compare 8bit value against 16bit value

Question

I was wondering what happens when an 8bit value is compared against a 16bit value.

I'll try to explain the problem by a code example:

bool result;
unsigned char a_8bit = 0xcd;
unsigned short b_16bit = 0xabcd;
result = a_8bit < b_16bit;

Possible results can be:

a_8bit is casted to unsigned short implicitly and compared to b_16bit as a 16bit value. Result is true
b_16bit is casted to unsigned char implicitly and compared to a_8bit as an 8bit value. Result is false

Does anybody has a clue what the compiler will do with this piece of code? Sure, i can try it out, but are there different interpretations by different compilers of this code?

arithmetic conversion of both to int. and then casted to bool. but the size of int is implementation defined — Koushik Shetty, Nov 25 '13 at 11:39
@Koushik There's no cast to `bool`. The results of `<` already have type `bool`. — James Kanze, Nov 25 '13 at 12:01

score 2 · Accepted Answer · answered Nov 25 '13 at 11:40

2

¹ A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion rank (4.13) is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int. [§ 4.5]

So, compiler can promote both of them to unsigned int and then do the comparison.

answered Nov 25 '13 at 11:40

masoud

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is it not int since int can represent the values of both the operands? – Koushik Shetty Nov 25 '13 at 11:44
Compiler can choose between `unsigned int` and `unsigned short` since both are able to represent two values, however the standard is more specific for `int`. – masoud Nov 25 '13 at 11:47
@MM. _IF_ the size of `int` is larger than the size of `short`. But on a typical 32 (or 64) bit machine, yes. They will both be converted to `int` before the comparison. – James Kanze Nov 25 '13 at 12:00
@JamesKanze: I was confused. Thanks for making it clear to me. – masoud Nov 25 '13 at 12:02

score 1 · Answer 2 · answered Nov 25 '13 at 11:39

1

The first, although to be precise, both are converted to unsigned and then compared.

answered Nov 25 '13 at 11:39

john

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score 0 · Answer 3 · answered Nov 25 '13 at 11:42

0

Normal integer promotion rules apply, see for example: http://www.idryman.org/blog/2012/11/21/integer-promotion/

answered Nov 25 '13 at 11:42

teroi

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compare 8bit value against 16bit value

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