Converts a char to unsigned int - C

Question

I don't understand the output of this little main :

 int main(int argc, char **argv) {
     char c = '\336';
     unsigned int u = (unsigned int) c;
     printf("%d\n",u); /* I'm waiting for 222  but no ...*/
     exit(EXIT_SUCCESS);
}

Why is this print like my variable is signed int ? How can I to have the value 222 which I want ?

Thank you a lot!

First, you should say what is the actual output. Second, `char` could be unsigned and if it has 8-bits (which often does), then 222 won't fit in it. — Shahbaz, Nov 15 '13 at 10:32
So which I can do to have the value 222 ? I try some things but doesn't working — alifirat, Nov 15 '13 at 10:36
@Shahbaz: 222 would fit in an unsigned 8-bit `char`, but it wouldn't fit in a signed 8-bit `char`. — dreamlax, Nov 15 '13 at 10:36
Also, you should use `%u` instead of `%d` if you are providing an `unsigned int`. — dreamlax, Nov 15 '13 at 10:38

score 0 · Answer 1 · edited May 23 '17 at 11:50

The standard does not specify if char is signed or unsigned by default. So, this is a implementation dependent, and on your machine it seemes it is signed, and '\336' as in this case value -34.

In order to achive what you are expecting, you must:

define c as unisgned char
use %u specifier in printf.

For further explanations, you may take a look here: Is char signed or unsigned by default?

Converts a char to unsigned int - C

1 Answers1