Printing hexadecimal characters in C
Solution 1
You are seeing the ffffff
because char
is signed on your system. In C, vararg functions such as printf
will promote all integers smaller than int
to int
. Since char
is an integer (8-bit signed integer in your case), your chars are being promoted to int
via sign-extension.
Since c0
and 80
have a leading 1-bit (and are negative as an 8-bit integer), they are being sign-extended while the others in your sample don't.
char int
c0 -> ffffffc0
80 -> ffffff80
61 -> 00000061
Here's a solution:
char ch = 0xC0;
printf("%x", ch & 0xff);
This will mask out the upper bits and keep only the lower 8 bits that you want.
Solution 2
Indeed, there is type conversion to int. Also you can force type to char by using %hhx specifier.
printf("%hhX", a);
In most cases you will want to set the minimum length as well to fill the second character with zeroes:
printf("%02hhX", a);
ISO/IEC 9899:201x says:
7 The length modifiers and their meanings are: hh Specifies that a following d, i, o, u, x, or X conversion specifier applies to a signed char or unsigned char argument (the argument will have been promoted according to the integer promotions, but its value shall be converted to signed char or unsigned char before printing); or that a following
Solution 3
You can create an unsigned char:
unsigned char c = 0xc5;
Printing it will give C5
and not ffffffc5
.
Only the chars bigger than 127 are printed with the ffffff
because they are negative (char is signed).
Or you can cast the char
while printing:
char c = 0xc5;
printf("%x", (unsigned char)c);
Solution 4
You can use hh
to tell printf
that the argument is an unsigned char. Use 0
to get zero padding and 2
to set the width to 2. x
or X
for lower/uppercase hex characters.
uint8_t a = 0x0a;
printf("%02hhX", a); // Prints "0A"
printf("0x%02hhx", a); // Prints "0x0a"
Edit: If readers are concerned about 2501's assertion that this is somehow not the 'correct' format specifiers I suggest they read the printf
link again. Specifically:
Even though %c expects int argument, it is safe to pass a char because of the integer promotion that takes place when a variadic function is called.
The correct conversion specifications for the fixed-width character types (int8_t, etc) are defined in the header
<cinttypes>
(C++) or<inttypes.h>
(C) (although PRIdMAX, PRIuMAX, etc is synonymous with %jd, %ju, etc).
As for his point about signed vs unsigned, in this case it does not matter since the values must always be positive and easily fit in a signed int. There is no signed hexideximal format specifier anyway.
Edit 2: ("when-to-admit-you're-wrong" edition):
If you read the actual C11 standard on page 311 (329 of the PDF) you find:
hh: Specifies that a following
d
,i
,o
,u
,x
, orX
conversion specifier applies to asigned char
orunsigned char
argument (the argument will have been promoted according to the integer promotions, but its value shall be converted tosigned char
orunsigned char
before printing); or that a followingn
conversion specifier applies to a pointer to asigned char
argument.
Solution 5
You are probably storing the value 0xc0 in a char
variable, what is probably a signed type, and your value is negative (most significant bit set). Then, when printing, it is converted to int
, and to keep the semantical equivalence, the compiler pads the extra bytes with 0xff, so the negative int
will have the same numerical value of your negative char
. To fix this, just cast to unsigned char
when printing:
printf("%x", (unsigned char)variable);
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Rayne
Updated on November 07, 2020Comments
-
Rayne over 3 years
I'm trying to read in a line of characters, then print out the hexadecimal equivalent of the characters.
For example, if I have a string that is
"0xc0 0xc0 abc123"
, where the first 2 characters arec0
in hex and the remaining characters areabc123
in ASCII, then I should getc0 c0 61 62 63 31 32 33
However,
printf
using%x
gives meffffffc0 ffffffc0 61 62 63 31 32 33
How do I get the output I want without the
"ffffff"
? And why is it that only c0 (and 80) has theffffff
, but not the other characters?-
burito almost 6 yearsThe string that matches your array of bytes would be...
"\xc0\xc0abc123"
-
-
lvella over 12 yearsMy solution using a cast to
unsigned char
is one instruction smaller in gcc4.6 for x86-64... -
Bob Stein over 10 years+1 the real best answer, explicit typing as close to the data declaration as possible (but not closer).
-
2501 almost 8 yearsMaybe I can help. This is (technically) undefined behavior because specifier
x
requires an unsigned type, but ch is promoted to int. The correct code would simply cast ch to unsigned, or use a cast to unsigned char and the specifier:hhx
. -
2501 almost 8 yearsSpecifiers are not correct for the type uint8_t. Fixed width types use special print specifiers. See:
inttypes.h
-
Timmmm almost 8 yearsYeah but all varargs integers are implicitly promoted to int.
-
2501 almost 8 yearsThat may be, but as far as C is defined, behavior is undefined if you don't use the correct specifier.
-
Timmmm almost 8 yearsBut %x is the correct specifier. (
char
andunsigned char
are promoted toint
)[en.cppreference.com/w/cpp/language/variadic_arguments]. You would only need to use the PRI specifiers for things that don't fit in your platformint
- e.g.unsigned int
. -
2501 almost 8 years
%x
is correct for unsigned int not int. Types char and unsigned char are promoted to int. In addition there is no guarantee that uint8_t is defined as unsigned char. -
Timmmm almost 8 yearsSee my edit. There is no signed hex specifier. It is perfectly value to promote
uint8_t
toint
. QED. -
2501 almost 8 yearsYou didn't address my comments. There is no signed hex specifier. Exactly, int cannot be printed using specifier
x
. Only unsigned int.The quote you posted is not from the C Standard. Therefore it is not relevant. -
2501 almost 8 yearsAs I have already written, you assume that uint8_t is always defined as unsigned char. This is not the case as C makes no such guarantee.
-
Gustavo Meira almost 7 yearsIf I have
printf("%x", 0)
, nothing is printed. -
user2262111 almost 5 yearsIt doesn't print anything because the minimum is set to 0. To fix this, try
printf("%.2x", 0);
which will boost the minimum characters drawn to 2. To set a max, prepend the . with a number. For example, you can force only 2 characters drawn by doingprintf("%2.2x", 0);
-
maxschlepzig about 4 yearsAny reason why
printf("%x", ch & 0xff)
should be better than just usingprintf("%02hhX", a)
as in @brutal_lobster's answer?