Try
Code:#include <stdio.h> #include <string.h> int parse ( char *input, char *output[], int maxtokens ) { char *seps = "|\n"; int i; char *p; for ( i = 0, p = strtok(input,seps) ; i < maxtokens && p != NULL ; i++, p = strtok(NULL,seps) ) { output[i] = p; } return i; } int main ( ) { char input[] = "command|MAC|field|field|field\n"; char *output[10]; int len = parse ( input, output, 10 ); int i; for ( i = 0 ; i < len ; i++ ) { printf( "%s\n", output[i] ); } return 0; }
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
thanks!! this works.. tommorow I'll try to work this out! If I have any questions you'll hear from me again
function foo() below returns a 2d character array.Originally Posted by Dave_Sinkula
or like thisCode:#include <stdio.h> char** foo() { char **token; token = malloc(8 * sizeof(char**)); token[ 0] = "002^Resultaat op aanmelden"; token[ 1] = "20050915"; token[ 2] = "121533"; token[ 3] = "INT^TEMP^1^S"; token[ 4] = "BOOL^LED^1^A"; token[ 5] = "BOOL^LED^2^A"; token[ 6] = ""; return token; } int main() { char** tokens = foo(); char** ptr = tokens; while(*ptr != 0) { printf("%s\n",*ptr); ptr++; fflush(stdout); } free(tokens); return 0; }
Code:int main() { int i; char** tokens = foo(); for(i = 0; tokens[i] != 0; ++i) { printf("%s\n",tokens[i]); } free(tokens); return 0; }
No it doesn't. It returns a pointer to a pointer. A pointer to a pointer is not a two dimensional array. Perhaps you'd benifit from a FAQ. They're not the same thing. We often times use multiple pointers to simulate multidimensional arrays, but they're not the same thing. For one, with your "multi dimension array" that you're returning, it's perfectly legal to have it like this:This obviously can't be a two dimensional array, because each row can be any length I want it to be.Code:char **twodee; int x; twodee = malloc( ROWS * sizeof *twodee ); for( x = 0; x < ROWS; x++ ) twodee[ x ] = malloc( (rand() % 1000 + 1) * sizeof **twodee );
Furthermore, your call to free is wrong in both examples. Rather, if you were doing anything other than assigning string literals, it would be.
Quzah.
Last edited by quzah; 10-19-2005 at 04:12 PM.
Hope is the first step on the road to disappointment.
If you'll notice, I answered the question as asked.
Whereas you did not.You can't.No it does not. But I do know how these kinds of array simulation are done. I just don't know if it's particularly suited to what robin171 might need.
Last edited by Dave_Sinkula; 10-19-2005 at 04:14 PM. Reason: Dammit! I am really getting pokey. :(
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
The call to free is indeed correct in my example. It did only one malloc(), therefore it needs only one free().
2d arrays can be declared in ways -- argv is a good example. It can be declared as "char **argv", or "char *argv[]" and both are identical. However you do it, it is a 2d array of strings.
Last edited by Ancient Dragon; 10-20-2005 at 07:17 AM.
Except argv isn't a 2d array of strings.the array members argv[0] through argv[argc-1] inclusive shall contain pointers to strings
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
thnx for the replies, I'm working on the example of Salem and it works pretty good. Maybe I need some more help later, for now I can work with this
