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p=(char *)malloc((sizeof(float))["\000\006\010\013\015\100"])

what's this?
Jun 27 '08 #1
11 2367
fuzhen said:
what's this?
Checking the subject line, I guess you are referring to this:

p=(char *)malloc((sizeof(float))["\000\006\010\013\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\013\015\100"])

and (3) is just 3, so that gives us:

p=(char *)malloc(3["\000\006\010\013\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(*("\000\006\010\013\015\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\013\015\100" is an array with values { 0, 6, 8, 11, 13, 64,
0 }, so "\000\006\010\013\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
Jun 27 '08 #2

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?
Jun 27 '08 #3
MisterE said:
>
>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?
No - confusing the newbies is about its limit, really.

--
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
Jun 27 '08 #4
MisterE wrote:
>
>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?
/Something/ like this, yes, depending on how close a similarity
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

Jun 27 '08 #5
"MisterE" <Mi*****@nimga.comwrites:
>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?
This may save you a couple of bytes.

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 Řyvind
... _`\(, http://www.darkside.no/olr/
... (_)/(_) ... biciclare necesse est ...
Jun 27 '08 #6
fuzhen wrote:
what's this?
Doodles from a sick programmer.
Jun 27 '08 #7

"fuzhen" <fu****@gmail.comschreef in bericht
news:b4**********************************@w34g2000 prm.googlegroups.com...
what's this?
in what context was this used?

Jun 27 '08 #8
Sjouke Burry wrote:
fuzhen wrote:
>what's this?
Doodles from a sick programmer.
Best answer I've seen.

--
Eric Sosman
es*****@ieee-dot-org.invalid
Jun 27 '08 #9
On Jun 11, 4:30 pm, Richard Heathfield <r...@see.sig.invalidwrote:
fuzhen said:
what's this?

Checking the subject line, I guess you are referring to this:

p=(char *)malloc((sizeof(float))["\000\006\010\013\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\013\015\100"])

and (3) is just 3, so that gives us:

p=(char *)malloc(3["\000\006\010\013\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(*("\000\006\010\013\015\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\013\015\100" is an array with values { 0, 6, 8, 11, 13, 64,
0 }, so "\000\006\010\013\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
This is the first time I am seeing array literals declared in this
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).
Jun 27 '08 #10
rahul wrote:
Richard Heathfield <r...@see.sig.invalidwrote:
Checking the subject line, I guess you are referring to this:

p=(char *)malloc((sizeof(float))["\000\006\010\013\015\100"])
<snip>
>
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
Jun 27 '08 #11
On Jun 12, 10:07 am, Peter Nilsson <ai...@acay.com.auwrote:
rahul wrote:
Richard Heathfield <r...@see.sig.invalidwrote:
Checking the subject line, I guess you are referring to this:
p=(char *)malloc((sizeof(float))["\000\006\010\013\015\100"])
<snip>
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
Thanks 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
Jun 27 '08 #12

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