how to get a part of a char array in c

Question

I am new in C and I have the following simple code. I know that with the strncpy I can copy characters from string.

#include <stdio.h>
#include <string.h>

int main ()
{
  char str1[]= "To be or not to be";
  char str2[40];

  strncpy ( str2, str1, 5 );
  str2[5] = '\0';   /* null character manually added */
  puts (str2);

  return 0;
}

The output of this code is

To be

If I want the result to be ´or not to´ how can I read these characters? From 7-15 in this case?

stop using c-style strings in c++ and use the std::string. it has find_first_of or at() which can help you — Koushik Shetty
– Koushik Shetty, Commented May 29, 2013 at 6:36
I need an answer for this example..So in C..Sorry about that — dali1985
– dali1985, Commented May 29, 2013 at 6:36

Vijay · Accepted Answer · 2013-05-29 06:42:04Z

10

Use :

  strncpy ( str2, str1+6, 9);
  str2[9] = '\0';   /* null character manually added */

edited May 29, 2013 at 6:42

answered May 29, 2013 at 6:36

Vijay

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Comments

Dayal rai · Accepted Answer · 2013-05-29 06:50:36Z

4

No need to count manually ,you can use standard library function for this.Try like following

strncpy ( str2, str1+strlen("To be "), strlen("or not to") );
str2[strlen("or not to")] = '\0';

edited May 29, 2013 at 6:50

answered May 29, 2013 at 6:45

Dayal rai

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Comments

Bart Friederichs · Accepted Answer · 2013-05-29 06:36:51Z

0

Simple way to do that, offset str1:

strncpy ( str2, str1 + 6, 9 );

answered May 29, 2013 at 6:36

Bart Friederichs

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