Question

I have code which goes like this. Could you tell me why it is not behaving as I would expect it to be?

/*
 * test.cpp
 *
 *  Created on: Dec 6, 2012
 *      Author: sandeep
 */

#include<iostream>
#include<string.h>
using namespace std;

int main()
{
    int i=0;
    string s="hello A B:bye A B";
    char *input;
    input=new char(s.size());
    for(i=0;i<=s.size();i++)
        input[i]=s[i];
    char *tokenized1[2],*tokenized2[3];
    tokenized1[0]=strtok(input,":");
    tokenized1[1]=strtok(NULL,":");
    i=0;
    char *lstring;
    while(i<2)
    {
        lstring=new char(strlen(tokenized1[i]));
        memcpy(lstring,tokenized1[i],strlen(tokenized1[i])+1);
        cout<<tokenized1[0]<<"  "<<tokenized1[1]<<endl;
        tokenized2[0]=strtok(lstring," ");
        tokenized2[1]=strtok(NULL," ");
        tokenized2[2]=strtok(NULL," ");
        char c=tokenized2[0][0];
        cout<<c<<endl;
        cout<<tokenized2[0]<<"  "<<tokenized2[1]<<"  "<<tokenized2[2]<<endl;
        i++;
    }

}

and the output is this.

hello A B  by
h
hello  A  B
hello A B  by
b
by  

There are some junk values at the end of 1st, 4th and 6th line of output.
Why tokenized1[1] got altered when I I did memcopy of tokenized1[0]? and how to solve this?

Answer 1

There’s a couple of bugs in the following new call. You need to use square brackets; also, the argument is off by one.

lstring=new char[strlen(tokenized1[i]) + 1];

Without the square brackets, your are allocating space for one character. As a result, the memcpy() writes past the allocated memory.

edit: I just noticed the other new, which will also need to be fixed:

input=new char[s.size() + 1];

Finally, s[i] reads past the end of the string in:

for(i=0;i<=s.size();i++)
    input[i]=s[i];

There could well be other bugs, not to mention memory leaks…