C++ Unexpected Integer Promotion

Question

I was writing some code recently that was actually supposed to test other code, and I stumbled upon a surprising case of integer promotion. Here's the minimal testcase:

#include <cstdint>
#include <limits>

int main()
{
    std::uint8_t a, b;
    a = std::numeric_limits<std::uint8_t>::max();
    b = a;

    a = a + 1;

    if (a != b + 1)
        return 1;
    else
        return 0;
}

Surprisingly this program returns 1. Some debugging and a hunch revealed that b + 1 in the conditional was actually returning 256, while a + 1 in assignment produced the expected value of 0.

Section 8.10.6 (on the equality/ineuqlity operators) of the C++17 draft states that

If both operands are of arithmetic or enumeration type, the usual arithmetic conversions are performed on both operands; each of the operators shall yield true if the specified relationship is true and false if it is false.

What are "the usual arithmetic conversions", and where are they defined in the standard? My guess is that they implicitly promote smaller integers to int or unsigned int for certain operators (which is also supported by the fact that replacing std::uint8_t with unsigned int yields 0, and further in that the assignment operator lacks the "usual arithmetic conversions" clause).

Answer 1

What are "the usual arithmetic conversions", and where are they defined in the standard?

[expr.arith.conv]/1

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

• (1.1) If either operand is of scoped enumeration type, no conversions are performed; if the other operand does not have the same type, the expression is ill-formed.

• (1.2) If either operand is of type long double, the other shall be converted to long double.

• (1.3) Otherwise, if either operand is double, the other shall be converted to double.

• (1.4) Otherwise, if either operand is float, the other shall be converted to float.

• (1.5) Otherwise, the integral promotions ([conv.prom]) shall be performed on both operands.⁵⁹ Then the following rules shall be applied to the promoted operands:

  • (1.5.1) If both operands have the same type, no further conversion is needed.

  • (1.5.2) Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank shall be converted to the type of the operand with greater rank.

  • (1.5.3) Otherwise, if the operand that has unsigned integer type has rank greater than or equal to the rank of the type of the other operand, the operand with signed integer type shall be converted to the type of the operand with unsigned integer type.

  • (1.5.4) Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, the operand with unsigned integer type shall be converted to the type of the operand with signed integer type.

  • (1.5.5) Otherwise, both operands shall be converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

⁵⁹⁾ As a consequence, operands of type bool, char8_­t, char16_­t, char32_­t, wchar_­t, or an enumerated type are converted to some integral type.

For uint8_t vs int (for operator+ and operator!= later), #1.5 is applied, uint8_t will be promoted to int, and the result of operator+ is int too.

On the other hand, for unsigned int vs int (for operator+), #1.5.3 is applied, int will be converted to unsigned int, and the result of operator+ is unsigned int.

Answer 2

Your guess is correct. Operands to many operators in C++ (e.g., binary arithmetic and comparison operators) are subject to the usual arithmetic conversions. In C++17, the usual arithmetic conversions are specified in [expr]/11. I'm not going to quote the whole paragraph here because it's rather large (you can just click on the link), but for integral types, the usual arithmetic conversions boil down to integral promotions being applied followed by effectively some more promoting in the sense that if the types of the two operands after the initial integral promotions are not the same, the smaller type is converted to the larger one of the two. The integral promotions basically mean that any type smaller than an int will be promoted to int or unsigned int, whichever of the two can represent all possible values of the original type, which is mainly what is causing the behavior in your example.

As you have already figured out yourself, in your code, the usual arithmetic conversions happen in a = a + 1; and, most noticeably, in the condition of your if

if (a != b + 1)
    …

where they cause b to be promoted to int, making the result of b + 1 to be of type int, as well as a being promoted to int and the !=, thus, happening on values of type int, which causes the condition to be true rather than false…