Re: [algogeeks] Re: C Doubts
@abhi we can do it without passing the address to a function. just store the address of const variable in another variable and change the value... On Mon, Jul 18, 2011 at 11:56 AM, Abhi abhi123khat...@gmail.com wrote: Perhaps the only way to alter the value of a variable declared constant in C is through passing its address to a function and changing the value henceforth though it generates a warning of 'discarding the qualifiers'. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/SbHlNcA7Fj8J. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: C Doubts
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of sizeof
(struct)
can any one explain me why the padding concept does not work here ??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.com wrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable) and
cannot be changed in any of the following assignment statements to the const
variable.
2.
Here for the structure struct s1 since the entire structure is ending
within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable aligned in
even boundaries and so the structure after aligning will look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
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Re: [algogeeks] Re: C Doubts
sizeof long double is 12. So padding concept is perfectly working
On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar aditya.kumar130...@gmail.com
wrote:
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of sizeof
(struct)
can any one explain me why the padding concept does not work here ??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.com wrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable) and
cannot be changed in any of the following assignment statements to the const
variable.
2.
Here for the structure struct s1 since the entire structure is ending
within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable aligned in
even boundaries and so the structure after aligning will look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
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COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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Re: [algogeeks] Re: C Doubts
@prateek . can you explain me ?? i dint get padding logic in this example of
mine.
On Mon, Jul 18, 2011 at 12:30 AM, sagar pareek sagarpar...@gmail.comwrote:
sizeof long double is 12. So padding concept is perfectly working
On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of sizeof
(struct)
can any one explain me why the padding concept does not work here ??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.com wrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable) and
cannot be changed in any of the following assignment statements to the const
variable.
2.
Here for the structure struct s1 since the entire structure is ending
within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable aligned in
even boundaries and so the structure after aligning will look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
--
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COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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Re: [algogeeks] Re: C Doubts
@Aditya
Here is the padding effect :
Address of char : starts anywhere
Address of long double : starts at 11 address locations from char variable
-- 1+11+12=24 bytes
On Mon, Jul 18, 2011 at 1:10 AM, sagar pareek sagarpar...@gmail.com wrote:
@aditya
actually first see your post, you have written o/p=24 accordingly padding
done is perfect. But actually its printing 16.
So now question arises of padding
and its pareek not prateek :)
On Mon, Jul 18, 2011 at 12:38 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
@prateek . can you explain me ?? i dint get padding logic in this example
of mine.
On Mon, Jul 18, 2011 at 12:30 AM, sagar pareek sagarpar...@gmail.comwrote:
sizeof long double is 12. So padding concept is perfectly working
On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of sizeof
(struct)
can any one explain me why the padding concept does not work here ??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.com wrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable) and
cannot be changed in any of the following assignment statements to the
const
variable.
2.
Here for the structure struct s1 since the entire structure is ending
within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable aligned
in even boundaries and so the structure after aligning will look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
--
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COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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NIT ALLAHABAD
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Senior Co-ordinator, Publicity
CSI, NSIT Students' Branch
NSIT, New Delhi, India
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Re: [algogeeks] Re: C Doubts
@pareek..my compiler gives 24 . newazz if ansa is 16 acc to you then it
follows padding principle perfectly. since memory cycle invloves 1 word
hence char will take 1 byte nd 3 bytes will be padded up . rest 12 bytes
will come from long double so 4+12=16 bytes :)
n ya sry abt d name .
On Mon, Jul 18, 2011 at 1:10 AM, sagar pareek sagarpar...@gmail.com wrote:
@aditya
actually first see your post, you have written o/p=24 accordingly padding
done is perfect. But actually its printing 16.
So now question arises of padding
and its pareek not prateek :)
On Mon, Jul 18, 2011 at 12:38 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
@prateek . can you explain me ?? i dint get padding logic in this example
of mine.
On Mon, Jul 18, 2011 at 12:30 AM, sagar pareek sagarpar...@gmail.comwrote:
sizeof long double is 12. So padding concept is perfectly working
On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of sizeof
(struct)
can any one explain me why the padding concept does not work here ??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.com wrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable) and
cannot be changed in any of the following assignment statements to the
const
variable.
2.
Here for the structure struct s1 since the entire structure is ending
within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable aligned
in even boundaries and so the structure after aligning will look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
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NIT ALLAHABAD
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Re: [algogeeks] Re: C Doubts
@Sagar
Memory sizes of long double variables are compiler and system configuration
dependent. So obviously, in accordance with your compiler, the size of long
double is 8 bytes.
On Mon, Jul 18, 2011 at 1:22 AM, Nikhil Gupta nikhilgupta2...@gmail.comwrote:
@Aditya
Here is the padding effect :
Address of char : starts anywhere
Address of long double : starts at 11 address locations from char variable
-- 1+11+12=24 bytes
On Mon, Jul 18, 2011 at 1:10 AM, sagar pareek sagarpar...@gmail.comwrote:
@aditya
actually first see your post, you have written o/p=24 accordingly padding
done is perfect. But actually its printing 16.
So now question arises of padding
and its pareek not prateek :)
On Mon, Jul 18, 2011 at 12:38 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
@prateek . can you explain me ?? i dint get padding logic in this example
of mine.
On Mon, Jul 18, 2011 at 12:30 AM, sagar pareek sagarpar...@gmail.comwrote:
sizeof long double is 12. So padding concept is perfectly working
On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of
sizeof (struct)
can any one explain me why the padding concept does not work here ??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.com wrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable) and
cannot be changed in any of the following assignment statements to the
const
variable.
2.
Here for the structure struct s1 since the entire structure is ending
within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable aligned
in even boundaries and so the structure after aligning will look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
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COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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CSI, NSIT Students' Branch
NSIT, New Delhi, India
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CSI, NSIT Students' Branch
NSIT, New Delhi, India
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Re: [algogeeks] Re: C Doubts
@Nikhil
why is Address of long double : starts at 11 address locations from char
variable ??
is shud start from 3rd adress location from char variable bcoz memory cycle
involves a word so are you padding 11bytes ??
On Mon, Jul 18, 2011 at 1:24 AM, Nikhil Gupta nikhilgupta2...@gmail.comwrote:
@Sagar
Memory sizes of long double variables are compiler and system configuration
dependent. So obviously, in accordance with your compiler, the size of long
double is 8 bytes.
On Mon, Jul 18, 2011 at 1:22 AM, Nikhil Gupta
nikhilgupta2...@gmail.comwrote:
@Aditya
Here is the padding effect :
Address of char : starts anywhere
Address of long double : starts at 11 address locations from char variable
-- 1+11+12=24 bytes
On Mon, Jul 18, 2011 at 1:10 AM, sagar pareek sagarpar...@gmail.comwrote:
@aditya
actually first see your post, you have written o/p=24 accordingly padding
done is perfect. But actually its printing 16.
So now question arises of padding
and its pareek not prateek :)
On Mon, Jul 18, 2011 at 12:38 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
@prateek . can you explain me ?? i dint get padding logic in this
example of mine.
On Mon, Jul 18, 2011 at 12:30 AM, sagar pareek
sagarpar...@gmail.comwrote:
sizeof long double is 12. So padding concept is perfectly working
On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of
sizeof (struct)
can any one explain me why the padding concept does not work here ??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.comwrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable)
and
cannot be changed in any of the following assignment statements to the
const
variable.
2.
Here for the structure struct s1 since the entire structure is ending
within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable aligned
in even boundaries and so the structure after aligning will look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
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SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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Nikhil Gupta
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CSI, NSIT Students' Branch
NSIT, New Delhi, India
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Nikhil Gupta
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NSIT, New Delhi, India
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You received this message because
Re: [algogeeks] Re: C Doubts
That is again compiler dependent. Usually when hardware configuration is
taken into account, the compiler uses padding of 3 bytes. But in some cases,
for the ease of hardware access and faster implementation, 11 bytes are
padded. Possibly depends on your system hardware's synchronization with the
compiler.
On Mon, Jul 18, 2011 at 1:27 AM, aditya kumar
aditya.kumar130...@gmail.comwrote:
@Nikhil
why is Address of long double : starts at 11 address locations from char
variable ??
is shud start from 3rd adress location from char variable bcoz memory cycle
involves a word so are you padding 11bytes ??
On Mon, Jul 18, 2011 at 1:24 AM, Nikhil Gupta
nikhilgupta2...@gmail.comwrote:
@Sagar
Memory sizes of long double variables are compiler and system
configuration dependent. So obviously, in accordance with your compiler, the
size of long double is 8 bytes.
On Mon, Jul 18, 2011 at 1:22 AM, Nikhil Gupta
nikhilgupta2...@gmail.comwrote:
@Aditya
Here is the padding effect :
Address of char : starts anywhere
Address of long double : starts at 11 address locations from char
variable -- 1+11+12=24 bytes
On Mon, Jul 18, 2011 at 1:10 AM, sagar pareek sagarpar...@gmail.comwrote:
@aditya
actually first see your post, you have written o/p=24 accordingly
padding done is perfect. But actually its printing 16.
So now question arises of padding
and its pareek not prateek :)
On Mon, Jul 18, 2011 at 12:38 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
@prateek . can you explain me ?? i dint get padding logic in this
example of mine.
On Mon, Jul 18, 2011 at 12:30 AM, sagar pareek
sagarpar...@gmail.comwrote:
sizeof long double is 12. So padding concept is perfectly working
On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of
sizeof (struct)
can any one explain me why the padding concept does not work here ??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.comwrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable)
and
cannot be changed in any of the following assignment statements to the
const
variable.
2.
Here for the structure struct s1 since the entire structure is
ending within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable
aligned in even boundaries and so the structure after aligning will
look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
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**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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To unsubscribe from this
Re: [algogeeks] Re: C Doubts
@nikhil my compiler gives sizeof long double =12
so aditya's concept is correct
On Mon, Jul 18, 2011 at 1:35 AM, aditya kumar
aditya.kumar130...@gmail.comwrote:
@Nikhil
thnks :)
On Mon, Jul 18, 2011 at 1:32 AM, Nikhil Gupta
nikhilgupta2...@gmail.comwrote:
That is again compiler dependent. Usually when hardware configuration is
taken into account, the compiler uses padding of 3 bytes. But in some cases,
for the ease of hardware access and faster implementation, 11 bytes are
padded. Possibly depends on your system hardware's synchronization with the
compiler.
On Mon, Jul 18, 2011 at 1:27 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
@Nikhil
why is Address of long double : starts at 11 address locations from char
variable ??
is shud start from 3rd adress location from char variable bcoz memory
cycle involves a word so are you padding 11bytes ??
On Mon, Jul 18, 2011 at 1:24 AM, Nikhil Gupta nikhilgupta2...@gmail.com
wrote:
@Sagar
Memory sizes of long double variables are compiler and system
configuration dependent. So obviously, in accordance with your compiler,
the
size of long double is 8 bytes.
On Mon, Jul 18, 2011 at 1:22 AM, Nikhil Gupta
nikhilgupta2...@gmail.com wrote:
@Aditya
Here is the padding effect :
Address of char : starts anywhere
Address of long double : starts at 11 address locations from char
variable -- 1+11+12=24 bytes
On Mon, Jul 18, 2011 at 1:10 AM, sagar pareek
sagarpar...@gmail.comwrote:
@aditya
actually first see your post, you have written o/p=24 accordingly
padding done is perfect. But actually its printing 16.
So now question arises of padding
and its pareek not prateek :)
On Mon, Jul 18, 2011 at 12:38 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
@prateek . can you explain me ?? i dint get padding logic in this
example of mine.
On Mon, Jul 18, 2011 at 12:30 AM, sagar pareek
sagarpar...@gmail.com wrote:
sizeof long double is 12. So padding concept is perfectly working
On Mon, Jul 18, 2011 at 12:26 AM, aditya kumar
aditya.kumar130...@gmail.com wrote:
struct st
{
char ch1;
long double ld;
}s;
printf(%d,sizeof(s));
//output : 24 (for 32-bit compiler)
-as i have mentioned above the behaviour is undefined in case of
sizeof (struct)
can any one explain me why the padding concept does not work here
??
On Mon, Jul 18, 2011 at 12:13 AM, Parthiban jega...@gmail.comwrote:
@Abhi:
Answers:
1. whenever a 'const' qualifier is added previously to a variable
declaration it means that the value of the variable is automatically
initialized to '0'(because of the 'auto' type of the const variable)
and
cannot be changed in any of the following assignment statements to
the const
variable.
2.
Here for the structure struct s1 since the entire structure is
ending within 8 bytes no padding is done which means
s1: [char a]
1byte
but for the structure struct s2 consider the following:
s2: [ char a
1byte
-- int a---
--4bytes]
so here the concept of padding comes to make all the variable
aligned in even boundaries and so the structure after aligning will
look as:
s2: [ -- char b
--4byte-
-- int a---
--4bytes]
so the size of strcut s2 will be 8bytes..
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**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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