integral type conversions

integral type conversions

[Perry Smith]

I can't understand the C++ spec.  It seems more vague than the old K&R.

If the source is a signed integral type and the destination is an  
unsigned integral type that is larger, are the additional bits done  
via sign extension or zero extension?

e.g.

int i;

unsigned long l = i;

Is i sign extended from int to long and then converted to unsigned  
long or is it converted to unsigned int and then zero extended to  
unsigned long?

(assume that long is bigger in this case).  I'm actually going from a  
long to an unsigned long long but I assume the same algorithm is used.

I guess, to be safe, I could do:

unsigned long l = static_cast<unsigned int>(i);

Thank you,
Perry

Re: integral type conversions

[John Love-Jensen]

Hi Perry,

> If the source is a signed integral type and the destination is an
> unsigned integral type that is larger, are the additional bits done
> via sign extension or zero extension?

On 2's complement machines (the only sort I've ever used), the additional
bits are done via sign extension.

This behavior should be identical to the behavior in C.

Side note: I use the Java convention of converting a char to an int:

char c = '\xAA';
int i;
i = c & 0xFF;

Works the reliably on plain-char-is-signed and plain-char-is-unsigned
machines.  Presuming the architecture is 8-bit-bytes (such as all my
platforms are).  And I'm not sure if there's caveats for 1's complement
machines to worry about.

Sincerely,
--Eljay