Practical difference between int and char

Question

I have to analyse the output of these code fragments:

int x, y;
x = 200; y = 100;
x = x+y; y = x-y; x = x-y;
printf ("%d %d\n", x, y);

char x, y;
x = 200; y = 100;
x = x+y; y = x-y; x = x-y;
printf ("%d %d\n", x, y);

So, I know now that int stands for integer and char for character; I've read about the differences and if I put in the printf the %d, it returns in the form of digits, and %c, in the form of a character.

The ASCII character code for 'A' is 65 for example, but why does the second function print 100 -56, instead of 100 200?

Your `char` isn't unsigned by default. Thus, it wraps around. — Richard J. Ross III, Apr 08 '13 at 01:32
A C char is 8 bits, so 127 is the largest positive number you can have. — Hot Licks, Apr 08 '13 at 01:34
To elaborate, it can be signed or unsigned, and which one `char` itself is varies by implementation. You have to use a specific type if you want that on every implementation. Other types, such as `int`, default to `signed`. — chris, Apr 08 '13 at 01:34
@BrettDuncavage, The problem is that it doesn't always work. — chris, Apr 08 '13 at 02:00
@BrettDuncavage Not so nice since it produces undefined behavior for many values. There are alternatives that don't, e.g., using exclusive-or. — Jim Balter, Apr 08 '13 at 02:12
@chris - Yeah, `char` has historically had an unstable definition, and C must still (sort of) accommodate (now largely extinct) hardware that has different-sized "bytes". So anything you say about `char` comes with a bucket-load of caveats. — Hot Licks, Apr 08 '13 at 03:05

score 11 · Accepted Answer · edited May 15 '16 at 18:14

11

On the platform used in the question, the type char seems to be 1 byte (8 bits) size and is a signed type with 1 sign bit and 7 value bits (and using 2's complement arithmetic). It stores values from -128 to 127. So, this is what's happening to x and y:

x = 200 => x takes value -56
y = 100 => y takes value 100
x = x+y => x takes value 44
y = x-y => y takes value -56
x = x-y => x takes value 100

edited May 15 '16 at 18:14

Jonathan Leffler

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answered Apr 08 '13 at 01:51

ssantos

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Why not -127 to 128? btw, thank you for your clarification, makes sense now. – Souza Apr 08 '13 at 02:03
@Souza, Use the leftmost bit as the sign bit and keep adding 1 on paper. That's not guaranteed by C (only a range of [-127,127] is guaranteed for `signed char`), but that's what is really common today. The term to look for is two's complement. – chris Apr 08 '13 at 02:04
5

chars aren't necessarily 8 bits, no are they necessarily signed. It would be better if your answer were prefixed with "Apparently, in the C environment you are using, ..." – Jim Balter Apr 08 '13 at 02:06

score 9 · Answer 2 · edited May 15 '16 at 18:11

9

C has a variety of integer types: char (at least 8 bits), short (at least 16 bits), int (at least 16 bits), long (at least 32 bits). There are unsigned varieties of those. If you assign a value that is too large to a plain type, the results are undefined (you should never do that, the compiler may assume you never do, and not check at all). In the unsigned case, they "wrap around". But note that the sizes are not guaranteed, just their minimal sizes. There have been machines in which all were 32 bits wide.

edited May 15 '16 at 18:11

Jonathan Leffler

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answered Apr 08 '13 at 01:40

vonbrand

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@chris That has nothing to do with this answer. A `char` can be 32 bits wide and `sizeof(char)` is *still* 1. – Jim Balter Apr 08 '13 at 02:08
@JimBalter, Yeah, I was just thinking about that. Even though I know it's not true, my mind rushed to one byte being eight bits. – chris Apr 08 '13 at 02:09

Practical difference between int and char

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