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Active9 months ago
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 are c0
in hex and the remaining characters are abc123
in ASCII, then I should get However,
printf
using %x
gives meHow do I get the output I want without the
Prince John Wesley'ffffff'
? And why is it that only c0 (and 80) has the ffffff
, but not the other characters?54k1010 gold badges7575 silver badges8787 bronze badges
RayneRayne5,2181313 gold badges3333 silver badges3939 bronze badges
7 Answers
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.Here's a solution:
This will mask out the upper bits and keep only the lower 8 bits that you want.
MysticialH Force Keygen For Hexagonal
Mysticial398k4040 gold badges303303 silver badges312312 bronze badges
Indeed, there is type conversion to int.Also you can force type to char by using %hhx specifier.
In most cases you will want to set the minimum length as well to fill the second character with zeroes: Autodesk 3ds max 2019 keygen.
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
Ciro Santilli 新疆改造中心996ICU六四事件167k3636 gold badges639639 silver badges517517 bronze badges
brutal_lobsterbrutal_lobster
You can create an unsigned char:
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
Peter Mortensenchar
while printing:14.4k1919 gold badges8888 silver badges117117 bronze badges
Hicham from CppDepend TeamHicham from CppDepend Team
You are probably storing the value 0xc0 in a
lvellalvellachar
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:6,64166 gold badges3535 silver badges8181 bronze badges
H Force Keygen For Hexadecimal
You can use
hh
to tell printf
that the argument is an unsigned char. Figure drawing. Use 0
to get zero padding and 2
to set the width to 2. x
or X
for lower/uppercase hex characters.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
TimmmmTimmmmd
, 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 n
conversion specifier applies to a pointer to a signed char
argument.40.5k3838 gold badges219219 silver badges276276 bronze badges
You are probably printing from a signed char array. Either print from an unsigned char array or mask the value with 0xff: e.g. ar[i] & 0xFF. The c0 values are being sign extended because the high (sign) bit is set.
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Peter Mortensen14.4k1919 gold badges8888 silver badges117117 bronze badges
ObscureRobotObscureRobot6,87222 gold badges2323 silver badges3535 bronze badges