Problem with array of strings !!

[prashant8809]

Hi,

Please excuse me for asking such a trivial question. I am very new to C programming and it would be very helpful if someone answers my question.

Here is my program ...

#include"scanner.h"
#include<stdio.h>
#include"scanner.c"
#include<stdlib.h>
#include<string.h>

#define MAX_TOKENS 10

int main(int argc, char **argv){
    int datatype, counter;
    int index = 0;
    char *my_argv[MAX_TOKENS];
    while((datatype = yylex()) != 6)
    {    
        printf("\nData :%s",yytext);
        printf("\nData type :%d\n",datatype);
        my_argv[index++] = yytext;
    }
    printf("\n%s",my_argv[0]);
    printf("\n%s",my_argv[1]);
}

Here yytext is of type char*. I store it in my_argv[]. Consider yytex returns "abc" and then "def". I should be getting the output as

abc
def

but instead I get

abc def
def

Can anyone explain me this !!

Thanks in advance.

[grumpy]

The offending line is

    my_argv[index++] = yytext;

which simply copies a pointer.

Copying pointers does not copy what they point at.

1) ensure there is a buffer to be copied to.
2) copy the required string to that buffer.

You have completed neither of those steps. A simple way to achieve the first is to make my_argv into a 2-dimensional array (aka an array of arrays), rather than an array of pointers. To copy a string (step 2) look up the function strcpy().

Also, never .... repeat never .... #include a .c file from another .c file. Simply compile the files separately, and link the objects together.

Also, the offending line has two side effects (sets a value to my_argv[index] and then increment index). Unless you know exactly what you are doing - and, by your question, you don't - always make each statement have exactly one side effect.

    my_argv[index] = yytext;
    ++index;

(note you still need to fix the first line with this adjustment).