When need to buffer in memory some raw data, for example from a stream, is it better to use an array of char
or of unsigned char
? I always used char
but at work are saying it is better unsigned char
and I don't know why.
The basic ASCII values are in range 0 to 127. The rest part of the ASCII is known as extended ASCII. Using char or signed char we cannot store the extended ASCII values. By using the unsigned char, we can store the extended part as its range is 0 to 255.
An unsigned type can only represent postive values (and zero) where as a signed type can represent both positive and negative values (and zero). In the case of a 8-bit char this means that an unsigned char variable can hold a value in the range 0 to 255 while a signed char has the range -128 to 127.
The unsigned char type can only store nonnegative integer values , it has a minimum range between 0 and 127 , as defined by the C standard. The signed char type can store , negative , zero , and positive integer values . It has a minimum range between -127 and 127 , as defined by the C standard .
unsigned char is a character datatype where the variable consumes all the 8 bits of the memory and there is no sign bit (which is there in signed char). So it means that the range of unsigned char data type ranges from 0 to 255.
UPDATE: C++17 introduced std::byte
, which is more suited to "raw" data buffers than using any manner of char
.
For earlier C++ versions:
unsigned char
emphasises that the data is not "just" text
if you've got what's effectively "byte" data from e.g. a compressed stream, a database table backup file, an executable image, a jpeg... then unsigned
is appropriate for the binary-data connotation mentioned above
unsigned
works better for some of the operations you might want to do on binary data, e.g. there are undefined and implementation defined behaviours for some bit operations on signed types, and unsigned
values can be used directly as indices in arrays
you can't accidentally pass an unsigned char*
to a function expecting char*
and have it operated on as presumed text
in these situations it's usually more natural to think of the values as being in the range 0..255, after all - why should the "sign" bit have a different kind of significance to the other bits in the data?
if you're storing "raw data" that - at an application logic/design level happens to be 8-bit numeric data, then by all means choose either unsigned
or explicitly signed
char
as appropriate to your needs
As far as the structure of the buffer is concerned, there is no difference: in both cases you get an element size of one byte, mandated by the standard.
Perhaps the most important difference that you get is the behavior that you see when accessing the individual elements of the buffer, for example, for printing. With char
you get implementation-defined signed or unsigned behavior; with unsigned char
you always see unsigned behavior. This becomes important if you want to print the individual bytes of your "raw data" buffer.
Another good alternative for use for buffers is the exact-width integer uint8_t
. It is guaranteed to have the same width as unsigned char
, its name requires less typing, and it tells the reader that you are not intended to use the individual elements of the buffer as character-based information.
Internally, it is exactly the same: Each element is a byte. The difference is given when you operate with those values.
If your values range is [0,255] you should use unsigned char
but if it is [-128,127] then you should use signed char
.
Suppose you are use the first range (signed char
), then you can perform the operation 100+100
. Otherwise that operation will overflow and give you an unexpected value.
Depending on your compiler or machine type, char
may be unsigned or signed by default:
Is char signed or unsigned by default?
Thus having char
the ranges described for the cases above.
If you are using this buffer just to store binary data without operating with it, there is no difference between using char
or unsigned char
.
EDIT
Note that you can even change the default char
for the same machine and compiler using compiler's flags:
-funsigned-char Let the type char be unsigned, like unsigned char.
Each kind of machine has a default for what char should be. It is either likeunsigned char by default or like signed char by default. Ideally, a portable program should always use signed char or unsigned char when it depends on the signedness of an object. But many programs have been written to use plain char and expect it to be signed, or expect it to be unsigned, depending on the machines they were written for. This option, and its inverse, let you make such a program work with the opposite default.
The type char is always a distinct type from each of signed char or unsigned char, even though its behavior is always just like one of those two.
As @Pablo said in his answer, the key reason is that if you're doing arithmetic on the bytes, you'll get the 'right' answers if you declare the bytes as unsigned char
: you want (in Pablo's example) 100 + 100 to add to 200; if you do that sum with signed char
(which you might do by accident if char
on your compiler is signed) there's no guarantee of that – you're asking for trouble.
Another important reason is that it can help document your code, if you're explicit about what datatypes are what. It's useful to declare
typedef unsigned char byte
or even better
#include <stdint.h>
typedef uint8_t byte
Using byte
thereafter makes it that little bit clearer what your program's intent is. Depending on how paranoid your compiler is (-Wall
is your friend), this might even cause a type warning if you give a byte*
argument to a char*
function argument, thus prompting you to think slightly more carefully about whether you're doing the right thing.
A 'character' is fundamentally a pretty different thing from a 'byte'. C happens to blur the distinction (because at C's level, in a mostly ASCII world, the distinction doesn't matter in many cases). This blurring isn't always helpful, but it's at least good intellectual hygiene to keep the difference clear in your head.
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