Hi, guys,I met a problem, Please look at the problem below:
int* bit = (int*)malloc(10 000*sizeof(int) );
memset(bit, 1, 10000*sizeof(in t));
printf("%d %d %d\n", bit[1],bit[2], bit[9999]);
Output: 16843009 16843009 16843009
Obviously I set the bit[0] to bit[9999] to 1, but it outputs are not
1's.
Why is this happen? Do I misused memset or sth else?
Any help is appreciated. Thanks!
Dec 4 '05
17  6361 
Alex Fraser wrote: AFAIK you can't portably set them to zero either, except possibly the
fixed-size ones added in C99 (uint32_t etc).
For character sets, a byte with all bits zero,
is a valid null character.
--
pete
On 4 Dec 2005 04:57:16 -0800, in comp.lang.c , "Frederick Ding"
<di*******@gmai l.com> wrote: Thanks, guys!
I'm very happy that so many kind people help me.
I've learned a lot from you, THANKS!
Yet I still have two questions:
1. malloc(10000 * sizeof *bit);
^^^^^^^^^^^^
sizeof *bit what's this, can sizeof be used like this?
sizeof is an operator, not a function. Its operand must be either a
type or an object. If the operand is an object, it doesn't need the
parens.
where can I find some reference to this?
The ISO C standard
6.5.3.4 The sizeof operator
Constraints
1 The sizeof operator shall not be applied to an expression that has
function type or an incomplete type, to the parenthesized name of such
a type, or to an expression that designates a bit-field member.
Semantics
2 The sizeof operator yields the size (in bytes) of its operand, which
may be an expression or the parenthesized name of a type.
I think sizeof is a function that calculate a type's size, so it
should be used with ( ),
No, its an operator
and the * between "sizeof" and "bit" is a pointer or a multiply?
No, its the indirection operator, same as in
int *bit;
2. I once used memset like this, and it looked like works:
memset(semi,-1,sizeof(float) *16);
Output: -1.#QNAN0 -1.#QNAN0
It seemed that it was "-1" and I omitted the topic I just asked.
Now the question is "-1.#QNAN0" means what?
it means your number is Not A Number. Clearly setting all bytes of a
float to -1 on your system causes the float to be a non-number.
--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt >
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Frederick Ding a écrit : Thanks, guys!
I'm very happy that so many kind people help me.
I've learned a lot from you, THANKS!
Yet I still have two questions:
1. malloc(10000 * sizeof *bit);
^^^^^^^^^^^^
sizeof *bit what's this, can sizeof be used like this? wow,
I can not find this expression in all of my C books.
It's a short way for sizeof p[0] that can be written *(p + 0), hence
*(p) hence finally *p.
It means 'the size of the first element of the pointed array'.
where can I find some reference to this?
Pointer arithmetics.
I think sizeof is a function that calculate a type's size, so it
Nope. sizeof is a unary operator. It works like this :
size_t size = sizeof object
size_t size = sizeof (type)
should be used with ( ),
It depends on the context. Note that an excess of parenthesis is
harmless (kinda belt and suspensers strategy !).
and the * between "sizeof" and "bit" is a pointer or a multiply?
None of them. It's the indirection operator (or dereference operator).
2. I once used memset like this, and it looked like works:
memset(semi,-1,sizeof(float) *16);
Don't use memset() on floating points. The result is implementation
dependent or more likely undefined.
memset(final,-1,sizeof(float) *16);
memset(cup,-1,sizeof(float) *16);
printf("%f %f\n",semi[0],final[15]);
Output: -1.#QNAN0 -1.#QNAN0
It seemed that it was "-1" and I omitted the topic I just asked.
Now the question is "-1.#QNAN0" means what?
what is the #QNAN0?
NAN means Not A Number. IOW, the internal format is undefined. Once
again, don't to that. What did you intent to do ?
--
A+
Emmanuel Delahaye
On Sun, 04 Dec 2005 14:51:21 +0100, in comp.lang.c , Emmanuel Delahaye
<em***@YOURBRAn oos.fr> wrote: Frederick Ding a écrit : memset(final,-1,sizeof(float) *16);
Output: -1.#QNAN0 -1.#QNAN0
NAN means Not A Number. IOW, the internal format is undefined.
See for example:
http://stevehollasch.com/cgindex/coding/ieeefloat.html
if you examine the bitpattern your memset created, and compare it to
the table near the bottom, all will be revealed.
Once again, don't to that. What did you intent to do ?
I imagine he wante to set all his floats to value -1. To the OP:
memset does /exactly/ what it says on the tin - sets n bytes of memory
to value m. It does NOT set the value of the variable to n.
--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt >
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"Malcolm" <re*******@btin ternet.com> wrote in message
news:dm******** *@nwrdmz02.dmz. ncs.ea.ibs-infra.bt.com... "Alex Fraser" <me@privacy.net > wrote It sets every byte to a given value. The only practical use, really,
is to set memory to zero, unless, as you say, you happen to want a
string of asterisks or something.
You cannot set integers to a non-zero value.
AFAIK you can't portably set them to zero either, except possibly the
fixed-size ones added in C99 (uint32_t etc).
int array[100];
memset(array, 0, 100 * sizeof(int));
is guaranteed to produce an array to 100 integers of value 0, on an ANSI
system.
"All bits zero" implies "all padding bits zero" and I don't see why that
couldn't be a trap representation for int.
Alex
"pete" <pf*****@mindsp ring.com> wrote in message
news:43******** ***@mindspring. com... Alex Fraser wrote: AFAIK you can't portably set [integers] to zero [with memset] either,
except possibly the fixed-size ones added in C99 (uint32_t etc).
For character sets, a byte with all bits zero,
is a valid null character.
Noted.
Alex
"Alex Fraser" <me@privacy.net > writes: "Malcolm" <re*******@btin ternet.com> wrote in message
news:dm******** *@nwrdmz02.dmz. ncs.ea.ibs-infra.bt.com... "Alex Fraser" <me@privacy.net > wrote >> It sets every byte to a given value. The only practical use, really,
>> is to set memory to zero, unless, as you say, you happen to want a
>> string of asterisks or something.
>> You cannot set integers to a non-zero value.
>
> AFAIK you can't portably set them to zero either, except possibly the
> fixed-size ones added in C99 (uint32_t etc).
int array[100];
memset(array, 0, 100 * sizeof(int));
is guaranteed to produce an array to 100 integers of value 0, on an ANSI
system.
"All bits zero" implies "all padding bits zero" and I don't see why that
couldn't be a trap representation for int.
The response to Defect Report #263 says:
For any integer type, the object representation where all the bits
are zero shall be a representation of the value zero in that type.
This is published in Technical Corrigendum 2, and appears in n1124
(which includes the C99 standard with TC1 and TC2 merged in).
--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
"Keith Thompson" <ks***@mib.or g> wrote in message
news:ln******** ****@nuthaus.mi b.org... "Alex Fraser" <me@privacy.net > writes: "Malcolm" <re*******@btin ternet.com> wrote in message
news:dm******** *@nwrdmz02.dmz. ncs.ea.ibs-infra.bt.com...
[snip] memset(array, 0, 100 * sizeof(int));
is guaranteed to produce an array to 100 integers of value 0, on an
ANSI system.
"All bits zero" implies "all padding bits zero" and I don't see why
that couldn't be a trap representation for int.
The response to Defect Report #263 says:
For any integer type, the object representation where all the bits
are zero shall be a representation of the value zero in that type.
This is published in Technical Corrigendum 2, and appears in n1124
(which includes the C99 standard with TC1 and TC2 merged in).
Thank you.
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