How Do Promotion Rules Work When the Signedness on Either Side of a Binary Operator Differ

How do promotion rules work when the signedness on either side of a binary operator differ?

This is outlined explicitly in §5/9:

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:

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

Otherwise, if either operand is double, the other shall be converted to double.

Otherwise, if either operand is float, the other shall be converted to float.

Otherwise, the integral promotions shall be performed on both operands.

Then, if either operand is unsigned long the other shall be converted to unsigned long.

Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int.

Otherwise, if either operand is long, the other shall be converted to long.

Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

[Note: otherwise, the only remaining case is that both operands are int]

In both of your scenarios, the result of operator+ is unsigned. Consequently, the second scenario is effectively:

int result = static_cast<int>(us + static_cast<unsigned>(neg));

Because in this case the value of us + neg is not representable by int, the value of result is implementation-defined – §4.7/3:

If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined.

Strange type deduction

Due to the usual arithmetic conversions if two operands have the same conversion rank and one of the operands has unsigned integer type then the type of the expression has the same unsigned integer type.

From the C++ 17 Standard (5 Expressions, p.#10)

— 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.

Pay attention to that the conversion rank of the type unsigned int is equal to the rank of the type int (signed int). From the C++ 17 Standard (4.13 Integer conversion rank, p.#1)

— The rank of any unsigned integer type shall equal the rank of the
corresponding signed integer type

A more interesting example is the following. Let's assume that there are two declarations

unsigned int x = 0;
long y = 0;

and the width of the both types is the same and equal for example to 4 bytes. As it is known the rank of the type long is greater than the rank of the type unsigned int. A question arises what id the type of the expression

x + y

The type of the expression is unsigned long.:)

Here is a demonstrative program but instead of the types long and unsigned int there are used the types long long and unsigned long.

#include <iostream>
#include <iomanip>
#include <type_traits>

int main() 
{
    unsigned long int x = 0;
    long long int y = 0;

    std::cout << "sizeof( unsigned long ) = " 
              << sizeof( unsigned long )
              << '\n';

    std::cout << "sizeof( long long ) = " 
              << sizeof( long long )
              << '\n';

    std::cout << std::boolalpha 
              << std::is_same<unsigned long long, decltype( x + y )>::value
              << '\n';

    return 0;
}

The program output is

sizeof( unsigned long ) = 8
sizeof( long long ) = 8
true

That is the type of the expression x + y is unsigned long long though neither operand of the expression has this type.

C++ Integer Overflow and Promotion

a) From §5/9:

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:

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

Otherwise, if either operand is double, the other shall be converted to double.

Otherwise, if either operand is float, the other shall be converted to float.

Otherwise, the integral promotions (4.5) shall be performed on both operands.

Then, if either operand is unsigned long the other shall be converted to unsigned long.

Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int.

Otherwise, if either operand is long, the other shall be converted to long.

Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

[Note: otherwise, the only remaining case is that both operands are int]

Therefore, since j is unsigned, i is promoted to unsigned and the addition is performed using unsigned int arithmetic.

b) This is UB. The result of the addition is unsigned int (as per (a)), and thus you overflow the int in the assignment.

c) From §4.5/1:

An rvalue of type char, signed char, unsigned char, short int, or unsigned short int can be converted to an rvalue of type int if int can represent all the values of the source type; otherwise, the source rvalue can be converted to an rvalue of type unsigned int.

Therefore, since a 4-byte int can represent any value in a 2-byte short or unsigned short, both are promoted to int (per §5.9's integral promotions rule), and then added as ints.

d) From §3.9.1/4:

Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2ⁿ where n is the number of bits in the value representation of that particular size of integer.

Therefore, UINT_MAX+1 is legal (not UB) and equal to 0.

Why is there a signedness issue when comparing uint16_t and unsigned int?

So what's going on, here ? Where does the int come from?

Integer promotion is going on here. On systems where std::uint16_t is smaller than int, it will be promoted to int when used as an operand (of most binary operations).

In a - b both operands are promoted to int and the result is int also. You compre this signed integer to 3u which is unsigned int. The signs differ, as the compiler warns you.

That warning doesn't show up if I write if ( a - b < static_cast<uint16_t>(3u) ) instead.

Here, the right hand operand is also promoted to int. Both sides of comparison are signed so there is no warning.

Can this actually result in an incorrect behavior?

if ( a - b < static_cast<uint16_t>(3u) ) does have different behaviour than a - b < static_cast<uint16_t>(3u). If one is correct, then presumably the other is incorrect.

Is there a less verbose way to silence it? (or a less verbose way to write a uint16_t literal?)

The correct solution depends on what behaviour you want to be correct.

P.S. You forgot to include the header that defines uint16_t.

Implicit type promotion rules

C was designed to implicitly and silently change the integer types of the operands used in expressions. There exist several cases where the language forces the compiler to either change the operands to a larger type, or to change their signedness.

The rationale behind this is to prevent accidental overflows during arithmetic, but also to allow operands with different signedness to co-exist in the same expression.

Unfortunately, the rules for implicit type promotion cause much more harm than good, to the point where they might be one of the biggest flaws in the C language. These rules are often not even known by the average C programmer and therefore cause all manner of very subtle bugs.

Typically you see scenarios where the programmer says "just cast to type x and it works" - but they don't know why. Or such bugs manifest themselves as rare, intermittent phenomena striking from within seemingly simple and straight-forward code. Implicit promotion is particularly troublesome in code doing bit manipulations, since most bit-wise operators in C come with poorly-defined behavior when given a signed operand.

Integer types and conversion rank

The integer types in C are char, short, int, long, long long and enum.

_Bool/bool is also treated as an integer type when it comes to type promotions.

All integers have a specified conversion rank. C11 6.3.1.1, emphasis mine on the most important parts:

Every integer type has an integer conversion rank defined as follows:

— No two signed integer types shall have the same rank, even if they have the same representation.

— The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision.

— The rank of long long int shall be greater than the rank of long int, which shall be greater than the rank of int, which shall be greater than the rank of short int, which shall be greater than the rank of signed char.

— The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any.

— The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width.

— The rank of char shall equal the rank of signed char and unsigned char.

— The rank of _Bool shall be less than the rank of all other standard integer types.

— The rank of any enumerated type shall equal the rank of the compatible integer type (see 6.7.2.2).

The types from stdint.h sort in here too, with the same rank as whatever type they happen to correspond to on the given system. For example, int32_t has the same rank as int on a 32 bit system.

Further, C11 6.3.1.1 specifies which types are regarded as the small integer types (not a formal term):

The following may be used in an expression wherever an int or unsigned int may
be used:

— An object or expression with an integer type (other than int or unsigned int) whose integer conversion rank is less than or equal to the rank of int and unsigned int.

What this somewhat cryptic text means in practice, is that _Bool, char and short (and also int8_t, uint8_t etc) are the "small integer types". These are treated in special ways and subject to implicit promotion, as explained below.

The integer promotions

Whenever a small integer type is used in an expression, it is implicitly converted to int which is always signed. This is known as the integer promotions or the integer promotion rule.

Formally, the rule says (C11 6.3.1.1):

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.

This means that all small integer types, no matter signedness, get implicitly converted to (signed) int when used in most expressions.

This text is often misunderstood as: "all small signed integer types are converted to signed int and all small, unsigned integer types are converted to unsigned int". This is incorrect. The unsigned part here only means that if we have for example an unsigned short operand, and int happens to have the same size as short on the given system, then the unsigned short operand is converted to unsigned int. As in, nothing of note really happens. But in case short is a smaller type than int, it is always converted to (signed) int, regardless of it the short was signed or unsigned!

The harsh reality caused by the integer promotions means that almost no operation in C can be carried out on small types like char or short. Operations are always carried out on int or larger types.

This might sound like nonsense, but luckily the compiler is allowed to optimize the code. For example, an expression containing two unsigned char operands would get the operands promoted to int and the operation carried out as int. But the compiler is allowed to optimize the expression to actually get carried out as an 8-bit operation, as would be expected. However, here comes the problem: the compiler is not allowed to optimize out the implicit change of signedness caused by the integer promotion because there is no way for the compiler to tell if the programmer is purposely relying on implicit promotion to happen, or if it is unintentional.

This is why example 1 in the question fails. Both unsigned char operands are promoted to type int, the operation is carried out on type int, and the result of x - y is of type int. Meaning that we get -1 instead of 255 which might have been expected. The compiler may generate machine code that executes the code with 8 bit instructions instead of int, but it may not optimize out the change of signedness. Meaning that we end up with a negative result, that in turn results in a weird number when printf("%u is invoked. Example 1 could be fixed by casting the result of the operation back to type unsigned char.

With the exception of a few special cases like ++ and sizeof operators, the integer promotions apply to almost all operations in C, no matter if unary, binary (or ternary) operators are used.

The usual arithmetic conversions

Whenever a binary operation (an operation with 2 operands) is done in C, both operands of the operator have to be of the same type. Therefore, in case the operands are of different types, C enforces an implicit conversion of one operand to the type of the other operand. The rules for how this is done are named the usual artihmetic conversions (sometimes informally referred to as "balancing"). These are specified in C11 6.3.18:

(Think of this rule as a long, nested if-else if statement and it might be easier to read :) )

6.3.1.8 Usual arithmetic conversions

Many operators that expect operands of arithmetic type cause conversions and yield result
types in a similar way. The purpose is to determine a common real type for the operands
and result. For the specified operands, each operand is converted, without change of type
domain, to a type whose corresponding real type is the common real type. Unless
explicitly stated otherwise, the common real type is also the corresponding real type of
the result, whose type domain is the type domain of the operands if they are the same,
and complex otherwise. This pattern is called the usual arithmetic conversions:

First, if the corresponding real type of either operand is long double, the other operand is converted, without change of type domain, to a type whose corresponding real type is long double.

Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double.
Otherwise, if the corresponding real type of either operand is float, the other operand is converted, without change of type domain, to a type whose corresponding real type is float.
Otherwise, the integer promotions are performed on both operands. Then the
following rules are applied to the promoted operands:

If both operands have the same type, then no further conversion is needed.

Otherwise, if both operands have signed integer types or both have unsigned
integer types, the operand with the type of lesser integer conversion rank is
converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
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, then
the operand with unsigned integer type is converted to the type of the
operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.

Notable here is that the usual arithmetic conversions apply to both floating point and integer variables. In the case of integers, we can also note that the integer promotions are invoked from within the usual arithmetic conversions. And after that, when both operands have at least the rank of int, the operators are balanced to the same type, with the same signedness.

This is the reason why a + b in example 2 gives a strange result. Both operands are integers and they are at least of rank int, so the integer promotions do not apply. The operands are not of the same type - a is unsigned int and b is signed int. Therefore the operator b is temporarily converted to type unsigned int. During this conversion, it loses the sign information and ends up as a large value.

The reason why changing type to short in example 3 fixes the problem, is because short is a small integer type. Meaning that both operands are integer promoted to type int which is signed. After integer promotion, both operands have the same type (int), no further conversion is needed. And then the operation can be carried out on a signed type as expected.

unsigned to signed conversion, what happens at the bit level?

The bit pattern doesn't change at all (on most architectures you're likely to encounter in practice). The difference is in the instructions generated by the compiler to manipulate the values.

Integer Overflow and the difference between pow() and multiplication

Here (Linux 64bits, gcc 5.2.1), 55201 is an integer literal of size 4, and the expression 55201 * 55201 seems to be stored in an integer of size 4 before being assigned to your long long int.

One option is storing the factor in another variable before multiplying, to increase the range.

int main(){
  long long int x, factor;
  factor = 55201;
  x = factor * factor;
  printf("%lld", x);
  return 0;
}

How Do Promotion Rules Work When the Signedness on Either Side of a Binary Operator Differ