integer promotion in c

Question

Let say I have a 32-bit machine.

I know during integer promotion the expressions are converted to:\\

• int if all

Answer 1

The integer promotion rules are defined in 6.3.1.8 Usual arithmetic conversions.

1.  si16  x, pt;
    si32  speed;
    u16 length;
    x = (speed*pt)/length;

2. x =  pt + length;

Ranking means effectively the number of bits from the type as defined by CAM in limits.h. The standards imposes for the types of lower rank in CAM to correspond types of lower rank in implementation.

For your code,

speed*pt

is multiplication between int32 and int 16, which means, it is transformed in

speed*(int16=>int32)pt

and the result tmp1 will be int32.

Next, it will continue

tmp1_int32 / length

Length will be converted from uint16 to int32, so it will compute tmp2 so:

tmp1_int32 / (uint16=>int32) length

and the result tmp2 will be of type int32.

Next it will evaluate an assignment expression, left side of 16 bits and the right side of 32, so it will cut the result so:

x = (int32=>int16) tmp2_int32

Your second case will be evaluated as

x = (int32=>int16) ( (int16=>int32) pt + (uint16=>int32) length )

In case an operator has both operands with rank smaller than the rank of int, the CAM allows to add both types if the operation does not overflow and then to convert the result to integer.

In other words, it is possible to covert INT16+INT16 either in

 INT16+INT16

or in

 (int32=>int16) ((int16=>int32) INT16 + (int16=>int32)INT16)

provided the addition can be done without overflow.