Question

Could someone explain the following codes

#include<stdio.h>
main()
{
    char c[]="abc\nabc";
    puts(c);
}

This code as expected generates :

abc
abc

But when i try to take the same string as an input from the user,

#include<stdio.h>
main()
{
    char c[]="abc\nabc";
    gets(c);             // i type in "abc\nabc" 
    puts(c);
}

This code generates :

abc\nabc

How can i make the program read the newline character correctly ?

Answer 1

Did you literally type \ then n?

If so, it literally placed a \ and then a n in your string, as if you did the following:

char c[] = "abc\\nabc"; /* note the escaped \ */

This is logically not a newline character, but rather a \ followed by a n.

If you would like to support escape sequences in user input, you’ll need to post-process any user input to create the appropriate escape sequences.

/* translate escape sequences inline */
for (i = 0, j = 0; c[i] != 0; ++i, ++j) {
   if (c[i] == '\\' && c[i+1] != 0) {
       switch(c[++i]) {
       case 'n':  c[j] = '\n'; break;
       case '\\': c[j] = '\\'; break;
       /* add the others you'd like to handle here */
       /* case 'a': ... */
       default:   c[j] = ' ';  break;
       }
   } else {
       c[j] = c[i];
   }
}

c[j] = 0;