I caught myself checking if the difference between two unsigned numbers was >= 0. I ran a test running Visual Studio 2022 Preview with the following code. In both cases the answer was true. That seems right to me as how could an unsigned number be considered negative?
However, when I changed all the types from UINT32
to UINT16
or UINT8
, the first comparison returned false. I suppose it is related to the native size. But shouldn't the result be the same regardless of size? (UINT64
seems to behave like UINT32
.)
#include <Windows.h>
#include <iostream>
using namespace std;
int main()
{
UINT32 a = 5;
UINT32 b = 10;
UINT32 c = 0;
if ((a - b) > 0)
{
cout << "\nTrue.";
}
else
{
cout << "\nFalse";
}
c = a - b;
if ((c) > 0)
{
cout << "\nTrue.";
}
else
{
cout << "\nFalse";
}
}
The issue arises because, when a
and b
are UINT16
or UINT8
types, they have ranks less than that of the int
type, so, by the "usual arithmetic conversion" rules, they are promoted to int
before the a - b
operation is performed, and the result of that operation is also of int
type.
From this draft C++17 Standard (bolding for emphasis is mine):
7.6 Integral promotions [conv.prom]
1 A prvalue of an integer type other than
bool
,char16_t
,char32_t
, orwchar_t
whose integer conversion rank (6.7.4) is less than the rank ofint
can be converted to a prvalue of typeint
ifint
can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int.
However, on your platform, the UINT32
type has a rank that is the same as that of int
; so, in that case, no promotion is performed, and the result of a - b
is a UINT32
(which cannot be negative).
If you have the relevant feature enabled (it is by default, IIRC), then the Visual Studio IDE will actually tell you what the issue is, if you declare an auto
local variable and initialize it with a - b
. The following shows the popup displayed, when hovering over the d
variable, for the case when using the UINT32
type:
However, when we change the type to UINT16
, we see that the a - b
expression is, indeed, evaluated as an int
: