Similar to the question Bitshift and integer promotion?, I have a question about integer promotion when using left bitshifts.
unsigned int test(void)
{
unsigned char value8;
unsigned int result;
value8 = 0x12;
result = value8 << 8;
return result;
}
In this case, will be the value8 first promote to unsiged int or is it implementation specific?
6.5.7 Bitwise shift operators … 3 Sematics …
The integer promotions are performed on each of the operands. The type of the result is
that of the promoted left operand. If the value of the right operand is negative or is
greater than or equal to the width of the promoted left operand, the behavior is undefined.
It says that the “The integer promotions are performed on each of the operands.”, but what is here the promotion rule?
I assume that it should be convert to int if lesser rank than int, but I can’t find it.
I ask this, as one compiler (Renesas nc30wa) doesn’t promote to int, so the result is always 0 for my sample.
On this platform, a char is 8 bit wide and int 16 bits.
The phrase “the integer promotions” is a very specific thing, found in (for C99) section
6.3.1.1 Booleans, characters, and integers:So assuming your
unsigned charcan be held in anint, it will be promoted to anint. On those rare platforms whereunsigned charis as wide as anint, it will promote to anunsigned int.This is only changed slightly in C11:
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.
Looking at your particular environment, the
M16C Series,R8C Family C Compiler Package V.5.45 C Compilerhas, in section2.1.4 nc30 Command Line Options, subsectionf. Generated code modification options:although I suspect
-fansiis probably a better choice since it covers a few other things as well.