| Re: - unsigned variables [message #41053 is a reply to message #41051] |
Thu, 23 September 2004 17:44  |
Longtime Lurker
Messages: 1
Registered: September 2004
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Junior Member |
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Holger Fleckenstein wrote:
> A strange behavior in IDL occured to me.
>
> In C++ if I do:
> unsigned short x=1;
> printf("%d",-x);
> I get
> -1
> like I would expect.
Whereas:
printf("%u",-x);
4294967295
printf("%hu", -x);
65535
like I would expect
%d simply tells C to output the bit pattern stored in the argument(s) as if
it were a *signed* integer
Similarly
printf("%f", -x);
gives
2.102785
on my little-endian machine. The bit pattern is identical in all four cases
all that changes is the way it is interpreted.
>
> In IDL however:
> x=1U
> print, -x
> gives
> 65535
> So it basically treats it like I had done:
> print, uint(-1)
But -x is a UINT just like x
help, -x
<Expression> UINT = 65535
So IDL is doing exactly what you ask it to
Consider
print, -x, format = '(F)'
65535.0000000000000000
>
> Does anybody have an explanation for this?
> Is this, because of a typecast before executing the print?
> (Can creat bugs, which are hard to localize.)
IDL outputs the true value of -x taking its type into account - any
formatting is applied to this value. C[++] outputs the value of the bits
interpreted according to the rules of the supplied conversion specifier.
The C behaviour has caught me out with code like
long long x = 1, y = 1;
printf("%d %d\n", x, y);
1 0
When I should have had
printf("%lld %lld\n", x, y);
1 1
IDL's behaviour seems preferable to me...
Paul
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