why this strtok doesn't work?

[Caven Yang]

Hello,

This is a part of c code, which I used in a shell script to do the math.

#include <iostream>
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <vector>

using namespace std;


int main(int argc,char **argv){
  char *tmp;
  vector<float> input;
  char in[128];

  strcpy(in,argv[1]);
  //cout << "arg # is " << argc << endl;
  //cout << "arg[1] is " << in << endl;
  tmp=strtok (in,"^"); 
  input.push_back(atof(tmp));
  cout << input[0] << "\t";
  int i=1;
  while( i<5 ){ // it is even worse if I use (tmp != NULL) or strcmp(tmp,NULL) here
    tmp=strtok (NULL,"^");
    i++;
    cout << atof(tmp) << "\t";
    //input.push_back(atof(tmp));
  }

  cout << endl;
  return 0;
}

I used g++ to compile it because I used cout here just for convenience.

Below is the result:

./a.out 1^2^3^4^5
1	2	3	4	5	
./a.out 1^2^3^4
Segmentation fault

it seems that when strtok() comes to the last token part, it can't do it. I don't understand that is why.


Thanks for any enlightenment.

[Peter87]

The problem is that you try to pass a null pointer to atof. You can simplify the program by using a for loop.

int main(int argc,char **argv){
	char *tmp;
	char in[128];

	strcpy(in,argv[1]);

	for (tmp=strtok(in,"^"); tmp != NULL; tmp=strtok(NULL,"^")) {
		cout << atof(tmp) << "\t";
	}

	cout << endl;
	return 0;
}