3

I have compiled the following C Program in Code::Blocks 10.05 on Windows 7.

int main() { unsigned char a=-5; signed char b=-5; char c=-5; printf("%d %d %d \n",a,b,c); printf("%u %u %u \n",a,b,c); }

I expected the following output

251 -5 -5 251 251 251

I have reasoned like this,

-5 is represented as 1111 1011 in 2's Complement Notation. Hence 251 will be printed.

But I got the following output.

251 -5 -5 251 4294967291 4294967291

I know 4294967291 is the 32 bit 2's Complement representation of -5.

But why the unsigned char a is printed as 251 instead of 4294967291?

Improve this question
3
  • 1
    printf() is variadic; it's arguments are promoted – as part of usual arithmetic conversions – to int if they are narrower than int. See e.g. the relevant C part of cppreference Commented Aug 1, 2015 at 6:12
  • 1
    You must match the format specifier to the type. %d should be used for all character types. (There are other specifiers but they are unnecessary). %u can only be used with unsigned char (out of the character types). Your %u line causes undefined behaviour. Commented Aug 1, 2015 at 6:29
  • 1
    This is quite similar to stackoverflow.com/questions/7084857/… but I am not sure if it is a dupe. The two questions lead to similar results but are worded quite differently. Commented Aug 1, 2015 at 6:30

4 Answers 4

Reset to default
4

printf is a variadic function, which causes (thanks to @TheParamagneticCroissant for the correction) the integral arguments to become promoted to int or unsigned int. Which according to the standard promotion rules will sign extend them appropriately. This is causing the numbers you are getting to be the results after this promotion process.

Instead you can use "%" PRIu8 to print 8 bit unsigned and "%" PRId8 to print 8 bit signed numbers. However beware that the arguments will still be promoted and then printf will downcast them again according to the format specifier, which may possibly still change that value but I am not sure of that from memory.

In addition char is not guaranteed in C to be 8 bit, it is better to include the <stdint.h> header and use the proper fixed width types.

A fixed version of your code would be as so:

#include <stdint.h> int main() { uint8_t a=-5; int8_t b=-5; printf("%" PRId8 "%" PRId8 "\n", a, b); printf("%" PRIu8 "%" PRIu8 "\n", a, b); }

Note I do not include an equivalent of a char as all the fixed width types are of defined sign so there is no direct equivalent of char, which is of undefined signedness.

Improve this answer
Sign up to request clarification or add additional context in comments.

5 Comments

If the downvoter would care to mention what I have done wrong I would be very grateful, I am tired and have no C compiler on me so am only coding from memory.
"The %u and %d format specifiers are used to print int sized signed and unsigned numbers. This means that your chars are being promoted to int and unsigned ints" – no, they are getting promoted regardless of the format specifier; they are being promoted because printf() is variadic. They are also only ever promoted to one specific type (which is either int or unsigned int depending on the size and range of int and char, and it is probably int on most modern machines.)
thanks for fixing it; I've retracted my DV. I also suggest we find a duplicate question, I'm almost entirely sure this question has been asked several times already.
@TheParamagneticCroissant Could you possibly take a glance at the fixed answer, I dont want to mislead the OP and am not 100% sure I am right even after the fix.
In addition I will take a look around and see if I can find a dupe
2

unsigned integral types (including unsigned char) work with modulo arithmetic. For unsigned char, that is modulo UCHAR_MAX + 1, where UCHAR_MAX is an implementation-defined value in <limits.h> with a typical value with most implementations of 255 (255 is also the minimum value that the standard permits for UCHAR_MAX). Adding 256 to -5 (i.e. to get a remainder between 0 and 255 in order to initialise a in your code) will explain why a has a value of 251.

As to why any of the values are printing at all, the values are being promoted to be of type int when passed to printf(). This promotion is unrelated to your format string but is what the %d format specifier expects, so the values print as you see.

Improve this answer

Comments

1

@Vality is slightly wrong in his answer. Because unsigned char must hold values up to at least 255, then if uint8_t exists, CHAR_BIT must be 8.

There's no reason to use the nasty PRI macros, just use this:

unsigned char a = -5; signed char b = -5; char c = -5; printf("%hhd %hhd %hhd \n", a, b, c); printf("%hhu %hhu %hhu \n", a, b, c);
Improve this answer

2 Comments

Im not quite sure about that, char I believe is allowed to be larger (say 12 bits), and uint8_t be another discrete type which is not otherwise given a name. Though I have not got a standard in front of me so cannot say for certain if that is permitted and if there any systems that do it. However if you do do it that way I would strongly suggest putting in an explicit check of CHAR_BIT and an #error. I agree though that your solution will work on all but the most bizare esoteric machines if not always.
@Vality No type can be smaller than char though. But I really don't care about those ancient systems, they won't even have a halfway-recent libc.
0

Because a holds the value 251, not the value -5.

Improve this answer

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.