Why char in C is performing different behaviour? - c

Question

I am a beginner and I am curious about char in C. I know that char takes only one character. If I inline initialize char like the below code:
#include <stdio.h>
int main() {
char c2 = 'ab';
printf("c2: %c\n", c2);
return 0;
}
It will output b. My first question is why b and why not a?
Again if I take input form console like the below code:
#include <stdio.h>
int main() {
char c3;
scanf("%c", &c3);
printf("c3: %c\n", c3);
return 0;
}
Now if I input ab then it will output a. Why now not giving output b as the first one (inline initialization)?
Please someone explain why this different behaviour is performing?

Answer 1

You can't fit 'ab' in a char, and trying to assign 'ab' to a char truncates it to the lower byte, 'b'.
Please, please turn your compiler warnings on and heed them; you can see these errors on Godbolt:
<source>: In function 'main':
<source>:4:15: warning: multi-character character constant [-Wmultichar]
4 | char c2 = 'ab';
| ^~~~
<source>:4:15: warning: overflow in conversion from 'int' to 'char' changes value from '24930' to '98' [-Woverflow]
98 is the ASCII value for b:
>>> chr(98)
'b'
For your second program, if you input 'ab', scanf with %c consumes 1 character, 'a'. (Other input remains in the input buffer and would be consumed by subsequent similar scanf calls.)

Answer 2

Compiling the first example may compile with warnings. e.g in gcc:
main.c: In function ‘main’:
main.c:4:15: warning: multi-character character constant [-Wmultichar]
4 | char c2 = 'ab';
| ^~~~
main.c:4:15: warning: overflow in conversion from ‘int’ to ‘char’ changes value from ‘24930’ to ‘98’ [-Woverflow]
Since:
The value of an integer character constant containing more than one character (e.g., 'ab'), [...] is implementation-defined.
The warnings also suggests the implementation behaviour: 24930 = 0x6162 where 0x62 is ASCII 'b'. So the implementation behaviour is a matter of byte-order, then assignment of an int to a char, which assigns the least significant byte.
The behaviour of the second example is no surprise - input is not assignment or initialisation, and it is not a C language behaviour, but a system I/O behaviour. Input is buffered and requesting a single character input will take the first character from the first-in-first-out (FIFO) input queue. The first character when you enter 'a' followed by 'b' is of course 'a'.
A second input request would retrieve the 'b' whereas in the initialisation example the 'b' is simply discarded by the assignment. Comparing I/O behaviour with C language assignment behaviour are not at all comparable. The input is a FIFO queue of single characters, whereas 'ab' is an int (in this implementation).

Answer 3

This is compiler-specific, so called implementation-defined behavior.
From the C standard 6.4.4.4/2:
An integer character constant is a sequence of one or more multibyte characters enclosed in single-quotes, as in 'x'.
Then 6.4.4.4/10, which happens to use the very same example as you:
An integer character constant has type int. The value of an integer character constant
containing a single character that maps to a single-byte execution character is the
numerical value of the representation of the mapped character interpreted as an integer.
The value of an integer character constant containing more than one character (e.g.,
'ab'), or containing a character or escape sequence that does not map to a single-byte
execution character, is implementation-defined. If an integer character constant contains
a single character or escape sequence, its value is the one that results when an object with
type char whose value is that of the single character or escape sequence is converted to
type int.
In the second example, the first character you place in stdin is the one read and the other is ignored. This has nothing to do with character constants but is simply how scanf works.