12.2 Integral Promotion

When there is more than one operand to an operator, they typically must be of exactly the same type. The compiler will automatically convert the operands, if necessary, so they do have the same type. The conversion is to a “larger” type so there is no loss of information; however, the change in type can cause different code behavior to what is sometimes expected. These form the standard type conversions.

Prior to these type conversions, some operands are unconditionally converted to a larger type, even if both operands to an operator have the same type. This conversion is called integral promotion and is part of Standard C behavior. The compiler performs these integral promotions where required, and there are no options that can control or disable this operation. If you are not aware that the type has changed, the results of some expressions are not what would normally be expected.

Integral promotion is the implicit conversion of enumerated types, signed or unsigned varieties of char, short int or bit-field types to either signed int or unsigned int. If the result of the conversion can be represented by an signed int, then that is the destination type, otherwise the conversion is to unsigned int.

Consider the following example.

unsigned char count, a=0, b=50;
if(a - b < 10)
  count++;

The unsigned char result of a - b is 206 (which is not less than 10), but both a and b are converted to signed int via integral promotion before the subtraction takes place. The result of the subtraction with these data types is -50 (which is less than 10) and hence the body of the if() statement is executed.

If the result of the subtraction is to be an unsigned quantity, then apply a cast. For example:

if((unsigned int)(a - b) < 10)

count++;

The comparison is then done using unsigned int, in this case, and the body of the if() would not be executed.

Another problem that frequently occurs is with the bitwise compliment operator, ~. This operator toggles each bit within a value. Consider the following code.

unsigned char count, c;
c = 0x55;
if( ~c == 0xAA)
  count++;

If c contains the value 0x55, it is often assumed that ~c will produce 0xAA. However, the result is 0xFFAA. So, the comparison in the above example would fail. The compiler may be able to issue a mismatched comparison error to this effect in some circumstances. Again, a cast could be used to change this behavior.

The consequence of integral promotion as illustrated above is that operations are not performed with char-type operands, but with int-type operands. However there are circumstances when the result of an operation is identical regardless of whether the operands are of type char or int. In these cases, the compiler will not perform the integral promotion so as to increase the code efficiency. Consider the following example.

unsigned char a, b, c;

a = b + c;

Strictly speaking, this statement requires that the values of b and c should be promoted to unsigned int, the addition performed, the result of the addition cast to the type of a, and then the assignment can take place. Even if the result of the unsigned int addition of the promoted values of b and c was different to the result of the unsigned char addition of these values without promotion, after the unsigned int result was converted back to unsigned char, the final result would be the same. If an 8-bit addition is more efficient than a 16-bit addition, the compiler will encode the former.

If in the above example, the type of a was unsigned int, then integral promotion would have to be performed to comply with the ANSI C standard.