Why does printf function ignore the latter \0? [duplicate] - c

Question

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What happens if I do printf("one\0two");?
(4 answers)
How to print a string with embedded nulls so that "(null)" is substituted for '\0'
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Closed 3 years ago.
I am stuck with some features that \0 has.
I know that \0 is a null character and it is a term to indicate that the formal is a string.
int j;
j = printf("abcdef\0abcdefg\0");
printf("%d", j);
return 0;
When I tried to print "abcdef\0abcdefg\0" out, C would only print string 'abcdef' and '6' instead of both 'abcdef' and 'abcdefg' which would sum up to 13. Why does this happen?

Answer 1

"abcdef\0abcdefg\0", the string literal, is effectively a static, const (for all intents an purposes) char array and so it has an associated size that the compiler maintains:
#include <stdio.h>
#define S "abcdef\0abcdefg\0"
//^string literals implicitly add a(nother) hidden \0 at the end
int main()
{
printf("%zu\n", sizeof(S)); //prints 16
}
But arrays are treated specially in C and passing them as a parameter to a function or almost any operator converts them to a pointer to their first element.
Pointers do not have an associated size.
When you pass a char const* to a function (e.g., printf), the function receives just one number--the address of the first element.
The way printf and most string functions in C obtain the size is by counting character until the first '\0'.
If you pass a pointer to the first element of a char array that has explicit embedded zeros in it, then for a function that counts until the first '\0', the string effectively ends at the first '\0'.

Answer 2

A string in C is a sequence of characters followed by a NUL character, which is '\0'. There is no separate "length" field.
When a string appears as a literal, such as "hello", what actually gets stored in memory is:
'h', 'e', 'l', 'l', 'o', '\0'
So you can see that if your string itself contains a '\0', as far as any of the C standard library functions are concerned, that's the end of the string.
Once printf see the first '\0' in your string, it stops printing and returns, because that's the end of the format string. printf has no way of knowing that there's another string after the '\0'. Maybe there is--or maybe there's just random other program data in memory after that point. It can't tell the difference.
If you want to actually print the '\0' characters, then you need to have some other way to track the "real" length of the string and use a function that accepts that length as a parameter. Alternately you could add the '\0' characters during the formatting process by specifying %c in the format string and passing 0 as the character value.

Answer 3

Here
j = printf("abcdef\0abcdefg\0"); /* printf stops printing once \0 encounters hence it prints abcdef */
printf() starts printing from base address of string literal "abcdef\0abcdefg\0" i.e from a until first \0 char encounters. So it prints abcdef.
-----------------------------------------------------------------
| a | b | c | d | e | f | \0 | a | b | c | d | e | f | f | g | \0 |
-----------------------------------------------------------------
0x100 0x101 ...............| 0x100 - assume this as base address of the string literal
| |
starts printing when printf sees
from 0x100 memory first \0
location it stops the printing & returns.
And then printf() returns number of printable characters i.e 6.
printf("%d", j); /* prints 6 */
From the manual page of printf
RETURN VALUE
Upon successful return, these functions return the number of
characters printed (excluding the null byte used to end output
to
strings).