Using a long variable to store a char and printing it

Question

I am storing a char in a long variable and trying to print it using printf.

long a = 'A';
printf("%c \n",a);

Considering default argument promotions the arguments get promoted to int but as long has higher rank than int it should not get promoted (or actually demoted) to int.

Now if it doesn't get promoted to int , in the printf statement isn't there a type mismatch as we are trying to print a long variable using %c. According to C standard this behavior should be undefined but the compiler is perfectly printing A as output without any error or warning. Why is it so ?

I am working on GCC codeblocks compiler.

Just because something is undefined behaviour doesn't mean it wont work. — Mitch, May 16 '20 at 17:06
So it means, it could or could not work but no one can be sure about that ? Right ? — LocalHost, May 16 '20 at 17:08
And the constant `'A'` is actually an `int` and not a `char` in C. — Andrew Henle, May 16 '20 at 17:08
Undefined behavior means that the behavior is not defined :) — klutt, May 16 '20 at 17:08
Also, you're getting no warning because you do not compile with `-Wall` — klutt, May 16 '20 at 17:18
UB, in your case, manifests by printing an `A`. Try `free(main)` and check that it prints an `A`. — pmg, May 16 '20 at 17:19
When you jam the wrong size screw into a machine, sometimes it works and sometimes it does not. — Eric Postpischil, May 16 '20 at 17:37
It works because your machine is little-endian, helping printf() to read the little end correctly. You will not have such luck on a big-endian machine. — Hans Passant, May 16 '20 at 18:04

klutt · Answer 1 · 2020-05-16T18:02:39.160

According to C standard this behavior should be undefined but the compiler is perfectly printing A as output without any error or warning. Why is it so ?

Undefined behavior means that the behavior is not defined. That means that you cannot really expect it to behave in a certain way. Some things are more likely than others, but UB basically means that anything can happen.

Plus, the reason you don't get warnings is that you have not activated warnings.

$ gcc -Wall main.c 
main.c: In function ‘main’:
main.c:6:14: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘long int’ [-Wformat=]
    6 |     printf("%c \n",a);
      |             ~^     ~
      |              |     |
      |              int   long int
      |             %ld

I'd recommend compiling with -Wall -Wextra -pedantic -std=c11 -Werror

Note that the warning says format ‘%c’ expects argument of type ‘int’ so passing an int to %c is perfectly fine. If you pass a char it will get promoted to int.

Undefined behavior: https://stackoverflow.com/a/4105123/6699433

if i try to print a `int` variable instead of `long int` using `%c` will this be an undefined behavior or is it valid ? — LocalHost, May 16 '20 at 17:48
@Noshiii This is valid. In fact, `%c` in `printf` expects an `int`. `char` is compatible to `int`. — RobertS supports Monica Cellio, May 16 '20 at 17:50

score 0 · Answer 2 · answered May 16 '20 at 17:16

0

The ASCII value of 'A' is casted and stored as a long number, the printf function casts it back to char as for %c, and prints back the 'A'

Actually, 'A' is just a number which is defined in a standardized way, so should be no problems.

Correct me if I'm wrong.

answered May 16 '20 at 17:16

`printf` is not required to do that cast – klutt May 16 '20 at 17:21
But anyway I think it implicitly casts.. C don't actually care about the data type unless for array and pointer stuff.. right? – May 16 '20 at 17:35
2

For `%c`, `printf` attempts to get an `int` argument, and it then converts (or behaves as if it does) that to an `unsigned char` and prints that. But when a `long` is passed, `printf` is not expected to get that and cast it. Its behavior in OP’s attempt may be the same as if that were done, but there are various ways that result could come to be, and it is incorrect to say that `printf` converts a stored `long` to a character. Additionally, the C standard leaves the specification of character values quite loose. – Eric Postpischil May 16 '20 at 17:36
1

@EricPostpischil Your comments are often too good to be only comments. This as answer surely would help the OP at most as it shows what actually and apparently seems to be happen. – RobertS supports Monica Cellio May 16 '20 at 18:11

Using a long variable to store a char and printing it

2 Answers2