Integer promotion with the operator <<

Question

Similar to the question Bitshift and integer promotion?, I have a question about integer promotion when using left bitshifts.

unsigned int test(void)
{
  u         


        

Answer 1

The phrase "the integer promotions" is a very specific thing, found in (for C99) section 6.3.1.1 Booleans, characters, and integers:

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanged by the integer promotions.

So assuming your unsigned char can be held in an int, it will be promoted to an int. On those rare platforms where unsigned char is as wide as an int, it will promote to an unsigned int.

This is only changed slightly in C11:

If an int can 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. All other types are unchanged by the integer promotions.

If a specific compiler doesn't follow this behaviour, then it's not really conforming. However, given that the compiler you listed is for embedded systems, it's not really surprising.

Many are built for specific purposes and conformance is not always high on the list of requirements. There may be compiler flags that will allow it to more closely conform to the standard.

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Looking at your particular environment, the M16C Series,R8C Family C Compiler Package V.5.45 C Compiler has, in section 2.1.4 nc30 Command Line Options, subsection f. Generated code modification options:

-fextend_to_int, -fETI: Performs operation after extending char-type data to the int type. Extended according to ANSI standards.

although I suspect -fansi is probably a better choice since it covers a few other things as well.