stores the pointer in static variable where did you last time left off , so on its 2nd call , when we pass the null , strtok() gets the pointer from the static variable .
If you provide the same string name , it again starts from beginning.
Moreover strtok() is destructive i.e. it make changes to the orignal string. so make sure you always have a copy of orignal one.
One more problem of using strtok() is that as it stores the address in static variables , in multithreaded programming calling strtok() more than once will cause an error. For this use strtok_r().
strtok replaces the characters in the second argument with a NULL and a NULL character is also the end of a string.
http://www.cplusplus.com/reference/clibrary/cstring/strtok/
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;
}
So, this is a code snippet to help better understand this topic.
Printing Tokens
Task: Given a sentence, s, print each word of the sentence in a new line.
char *s;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
//logic to print the tokens of the sentence.
for (char *p = strtok(s," "); p != NULL; p = strtok(NULL, " "))
{
printf("%s\n",p);
}
Input: How is that
Result:
How
is
that
Explanation: So here, "strtok()" function is used and it's iterated using for loop to print the tokens in separate lines.
The function will take parameters as 'string' and 'break-point' and break the string at those break-points and form tokens. Now, those tokens are stored in 'p' and are used further for printing.
strtok is replacing delimiter with'\0' NULL character in given string
CODE
#include<iostream>
#include<cstring>
int main()
{
char s[]="30/4/2021";
std::cout<<(void*)s<<"\n"; // 0x70fdf0
char *p1=(char*)0x70fdf0;
std::cout<<p1<<"\n";
char *p2=strtok(s,"/");
std::cout<<(void*)p2<<"\n";
std::cout<<p2<<"\n";
char *p3=(char*)0x70fdf0;
std::cout<<p3<<"\n";
for(int i=0;i<=9;i++)
{
std::cout<<*p1;
p1++;
}
}
OUTPUT
0x70fdf0 // 1. address of string s
30/4/2021 // 2. print string s through ptr p1
0x70fdf0 // 3. this address is return by strtok to ptr p2
30 // 4. print string which pointed by p2
30 // 5. again assign address of string s to ptr p3 try to print string
30 4/2021 // 6. print characters of string s one by one using loop
Before tokenizing the string
I assigned address of string s to some ptr(p1) and try to print string through that ptr and whole string is printed.
after tokenized
strtok return the address of string s to ptr(p2) but when I try to print string through ptr it only print "30" it did not print whole string. so it's sure that strtok is not just returning adress but it is placing '\0' character where delimiter is present.
cross check
1.
again I assign the address of string s to some ptr (p3) and try to print string it prints "30" as while tokenizing the string is updated with '\0' at delimiter.
2.
see printing string s character by character via loop the 1st delimiter is replaced by '\0' so it is printing blank space rather than ''