C strtok() split string into tokens but keep old data unaltered

问题:

I have the following code:

#include  #include   int main (void) { char str[] = "John|Doe|Melbourne|6270|AU";  char fname[32], lname[32], city[32], zip[32], country[32]; char *oldstr = str;      strcpy(fname, strtok(str, "|"));     strcpy(lname, strtok(NULL, "|"));     strcpy(city, strtok(NULL, "|"));     strcpy(zip, strtok(NULL, "|"));     strcpy(country, strtok(NULL, "|"));      printf("Firstname: %s\n", fname);     printf("Lastname: %s\n", lname);     printf("City: %s\n", city);     printf("Zip: %s\n", zip);     printf("Country: %s\n", country);     printf("STR: %s\n", str);     printf("OLDSTR: %s\n", oldstr);  return 0; } 

Execution output:

$ ./str Firstname: John Lastname: Doe City: Melbourne Zip: 6270 Country: AU STR: John OLDSTR: John 

Why can't I keep the old data nor in the str or oldstr, what am I doing wrong and how can I not alter the data or keep it?

回答1:

when you do strtok(NULL, "|") strtock find token and put null on place (replace token with \0) and modify string.

you str, becomes:

char str[] = John0Doe0Melbourne062700AU;    Str array in memory  +------------------------------------------------------------------------------------------------+ |'J'|'o'|'h'|'n'|0|'D'|'o'|'e'|0|'M'|'e'|'l'|'b'|'o'|'u'|'r'|'n'|'e'|0|'6'|'2'|'7'|'0'|0|'A'|'U'|0| +------------------------------------------------------------------------------------------------+                  ^  replace | with \0  (ASCII value is 0) 

Consider the diagram is important because char '0' and 0 are diffident (in string 6270 are char in figure parenthesised by ' where for \0 0 is as number)

when you print str using %s it print chars upto first \0 that is John

To keep your original str unchanged you should fist copy str into some tempstr variable and then use that tempstr string in strtok():

char str[] = "John|Doe|Melbourne|6270|AU"; char* tempstr = calloc(strlen(str)+1, sizeof(char)); strcpy(tempstr, str); 

Now use this tempstr string in place of str in your code.

回答2:

strtok requires an writeable input string and it modifies the input string. If you want to keep the input string you have to a make a copy of it first.

For example:

char str[] = "John|Doe|Melbourne|6270|AU"; char oldstr[32];  strcpy(oldstr, str);  // Use strncpy if you don't know                       // the size of str 

回答3:

Because oldstr is just a pointer, an assignment will not make a new copy of your string.

Copy it before passing str to the strtok:

          char *oldstr=malloc(sizeof(str));           strcpy(oldstr,str); 

Your corrected version:

#include  #include  #include int main (void) {     char str[] = "John|Doe|Melbourne|6270|AU";    char fname[32], lname[32], city[32], zip[32], country[32];    char *oldstr = malloc(sizeof(str));    strcpy(oldstr,str);      ...................     free(oldstr); return 0; } 

EDIT:

As @CodeClown mentioned, in your case, it's better to use strncpy. And instead of fixing the sizes of fname etc before hand, you can have pointers in their place and allocate the memory as is required not more and not less. That way you can avoid writing to the buffer out of bounds......

Another Idea: would be to assign the result of strtok to pointers *fname, *lname, etc.. instead of arrays. It seems the strtok is designed to be used that way after seeing the accepted answer.

Caution:In this way, if you change str further that would be reflected in fname,lname also. Because, they just point to str data but not to new memory blocks. So, use oldstr for other manipulations.

#include  #include  #include int main (void) {      char str[] = "John|Doe|Melbourne|6270|AU";     char *fname, *lname, *city, *zip, *country;     char *oldstr = malloc(sizeof(str));     strcpy(oldstr,str);     fname=strtok(str,"|");     lname=strtok(NULL,"|");     city=strtok(NULL, "|");     zip=strtok(NULL, "|");     country=strtok(NULL, "|");      printf("Firstname: %s\n", fname);     printf("Lastname: %s\n", lname);     printf("City: %s\n", city);     printf("Zip: %s\n", zip);     printf("Country: %s\n", country);     printf("STR: %s\n", str);     printf("OLDSTR: %s\n", oldstr);     free(oldstr); return 0; } 

回答4:

You just copy the pointer to the string, but not the string itself. Use strncpy() to create a copy.

char *oldstr = str; // just copy of the address not the string itself! 

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