%d prints an integer: it will print the ascii representation of your character. What you need is %c:

printf("%c", ch);

printf("%d", '\0'); prints the ascii representation of '\0', which is 0 (by escaping 0 you tell the compiler to use the ascii value 0.

printf("%d", sizeof('\n')); prints 4 because a character literal is an int, in C, and not a char.

If you try this:

#include<stdio.h>

void main()
{
 char name[]="siva";
 printf("name = %p\n", name);
 printf("&name[0] = %p\n", &name[0]);
 printf("name printed as %%s is %s\n",name);
 printf("*name = %c\n",*name);
 printf("name[0] = %c\n", name[0]);
}

Output is:

name = 0xbff5391b  
&name[0] = 0xbff5391b
name printed as %s is siva
*name = s
name[0] = s

So 'name' is actually a pointer to the array of characters in memory. If you try reading the first four bytes at 0xbff5391b, you will see 's', 'i', 'v' and 'a'

Location     Data
=========   ======

0xbff5391b    0x73  's'  ---> name[0]
0xbff5391c    0x69  'i'  ---> name[1]
0xbff5391d    0x76  'v'  ---> name[2]
0xbff5391e    0x61  'a'  ---> name[3]
0xbff5391f    0x00  '\0' ---> This is the NULL termination of the string

To print a character you need to pass the value of the character to printf. The value can be referenced as name[0] or *name (since for an array name = &name[0]).

To print a string you need to pass a pointer to the string to printf (in this case name or &name[0]).

In the format argument of printf, only the % and \ characters are special (no, " is not special and \" is unspecified per POSIX).

But, two important notes.

1. In most printf implementations¹, it's the byte values for \ and % that are special and the POSIX specification could even be interpreted as requiring it as it requires the printf utility to be an interface to the printf(3) C function and not wprintf(3) for instance (like it requires %.3s to truncate to 3 bytes and not 3 characters).

   In some character encodings including BIG5 and GB18030, there are hundreds of characters that contain the encoding of backslash, and to escape those for printf, you'd need to insert a \ before each 0x5c byte within the encoding of those characters!

   For instance in BIG5-HKSCS, as used for instance in the zh_HK.big5hkscs (Hong Kong) locale, all of Ěαжふ㘘㙡䓀䨵䪤么佢俞偅傜兝功吒吭园坼垥塿墦声娉娖娫嫹嬞孀尐岤崤幋廄惝愧揊擺暝枯柦槙檝歿汻沔涂淚滜潿瀙瀵焮燡牾狖獦珢珮琵璞疱癧礒稞穀笋箤糭綅縷罡胐胬脪苒茻莍蓋蔌蕚螏螰許豹贕赨跚踊蹾躡鄃酀酅醆鈾鎪閱鞸餐餤駹騱髏髢髿鱋鱭黠﹏𠗫𠰺𣘀𦖭𦰡𧃸𨜏𩄼𪀔 contain byte 0x5c (which is also the encoding of \).

   With most printf implementations, in that locale, printf 'αb' doesn't output αb but byte 0xa3 (the first byte of the encoding of α) followed by the BS character (the expansion of \b).

   $ LC_ALL=zh_HK.big5hkscs luit
   $ locale charmap
   BIG5-HKSCS
   $ printf 'αb' | LC_ALL=C od -tx1 -tc
   0000000  a3  08
           243  \b
   0000002
   

   Best is to avoid using (and even installing / making available) those locales as they cause all sorts of bugs and vulnerabilities of that sort.

2. Some printf implementations support options, and even those that don't are required to support -- as the option delimiter. So printf -- won't output -- but likely report an error about a missing format argument. So if you can't guarantee your format won't start with -, you have to use the -- option delimiter:

    printf -- "$escaped_format" x y...
   

In any case, if you want to print arbitrary strings, you'd use:

printf '%s\n' "$data" # with terminating newline
printf %s "$data"     # without

There's no character that is special in the string passed to %s (though note that with the exception of the printf builtin of zsh, you can't pass the NUL character in any of printf arguments).

Note that while the canonical way to enter a literal \ is with \\ and a literal % with %%, on ASCII-based systems, you can also use \134 and \45 and with some printf implementations \x5c, \x25, or \x{5c}, \x{25}, or (even on non-ASCII systems): \u005c, \u0025 or \u{5c}, \u{25}.

^{¹ yash's printf builtin being the only exception I am aware of.}