Question

For example, If I have,

 int a = 42;
 unsigned b = 10;
 int c = a + b;

For this statement, int c = a + b; Would a be first converted to an unsigned int or would it be b that will be converted to a signed int? Both unsigned int and signed have the same conversion rank so how do we know which one will be converted? Is there a Standard rule?

Answer 1

Short answer: Per C99 6.3.1.8-p1, as value will be converted to an unsigned int by, per C99 6.3.1.3-p2, having UINT_MAX+1 added to it until it falls in the range allowed by unsigned int. Since it is already in that range, no addition will be performed. By C99 6.3.1.3-p3, the results assigned back to int c would be implementation defined if (p1) and (p2) didn’t apply. But in this case note the “value” clause of 6.3.1.3-p1. The value (52) in this case can be represented by int, so it is not changed, and is defined.

C99 6.3.1.3 Signed and unsigned integers

1. When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.

2. Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.60)

3. Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.

C99 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.62)

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.