CODEWITHSUNDEEP
cprintf
Jessie
Jessie

Reputation: 185

How to print special characters explicitly in C?

When I use below code:

#include <stdio.h>

int main(void)
{
    printf("%s","Hello world\nHello world");
    return 0;
}

it prints as:

 Hello world
 Hello world

How can I prevent this and print it as raw string literal in C? I mean it should be displayed as it is in terminal window like below:

Hello world\nHello world

I know I can achieve this by using backslash for printf but is there any other C function or way to do this without backslashing? It would be helpful when reading files.

Upvotes: 13

Views: 55011

Answers (8)

Kingpin
Kingpin

Reputation: 1

Bro just use this simple code

#include <stdio.h>
int main()
{
     printf("Hello World\\nHello World");
     return 0;
}

And it should work pretty fine ! Happy Coding 😉

Upvotes: 0

shyed2001
shyed2001

Reputation: 7

  /// My experience Win 10 Code blocks GCC MinGW

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <unistd.h>
#include <windows.h>
///#include <threads.h>
#include <conio.h>
/// #include <dos.h>
#include <direct.h>

int main(void)

{
  /// This will give your desired result, turn string into Raw string :
  printf(R"(Hello world\nHello world)");
  printf(R"(Raw string support printing  *&^%$#@!~()_+-=,<.>/?:;"' )");
  printf("\n");
  printf(R"(.C with a Capital C file format does not support raw string )");
  printf("\n");
  printf(R"(.c with a small c file format does support raw string )");
  printf("\n");
  printf(R"( Raw string did not support \n new line )");
  printf("\n");

  printf(
      R"(More reading material at - https: // en.wikipedia.org/wiki/String_literal#Raw_strings;)");
  printf("\n");
  printf(
      R"(More reading material at - https: // en.wikipedia.org/wiki/String_literal;)");
  printf("\n");
  printf(
      R"(More reading material at - https://stackoverflow.com/questions/24850244/does-c-support-raw-string-literals;)");
  printf("\n");
  printf(
      R"(More reading material at - https: // learn.microsoft.com/en-us/cpp/c-language/c-string-literals?view=vs-2019)");
  printf("\n");
  printf(
      R"(More reading material at-https: // learn.microsoft.com/en-us/cpp/c-language/string-literal-concatenation?view=vs-2019)");
  printf("\n");
  /// Raw string.

    printf(R"(More reading material at - https://www.geeksforgeeks.org/const-qualifier-in-c/;)");
  printf("\n");
  
  
  return 0;
}

Upvotes: 0

Vidya
Vidya

Reputation: 402

Just use,putchar(specialCharName). It displays the entered special character.

Upvotes: 1

ВелоКастръ
ВелоКастръ

Reputation: 4453

Thank you the user @chunk for contributing to the improvement this answer.

Why did not you write general-purpose solution? It would keep you from many problems in the future.

char *
str_escape(char str[])
{
    char chr[3];
    char *buffer = malloc(sizeof(char));
    unsigned int len = 1, blk_size;

    while (*str != '\0') {
        blk_size = 2;
        switch (*str) {
            case '\n':
                strcpy(chr, "\\n");
                break;
            case '\t':
                strcpy(chr, "\\t");
                break;
            case '\v':
                strcpy(chr, "\\v");
                break;
            case '\f':
                strcpy(chr, "\\f");
                break;
            case '\a':
                strcpy(chr, "\\a");
                break;
            case '\b':
                strcpy(chr, "\\b");
                break;
            case '\r':
                strcpy(chr, "\\r");
                break;
            default:
                sprintf(chr, "%c", *str);
                blk_size = 1;
                break;
        }
        len += blk_size;
        buffer = realloc(buffer, len * sizeof(char));
        strcat(buffer, chr);
        ++str;
    }
    return buffer;
}

How it work!

int
main(const int argc, const char *argv[])
{
    puts(str_escape("\tAnbms\n"));
    puts(str_escape("\tA\v\fZ\a"));
    puts(str_escape("txt \t\n\r\f\a\v 1 \t\n\r\f\a\v tt"));
    puts(str_escape("dhsjdsdjhs hjd hjds "));
    puts(str_escape(""));
    puts(str_escape("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!\"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~ \t\n\r\f\a\v"));
    puts(str_escape("\x0b\x0c\t\n\r\f\a\v"));
    puts(str_escape("\x01\x02\x03\x04\x05\x06\x07\x08\t\n\x0b\x0c\r\x0e\x0f\x10\x11\x12\x13\x14"));
}

Output

\tAnbms\n
\tA\v\fZ\a
txt \t\n\r\f\a\v 1 \t\n\r\f\a\v tt
dhsjdsdjhs hjd hjds 

0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~ \t\n\r\f\a\v
\v\f\t\n\r\f\a\v
\a\b\t\n\v\f\r

This solution based on an information from the Wikipedia https://en.wikipedia.org/wiki/Escape_sequences_in_C#Table_of_escape_sequences and the answers other users of the stackoverflow.com.

Testing environment

$ lsb_release -a
No LSB modules are available.
Distributor ID: Debian
Description:    Debian GNU/Linux 8.6 (jessie)
Release:    8.6
Codename:   jessie
$ uname -a
Linux localhost 3.16.0-4-amd64 #1 SMP Debian 3.16.36-1+deb8u2 (2016-10-19) x86_64 GNU/Linux
$ gcc --version
gcc (Debian 4.9.2-10) 4.9.2
Copyright (C) 2014 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

Upvotes: 0

zwol
zwol

Reputation: 140569

There is no built-in mechanism to do this. You have to do it manually, character-by-character. However, the functions in ctype.h may help. Specifically, in the "C" locale, the function isprint is guaranteed to be true for all of the graphic characters in the basic execution character set, which is effectively the same as all the graphic characters in 7-bit ASCII, plus space; and it is guaranteed not to be true for all the control characters in 7-bit ASCII, which includes tab, carriage return, etc.

Here is a sketch:

#include <stdio.h>
#include <ctype.h>
#include <locale.h>

int main(void)
{
    int x;
    setlocale(LC_ALL, "C"); // (1)

    while ((x = getchar()) != EOF)
    {
        unsigned int c = (unsigned int)(unsigned char)x; // (2)

        if (isprint(c) && c != '\\')
            putchar(c);
        else
            printf("\\x%02x", c);
    }
    return 0;
}

This does not escape ' nor ", but it does escape \, and it is straightforward to extend that if you need it to.

Printing \n for U+000A, \r for U+000D, etc. is left as an exercise. Dealing with characters outside the basic execution character set (e.g. UTF-8 encoding of U+0080 through U+10FFFF) is also left as an exercise.

This program contains two things which are not necessary with a fully standards-compliant C library, but in my experience have been necessary on real operating systems. They are marked with (1) and (2).

1) This explicitly sets the 'locale' configuration the way it is supposed to be set by default.

2) The value returned from getchar is an int. It is supposed to be either a number in the range representable by unsigned char (normally 0-255 inclusive), or the special value EOF (which is not in the range representable by unsigned char). However, buggy C libraries have been known to return negative numbers for characters with their highest bit set. If that happens, the printf will print (for instance) \xffffffa1 when it should've printed \xa1. Casting x to unsigned char and then back to unsigned int corrects this.

Upvotes: 6

John Bode
John Bode

Reputation: 123458

If I understand the question, if you have a string containing control characters like newline, tab, backspace, etc., you want to print a text representation of those characters, rather than interpret them as control characters.

Unfortunately, there's no built-in printf conversion specifier that will do that for you. You'll have to walk through the string character by character, test each one to see if it's a control character, and write some text equivalent for it.

Here's a quick, lightly tested example:

#include <stdio.h>
#include <limits.h>
#include <ctype.h>
...
char *src="This\nis\ta\btest";

char *lut[CHAR_MAX] = {0};  // look up table for printable equivalents
                            // of non-printable characters
lut['\n'] = "\\n";
lut['\t'] = "\\t";
lut['\b'] = "\\b";
...
for ( char *p = src; *p != 0; p++ )
{
  if ( isprint( *p ) )
    putchar( *p );
  else
    fputs( lut[ (int) *p], stdout ); // puts adds a newline at the end,
                                     // fputs does not.
}
putchar( '\n' );

Upvotes: 0

yellowantphil
yellowantphil

Reputation: 1493

Something like this might be what you want. Run myprint(c) to print the character C or a printable representation of it:

#include <ctype.h>

void myprint(int c)
{
    if (isprint(c))
        putchar(c); // just print printable characters
    else if (c == '\n')
        printf("\\n"); // display newline as \n
    else
        printf("%02x", c); // print everything else as a number
}

If you're using Windows, I think all your newlines will be CRLF (carriage return, linefeed) so they'll print as 0d\n the way I wrote that function.

Upvotes: 1

Ivo Valchev
Ivo Valchev

Reputation: 265

What you're looking for is this:

#include <stdio.h>
int main(void)
{
    printf("%s","Hello world\\nHello world");
    return 0;
}

This would produce the following output: Hello world\nHello world

Upvotes: 0

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