what's this?
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
fuzhen said:
Checking the subject line, I guess you are referring to this:what's this?
p=(char *)malloc((sizeo f(float))["\000\006\010\0 13\015\100"])
to which the answer is that it's a badly written call to malloc. But, given
suitable furniture (a function wrapped around it, <stdlib.h>, etc - and a
semicolon wouldn't go amiss at the end there), it's perfectly legal, provided
sizeof(float) doesn't exceed 6 on the target system. On systems where it
does, the behaviour is undefined.
Let's start off by assuming sizeof(float) is 3 - which is a perfectly legal
value for sizeof(float), and has the merit of being a little unlikely, to say
the least.
Thus, given that assumption, the code reduces to:
p=(char *)malloc((3)["\000\006\010\0 13\015\100"])
and (3) is just 3, so that gives us:
p=(char *)malloc(3["\000\006\010\0 13\015\100"])
Now, a[i] and *(a + i) are guaranteed to be equivalent, so let's do some
substituting:
p=(char *)malloc(*(3+"\ 000\006\010\013 \015\100"))
Okay, x+y is the same as y+x, so:
p=(char *)malloc(*("\00 0\006\010\013\0 15\100"+3))
and *(a + i) is the same as a[i], so:
p=(char *)malloc("\000\ 006\010\013\015 \100"[3])
Now, "\000\006\010\0 13\015\100" is an array with values { 0, 6, 8, 11, 13, 64,
0 }, so "\000\006\010\0 13\015\100"[3] is element 3 of that array: element 0
is 0, element 1 is 6, element 2 is 8, element 3 is 11. So we can substitute
that back in again:
p=(char *)malloc(11)
and finally we can lose the spurious and utterly pointless cast, which gives:
p = malloc(11)
Is that sufficiently simple for you? (Remember to replace 11 with a different
value from the array if sizeof(float) doesn't happen to be 3 on your system.)
--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
Does something like this have a legitimate use at all?Is that sufficiently simple for you? (Remember to replace 11 with a
different
value from the array if sizeof(float) doesn't happen to be 3 on your
system.)
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
MisterE said:
No - confusing the newbies is about its limit, really.>
>>>Is that sufficiently simple for you? (Remember to replace 11 with a
>different
>value from the array if sizeof(float) doesn't happen to be 3 on your
>system.)
Does something like this have a legitimate use at all?
--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
MisterE wrote:
/Something/ like this, yes, depending on how close a similarity>
>>>Is that sufficiently simple for you? (Remember to replace 11 with a
>different
>value from the array if sizeof(float) doesn't happen to be 3 on your
>system.)
Does something like this have a legitimate use at all?
you insist on. `malloc` is, after all, a useful function.
--
"Never ask that question!" /Babylon 5/
Hewlett-Packard Limited Cain Road, Bracknell, registered no:
registered office: Berks RG12 1HN 690597 England
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
"MisterE" <MisterE@nimga. comwrites:
This may save you a couple of bytes.>>Is that sufficiently simple for you? (Remember to replace 11 with a
>different
>value from the array if sizeof(float) doesn't happen to be 3 on your
>system.)
Does something like this have a legitimate use at all?
Making it more readable by placing the table as a static before the
call would probably not cost you any bytes, this would, however only
work if this is used where you can have declarations, otherwise eg. in
a macro, this may be a solution.
The numbers look odd though, appearently only size 1 to 4 produces
sensible values, what's the context here?
--
... __o Øyvind
... _`\(, http://www.darkside.no/olr/
... (_)/(_) ... biciclare necesse est ...
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
fuzhen wrote:Doodles from a sick programmer.what's this?
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
"fuzhen" <fucore@gmail.c omschreef in bericht
news:b4f91191-dafe-4a63-84a7-82687fcfacfd@w3 4g2000prm.googl egroups.com...in what context was this used?what's this?
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
Sjouke Burry wrote:Best answer I've seen.fuzhen wrote:Doodles from a sick programmer.>what's this?
--
Eric Sosman
esosman@ieee-dot-org.invalid
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
On Jun 11, 4:30 pm, Richard Heathfield <r...@see.sig.i nvalidwrote:This is the first time I am seeing array literals declared in thisfuzhen said:
>>what's this?
Checking the subject line, I guess you are referring to this:
>
p=(char *)malloc((sizeo f(float))["\000\006\010\0 13\015\100"])
>
to which the answer is that it's a badly written call to malloc. But, given
suitable furniture (a function wrapped around it, <stdlib.h>, etc - and a
semicolon wouldn't go amiss at the end there), it's perfectly legal, provided
sizeof(float) doesn't exceed 6 on the target system. On systems where it
does, the behaviour is undefined.
>
Let's start off by assuming sizeof(float) is 3 - which is a perfectly legal
value for sizeof(float), and has the merit of being a little unlikely, to say
the least.
>
Thus, given that assumption, the code reduces to:
>
p=(char *)malloc((3)["\000\006\010\0 13\015\100"])
>
and (3) is just 3, so that gives us:
>
p=(char *)malloc(3["\000\006\010\0 13\015\100"])
>
Now, a[i] and *(a + i) are guaranteed to be equivalent, so let's do some
substituting:
>
p=(char *)malloc(*(3+"\ 000\006\010\013 \015\100"))
>
Okay, x+y is the same as y+x, so:
>
p=(char *)malloc(*("\00 0\006\010\013\0 15\100"+3))
>
and *(a + i) is the same as a[i], so:
>
p=(char *)malloc("\000\ 006\010\013\015 \100"[3])
>
Now, "\000\006\010\0 13\015\100" is an array with values { 0, 6, 8, 11, 13, 64,
0 }, so "\000\006\010\0 13\015\100"[3] is element 3 of that array: element 0
is 0, element 1 is 6, element 2 is 8, element 3 is 11. So we can substitute
that back in again:
>
p=(char *)malloc(11)
>
and finally we can lose the spurious and utterly pointless cast, which gives:
>
p = malloc(11)
>
Is that sufficiently simple for you? (Remember to replace 11 with a different
value from the array if sizeof(float) doesn't happen to be 3 on your system.)
>
--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
manner. This is a new addition in C99, isn't it? And this cryptic
thing is supposed to pass array literals ( just to quote one of the
possible scenarios).
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
rahul wrote:<snip>Richard Heathfield <r...@see.sig.i nvalidwrote:Checking the subject line, I guess you are referring to this:
p=(char *)malloc((sizeo f(float))["\000\006\010\0 13\015\100"])No. Neither string literals nor a[i] == i[a] is new.>
This is the first time I am seeing array literals declared in this
manner. This is a new addition in C99, isn't it?
--
Peter
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Re: p=(char *)malloc((sizeo f(float))["\000\006\ 010\013\015\100 "])
On Jun 12, 10:07 am, Peter Nilsson <ai...@acay.com .auwrote:Thanks Peter,rahul wrote:>Richard Heathfield <r...@see.sig.i nvalidwrote:Checking the subject line, I guess you are referring to this:<snip>p=(char *)malloc((sizeo f(float))["\000\006\010\0 13\015\100"])
>>This is the first time I am seeing array literals declared in this
manner. This is a new addition in C99, isn't it?
No. Neither string literals nor a[i] == i[a] is new.
>
--
Peter
I missed that part in the FAQ. I got my answer.
char *a = "hello";
printf ("%c", 2[a]);
The above snippet directly does
printf ("%c", 2["hello"]);
a[i] == i[a]. Hey...I knew that
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