To summarize, if the function is given arguments char *str1 = "Watch out the streets?";
char *str2 = "street?";, it should return a pointer in str1 to the first instance of "street", starting from s.
I am completely stuck and appreciate any help. I have a feeling my problem is with pointers. I'm a beginner as you might imagine.
I just edited the code to my latest attempt. I am very close, but cannot still find the idea to modify the first string.
Output image:
const char *qstr_strstr(const char *str1, const char *str2) {
int j = 0;
char newstr2[256];
while (str2[j]) {
if (str2[j] == '?') {
break;
} else {
newstr2[j] = str2[j];
}
j++;
}
newstr2[j] = '\0';
size_t n = strlen(newstr2);
while(*str1) {
if (!memcmp(str1++, newstr2, n)) {
return (str1 - 1);
}
}
return NULL;
}
You are heading in the right direction, and you are starting to think about stepping through stings correctly, but you are getting wrapped up in your use of any additional string -- there is no need for that.
When you are analyzing stings and substrings, all you care about is:
the index (position) within the original string (to iterate);
whether you have found the beginning of the substring yet (if so set flag indicating in word matching chars); and
whether you make it all the way to the end of the substring while your flag is set (if you do, you have found your substring). Otherwise, if there is a mismatch in characters before reaching the end, unset your flag and keep going. Repeat until you run out of str1.
Complicating your case here is the Red-Herring '?' character. You can either use it as a test for end-of-substring, or ignore it altogether. Your goal is to locate str2 "street" within str1 "... streets?". You would generally use ispunct() from ctype.h to identify '?' as a non-word character and not part of the string anyway. Here, you can simply check for '?' in a similar way that you would check for '\0' to mark end-of-string.
Putting it altogether and not using any of the string.h or ctype.h functions, you could do:
#include <stdio.h>
const char *qstr_strstr (const char *str1, const char *str2)
{
int i = 0, instr2 = 0, ndx = 0; /* str1 index, in-str2 flag, str2 index */
if (str1 == NULL || str2 == NULL) /* validte both str1 and str2 not NULL */
return NULL;
do { /* loop over each char in str1 */
/* if in str2 and (at end or at '?') */
if (instr2 && (!str2[ndx] || str2[ndx] == '?'))
return &str1[i-ndx]; /* return address in str1 to start of str2 */
/* if char in str1 and str2 equal */
if (str1[i] == str2[ndx]) {
ndx += 1; /* increment str2 index */
instr2 = 1; /* set in-str2 flag true (done each time) */
}
else /* otherwise chars not equal */
ndx = instr2 = 0; /* reset str2 index, set in-str2 flag false */
} while (str1[i++]); /* includes \0, allows match at end of both */
return NULL;
}
int main (void) {
const char *str1 = "Watch of the streets?",
*str2 = "street?",
*ptr = qstr_strstr (str1, str2);
if (ptr) {
size_t i = ptr - str1;
printf ("ptr : %s\nptr addr : %p\n---\n"
"&str1[%zu] : %s\n&str1[%zu] : %p\n",
ptr, (void*)ptr, i, str1 + i, i, (void*)(str1 + i));
}
}
Example Use/Output
Outputting the string and adderss retuned by qstr_strstr() as well as the address of that string and address located within the original str1 you would have:
$ ./bin/strstrawk
ptr : streets?
ptr addr : 0x4006c1
---
&str1[13] : streets?
&str1[13] : 0x4006c1
You can further copy just the substring street from the address in str1 or whatever else you need to do. Look things over and let me know if you have further questions.
Edit Based On Picture Added
If you want to output just the characters before the '?' from the pointer returned by qstr_strstr(), then you can do that trivially by using the precision modifier with the "%s" output conversion specifier in your printf format string and limit the number of characters output to just the number before the '?' in str2. Your format conversion specifier would look like "%.*s" where it expects 2 arguments, the first of type int being the number of characters, and the second the string.
In the case above the arguments would be:
(int)strcspn (str2, "?"), ptr
Where strcspn() returns the number of characters before the '?' in str2 (cast to int) and ptr the return from qstr_strstr(). Changing the code above you would have:
...
#include <string.h>
...
printf ("ptr : %.*s\nptr addr : %p\n---\n"
"&str1[%zu] : %s\n&str1[%zu] : %p\n",
(int)strcspn (str2, "?"), ptr, (void*)ptr,
i, str1 + i, i, (void*)(str1 + i));
(the remaining code is unchanged)
$ ./bin/strstrawk
ptr : street
ptr addr : 0x4006c1
---
&str1[13] : streets?
&str1[13] : 0x4006c1
Where the output using ptr as the string and the length from str2 is not "street".
Let me know if you have further questions.
As you can see below I have created a little program to concatenate 2 strings using C, as you may imagine this code doesn't work, I have already corrected it myself by using Array notation instead of pointers, and it works just fine, however I'm still not sure why is it that my code fails being almost a replica of my corrected code.
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
void concatena(char *str1, char *str2){
char *strAux;
int mover;
mover = 0;
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+2));
*(strAux) = '\0';
if(str1 == '\0')
*strAux = '\0';
else
while(str1 != '\0'){
*(strAux+mover++)=*(str1++);
}
if(str2 == '\0')
*strAux = '\0';
else
while(str2 != '\0'){
*(strAux+mover++)=*(str2++);
}
strAux='\0';
str1=strAux;
printf("%s", str1);
free(strAux);
}
I´m still a C beginner (And yes, I'm aware that there are libraries like string.h, I'm asking this for academic reasons) and I have been told that char pointers and arrays are the same thing, something that confuses the heck out of me.
Any help is greatly appreciated.
The first problem I see is with this section:
if(str2 == '\0')
*strAux = '\0';
Just before this code, you've filled up strAux with the string from str1.
Then, if str2 is empty, you suddenly put a null-terminator at the beginning of strAux, eliminating all the work you've done so far!
I think what you intend is:
if(*str2 == '\0')
*(strAux+mover) = '\0';
Its the same thing again after your loop for str2, you have the code:
strAux='\0';
Again, this puts a null-terminator at the start of strAux, effectively ending the newly created string before it even gets started.
Here's how I'd re-write your code:
void concatena(char *str1, char *str2){
char *strAux;
int mover = 0;
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+1)); // Changed to +1, NOT +2
*(strAux) = '\0'; // Start the string as (empty)
while(*str1 != '\0'){ // Copy the first string over.
*(strAux+mover++)=*(str1++);
}
while(*str2 != '\0'){ // Copy the second string over.
*(strAux+mover++)=*(str2++);
}
*(strAux+mover)='\0'; // End the new, combined string.
printf("%s", strAux); // Show the results.
free(strAux);
}
Accepting the same constraints, here is how I would (re)write your code. Unfortunately there is a specification shortcoming: should the concatenation occur to the first string passed? Or should a new string be created? Here are both methods:
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
char *concatena (const char *str1, const char *str2)
{
char *op, *newStr = (char*)malloc (strlen (str1) + strlen (str2) + 1);
if (!newStr)
{
fprintf (stderr, "concatena: error allocating\n");
return;
}
op = newStr; // set up output pointer
while (str1 && *str1) // copy first string
*op++ = *str1++;
while (str2 && *str2) // concatenate second string
*op++ = *str2++;
*op = '\000'; // add conventional NUL termination
return newStr;
}
void concatenb (char *str1, const char *str2)
{
char *op;
if (!str1)
{
fprintf (stderr, "concatenb: NULL string 1\n");
return;
}
op = &str1 [strlen (str1)]; // set output pointer at trailing NUL
while (str2 && *str2) // concatenate second string
*op++ = *str2++;
*op = '\000'; // add conventional NUL termination
}
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+2));
2 is not required, just 1 is sufficient for the termination character.
*(strAux) = '\0';
This should be happening only at the end of all your computation. Not in between the concatenation i.e.,
while(*str1 != '\0'){ // This loops copies the first string
// ^ Notice that you need to dereference to check for the termination character.
*(strAux+mover++)=*(str1++);
}
while(*str2 != '\0'){ // This loop copies the second string
*(strAux+mover++)=*(str2++);
}
// Finally adding termination character
*(strAux+mover) = '\0'; // since with mover you are keeping track of locations.
The amount of errors in your code is disheartening. You should probably pick up a good C book and start over.
First off, there's a library function that you can use to concatenate strings:
const unsigned int len = strlen(str1) + strlen(str2) + 1;
char * dst = malloc(len);
strncat(dst, str1, len);
strncat(dst, str2, len);
Now, if you insist on doing it manually, you have to get pointers and dereferencing right:
char * d = dst;
while (*str1 != 0) *dst++ = *str1++;
while (*str2 != 0) *dst++ = *str2++;
*dst = 0;
// d now points to the beginning of the concatenated string
The two loops check if the current character in the input string is nonzero, and if so, then they copy that character to the current character in the output string, and then both input and output pointer are advanced. (This is all done in one wash by use of the postfix ++ operator.) Finally, the last character is set to zero to create a new null-terminator.
In the process we modified all three pointers dst, str1 and str2. The latter two came in as input function arguments by copy, so that's fine. For returning the concatenated string we made a copy of dst before the loop, which we can return in the end.
/* 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 ''