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 causing 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 phenomenon 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
long long and
bool is also treated as an integer type when it comes to type promotions.
All integers have a specified conversion rank. C11 126.96.36.199, 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 intshall 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
— 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 188.8.131.52).
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 184.108.40.206 specifies which types that are regarded as the small integer types (not a formal term):
The following may be used in an expression wherever an
— An object or expression with an integer type (other than
unsigned int) whose integer conversion rank is less than or equal to the rank of
What this somewhat cryptic text means in practice, is that
short (and also
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 220.127.116.11):
intcan 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
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
With the exception of a few special cases like
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 🙂 )
18.104.22.168 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
- 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
- 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
- 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 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 –
unsigned int and
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.