This is governed by the usual arithmetic conversions. For simple cases, the general rule of thumb is that the type with "less" precision is converted to match the type with "more" precision, but it gets somewhat complex once you start mixing signed
and unsigned
.
In C99, this is described by section 6.3.1.8, which I include here for your convenience:
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.
I've highlighted the part that applies to your particular example.
The concept of integer conversion rank is defined in section 6.3.1.1, and it basically describes what you might expect (that types with less precision have a lower rank than types with more precision).
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