Why printf show false value of an hex number

Question

Code

char a; 
a = 0xf1;
printf("%x\n", a);

Output

fffffff1

printf() show 4 bytes, that exactly we have one byte in a.

What is the reason of this misbehavior?

How can i correct it?

Answer 1

printf is a variable argument function, so the compiler does its best but cannot check strict compliance between format specifier and argument type.

Here you're passing a char with a %x (integer, hex) format specifier.

So the value is promoted to a signed integer (because > 127: negative char and char is signed on most systems, on yours that's for sure)

Either:

• change a to int (simplest)
• change a to unsigned char (as suggested by BLUEPIXY) that takes care of the sign in the promotion
• change format to %hhx as stated in the various docs (note that on my gcc 6.2.1 compiler hhx is not recognized, even if hx is)

note that the compiler warns you before reaching printf that you have a problem:

gcc -Wall -Wpedantic test.c
test.c: In function 'main':
test.c:6:5: warning: overflow in implicit constant conversion [-Woverflow]
 a = 0xf1;

Answer 2

What is the reason of this misbehavior?

This question looks strangely similar to another I have answered; it even contains a similar value (0xfffffff1). In that answer, I provide some information required to understand what conversion happens when you pass a small value (such as a char) to a variadic function such as printf. There's no point repeating that information here.

If you inspect CHAR_MIN and CHAR_MAX from <limits.h>, you're likely to find that your char type is signed, and so 0xf1 does not fit as an integer value inside of a char.

Instead, it ends up being converted in an implementation-defined manner, which for the majority of us means it's likely to end up with one of the high-order bits becoming the sign bit. When these values are promoted to int (in order to pass to printf), sign extension occurs to preserve the value (that is, a char that has a value of -1 should be converted to an int that has a value of -1 as an int, so too is the underlying representation for your example likely to be transformed from 0xf1 to 0xfffffff1).

printf("CHAR_MIN .. CHAR_MAX: %d .. %d\n", CHAR_MIN, CHAR_MAX);
printf("Does %d fit? %s\n", '\xFF', '\xFF' >= CHAR_MIN && '\xFF' <= CHAR_MAX ? "Yes!"
                                                                             : "No!");

printf("%d %X\n", (char) -1, (char) -1); // Both of these get converted to int
printf("%d %X\n", -1, -1);               // ... and so are equivalent to these

How can i correct it?

Declare a with a type that can fit the value 0xf1, for example int or unsigned char.

Answer 3

You should use int a instead of char a because char is unsigned and can store only 1 byte from 0 to 255. And hex number need many storage to store it, and also int storage size is 2 or 4 bytes. So it's good to use int here to store hex number.