What happens when we add double quotes with in the string literals in C

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What happens when when the string literal contains double quotes with in it and why I am not getting error?

What I have tried:

#include<stdio.h>
int main()
{
    printf("%s","abcdefghijklmn""opqrstuvwxyz");
    return 0;
}

Posted 22-Apr-21 20:00pm

_-_-_-me

Updated 22-Apr-21 20:28pm

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OriginalGriff · Accepted Answer · 2021-04-22T20:09:00

You have to "escape" the double quote character, in the same way that you add a newline:

C++

printf("line one\nline two\n");

"\n" is a escape sequence that C recognises as a "special character" which represents a newline, and there are several of them but the most common are:

\n    Newline
\t    Tab
\\    A literal back slash
\'    A literal single quote
\"    A literal double quote

So to insert a double quote:

C++

printf("Joe says \"Hello world!\"");

Will print

Joes says "Hello world!"

on the console.

CPallini · Accepted Answer · 2021-04-22T20:28:00

Our Griff already showed you how to escape the double quotes. As an addition I'm trying to address the doubt

Quote:
why I am not getting error?

You're not getting an error because

C

"abcdefghijklmn""opqrstuvwxyz"

is a valid construct, namely a concatenation of string literals (see, for instance c - How does concatenation of two string literals work? - Stack Overflow[^]).

The following program spots the differences

C

#include <stdio.h>
  
int main()
{
  const char * escaped = "foo\"bar"; // insert a double quote in the string using the proper escape sequence
  const char * concatenated = "foo""bar"; // concatenate the two string literals.

  printf("escaped: %s\n", escaped);
  printf("concatenated: %s\n", concatenated);

  return 0;
}

output:

escaped: foo"bar                                                                                                                                           
concatenated: foobar

What happens when we add double quotes with in the string literals in C

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