15 Aug
2024
15 Aug
'24
10:56 p.m.
When signed and unsigned values (variables or literals) of the same size
are combined the compiler assumes that all of the values are signed.
This can yield a problem if the unsigned integer is large enough that
the most significant bit is set because this bit indicates sign.
for example:
uint8_t bValue = 128;
int8_t bValue1 = -128
Both have the same value in memory (0x80).
On 8/15/2024 3:52 PM, Paul Koning via cctalk wrote:
On Aug 15, 2024, at 4:41 PM, Sean Conner via
cctalk <cctalk(a)classiccmp.org> wrote:
It was thus said that the Great ben via cctalk once stated:
Obviously, "correctly" is in the eye of the beholder. You can do
size extension, signed or unsigned, on any computer. How complicated it is depends on the
machine.
For example, on VAX there are instructions for signed as well as unsigned promotion
(CVTxy and MOVZxy respectively). On PDP11, MOVB into a register does sign extension;
unsigned promotion requires two instructions but that's no big deal. And of course,
promotion to bigger types requires multiple instructions either way since you're now
dealing with multiple registers.
Unsigned promotion on a CDC 6600 is one instruction; signed requires three.
paul I don't know about the VAX,but my gripe is
the x86 and the 68000 don't
automaticaly promote smaller data types to larger ones. What little
programming I have done was in C never cared about that detail.
Now I can see way it is hard to generate good code in C when all the
CPU's are brain dead in that aspect.
char *foo, long bar;
... foobar = *foo + bar
is r1 = foo
r3 = * r1
r2 = bar
sex byte r3
sex word r3
r4 = r3 + r2
foobar = r3
what I want is
bar = * foo + bar
nice easy coding.
What CPUs did it correctly? And how did they handle signed
vs. unsigned
promotion?
unsigned char *ufoo;
unsigned long ubar;
ufoobar = *ufoo + ubar;
signed char *foo;
signed long bar;
foobar = *foo + bar;
-spc