#includestdio.h
int main()
{
char str[]={'a','b','c'};
char str1[]={abc};
printf(%d,sizeof(str));
printf(%d,sizeof(str1));
getchar();
}
This is giving 3 in case of str and 4 in case of str1 bcz str is array of
character and str1 is a string.
For understanding this point
@rahul According to C specification, half filled array will be filled with
value 0. In your example you are setting str[0] as 'g' and str[1] as 'k'.
So the compiler sets str[29] as 0. So you string str becomes
{'g', 'k', '\0', '\0', '\0', '\0', '\0', '\0', '\0', '\0'}
Confusion is arising
@sachin..thnx for explanation..got it..plz tell
#includestdio.h
int main()
{
char str[]={'a','b','c'};
char str1[]={abc};
printf(%d,sizeof(str));
printf(%d,sizeof(str1));
getchar();
}
why str has size 3 and str1 has 4...NUll should also come after c of
str???then y
char str[]=ab;
char str1[]={'a','b'};
sizeof(str) ...o/p is 3
sizeof(str1)o/p is 2..
Why so
plz explain...
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because of null char in 1st
On Sat, Oct 6, 2012 at 5:53 PM, rahul sharma rahul23111...@gmail.comwrote:
char str[]=ab;
char str1[]={'a','b'};
sizeof(str) ...o/p is 3
sizeof(str1)o/p is 2..
Why so
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For string, C appends '\0' internally. hence sizeof(str) returned the value
3.
str1 is char array with two character. hence sizeof(str1) returned the
value 2.
-- RK :)
On Sat, Oct 6, 2012 at 5:53 PM, rahul sharma rahul23111...@gmail.comwrote:
char str[]=ab;
char str1[]={'a','b'};
int main()
{
char str[10]={'g','k'};
char str1[10]=gh;
printf(%s,str);
printf(%s,str1);
getchar();
}
then how does this work???
str printing gk...then NULL is automatically appended in this also...plz
tell
On Sat, Oct 6, 2012 at 6:33 PM, Rathish Kannan
#includestdio.h
int main()
{
char str[10]={'g','k'};
char str1[10]=gh;
int i;
for(i=0;str1[i]!=NULL;i++)
printf(%c,str[i]);
getchar();
}
NUll is there in character array also...make clear me...
On Sat, Oct 6, 2012 at 9:22 PM, rahul sharma rahul23111...@gmail.comwrote:
int
violation of sequence point ruleoutput depends on compiler to compiler.
On Tue, Jul 3, 2012 at 1:22 PM, rahul sharma rahul23111...@gmail.comwrote:
#includestdio.h
#includeconio.h
int main()
{
int i;
i=5;
i=++i/i++;
printf(%d,i);
getch();
}
Why o/p is 1 and not
#includestdio.h
int main()
{
double p=fabs(24.9996);
double t=fabs(25.0003);
printf(%0.3lf %0.3lf\n,p,t);
if(fabs(p)==fabs(t))
printf(equal);// why this not executed?
}
why the equal not printed in this code...?
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printf(%0.3lf %0.3lf\n,p,t);
its just printing at your convenience . you are not changing the value of
p,t .
change this statement to
printf(%0.5lf %0.5lf\n,p,t);
On Tue, Jan 17, 2012 at 11:27 AM, dabbcomputers dabbcomput...@gmail.comwrote:
#includestdio.h
int main()
{
double
atul he is assigning the value later on.
i think format specifier . rounds up the number in last decimal place.
On Tue, Jan 17, 2012 at 11:53 AM, atul anand atul.87fri...@gmail.comwrote:
printf(%0.3lf %0.3lf\n,p,t);
its just printing at your convenience . you are not changing the value of
It's not a good way to compare two floats directly using =, try something
like below:
http://ideone.com/c65Vl
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i gues atul is crkt.
its jst printin by user convenience...when actually d two are not equal...
i guess the code...xplains well...dats y equal is not printed..out
#includestdio.hint main(){
double p=fabs(24.9996);
printf(%lf\n,p);
double t=fabs(25.0003);
printf(%lf\n,t);
@amol this is not the behaviour of printf, its totally about the typecasting
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It's painful to see printf being accused for the (un) expected
output..The declaration of printf means that any data type can be
passed to it as argument.Inherently what printf does is print the bytes in
meaningful form according to the first argument which is a string.So its
impossible for
According to KR the behavior of printf is undefined if u do not use
correct format specifiers for the variables.
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@amol I think it is not the behaviour of printf
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*#includestdio.h
int main()
{
float f=25.25;
printf(%d\n,f);
long int x=90;
printf(%f,x);
return 0;
}
*
above program gives output
*0
25.25*
shudn it be :
*25
90.*
??
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take care when ever you use %d and %f
%d is not flexible in handling float varibaledo not expect it to
typecast itself.
reason could be , you are trying to fit large data type i.e float to the
int whose range is less.
whereas in second case float can incorporate int type , so automatic
@arun: http://mindprod.com/jgloss/immutable.html
this might help you, in essense, if a compiler treats them as immutable,
the reason is to reduce the overhead of creating another contant literal
(as explain at the link, the string literals are the most commonly used)
this is from a java (or
depends on compiler i think..but most probably it compares the
addresses.
On Wed, Jan 4, 2012 at 12:20 PM, saurabh singh saurab...@gmail.com wrote:
@all.Your explanations work because probably all of you are using a
compiler that's behaving in the same way.Don't conclude from what
actually is there any reason as to why same address is returned to the
pointer when both pointers(p and q) are initialised to persons unlike
when p[] and q[] =persons?
On Tue, Sep 6, 2011 at 9:08 AM, Sandy sandy.wad...@gmail.com wrote:
String constants (literals) are saved into the .data
it's near to a common mis conception that string liberals are in data
sections of THE PROGRAM
PLEASE READ THE FILE
a.out.h
and find the difference between initialized data and non initialized data
On 9/6/11, Sandy sandy.wad...@gmail.com wrote:
String constants (literals) are saved into the
@all.Your explanations work because probably all of you are using a
compiler that's behaving in the same way.Don't conclude from what you
see...The compiler is free to store the constant strings the way it
wants.
Saurabh Singh
B.Tech (Computer Science)
MNNIT
does y goes to d of %*d and it print 5???i hav a doubt
On Fri, Oct 28, 2011 at 9:32 PM, amrit harry dabbcomput...@gmail.comwrote:
let this statement int x=100,y=5;printf(%*d,x,y);
in this line first x is assign to '*' and it become %100d
and it will padd 100 spaces before print. and if we
yes 5 will be printed with 99 extra padding spaces.
On Sat, Oct 29, 2011 at 4:16 PM, rahul sharma rahul23111...@gmail.comwrote:
does y goes to d of %*d and it print 5???i hav a doubt
On Fri, Oct 28, 2011 at 9:32 PM, amrit harry dabbcomput...@gmail.comwrote:
let this statement int
can u plz tell me what exactly %*d means?
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let this statement int x=100,y=5;printf(%*d,x,y);
in this line first x is assign to '*' and it become %100d
and it will padd 100 spaces before print. and if we use( %*d,x)
then x is assign to '*' and garbage value wud be printed.
On Fri, Oct 28, 2011 at 8:53 PM, annarao kataru
*explain the o/p...if i/p are 100 200 300
int main()
{
int a=1,b=2,c=3;
scanf(%d %*d %d,a,b,c);
printf(%d %d %d,a,b,c);
return(0);
}
*Thanking you
*With regards-
Raghav garg
Contact no. 9013201944
as expected value 100 goes to a,since %*d is variable field width specifier
so the input 200 goes for that,and the remaining input 300 goes to b
value of c is not change
so the output will be:
100 300 3
On Sun, Oct 9, 2011 at 12:22 AM, Raghav Garg rock.ragha...@gmail.comwrote:
*explain the
This is what you use if you want *scanf()* to eat some data but you don't
want to store it anywhere; you don't give *scanf()* an argument for this
conversion
On Sun, Oct 9, 2011 at 1:32 AM, shiva@Algo shiv.jays...@gmail.com wrote:
as expected value 100 goes to a,since %*d is variable field
number will be sored in 2'scomplement form .as 4=100 so its 1's comlement
form will be 011 and adding 1 to it will result in 011+1=100.so -4 will be
printed
On Wed, Sep 21, 2011 at 10:23 AM, kartik sachan kartik.sac...@gmail.comwrote:
@ravi i think ur concepts is correct the no is stored in
int main(){
struct {
int a:1;
int b:3;
}obj;
obj.b=12;
obj.a=7;
printf
http://www.opengroup.org/onlinepubs/009695399/functions/printf.html(%d
%d,obj.b,obj.a);return 0;}
can anybody explain the output
plzz also show
output is -4 -1
int a:1 signifies that only 1 bit will be stored for a and since we have
only 1 bit so it will serve as sign bit and we get obj.a as -1
similarly for int b:3, 3 bits will be used to store b. 12 is 1100. we save
only 3 bits i.e. 100 where msb signifies the sign i.e. no. is negative.
@ravi i think ur concepts is correct the no is stored in 2's formif
negative
thanks ravi
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on running,every time i get second a=30... any reasons for that???
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common guys its undefined acc to standard c
On Mon, Sep 19, 2011 at 12:36 PM, Siddhartha Banerjee
thefourrup...@gmail.com wrote:
on running,every time i get second a=30... any reasons for that???
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#includestdio.h
main()
{
int a ;
a=abc();
printf(\n %d this is fisrt a:,a);
a=abc();
printf(\n %d this is second a:,a);
}
int abc()
{
int i=23;
if(10)
return ;
i++;
return(++ i);
}
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#includestdio.h
main()
{
int a ;
a=abc();
printf(\n %d this is fisrt a:,a);
a=abc();
printf(\n %d this is second a:,a);
}
int abc()
{
int i=23;
if(10)
return ;
i++;
return(++ i);
}
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result is not defined by standard c. It doesnt produce compiler error just
because return type is int and no value is returned in if(10)
basically garbage
On Sun, Sep 18, 2011 at 12:23 PM, Bhavesh agrawal agr.bhav...@gmail.comwrote:
#includestdio.h
main()
{
int a ;
a=abc();
it is just printing some garbage value as u have specified return but not
the value.try putting some value with return then you'll understand
what's happening..
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MNNIT Allahabad
http://gplus.to/amolsharma99
@what's special in this ...
it is giving 0 in both cases.
basically a garbage value
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result is this
-1075990972 this is fisrt a:
30 this is second a
what is the meaning of a=30 in second call
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Error.
the 10 condition does not return anything, while the function should return
an int.
On Sun, Sep 18, 2011 at 6:39 AM, hurtlocker bhavesh24...@rediffmail.comwrote:
#includestdio.h
main()
{
int a ;
a=abc();
printf(\n %d this is fisrt a:,a);
a=abc();
0 this is fisrt a:
0 this is second a:
On Sun, Sep 18, 2011 at 12:09 PM, hurtlocker bhavesh24...@rediffmail.comwrote:
#includestdio.h
main()
{
int a ;
a=abc();
printf(\n %d this is fisrt a:,a);
a=abc();
printf(\n %d this is second a:,a);
}
int abc()
{
another que..
#includestdio.h
main()
{
int a;
int i=10;
printf(%d %d %d\n,i+++i,i,i---i);
printf(%d\n,i---i);
a=i---i;
printf(%d \n%d,a,i);
return 0;
}
output:
18 10 0
0
0
8
how ??
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again undefined in standard c :)
On Sun, Sep 18, 2011 at 1:49 PM, Bhavesh agrawal agr.bhav...@gmail.comwrote:
another que..
#includestdio.h
main()
{
int a;
int i=10;
printf(%d %d %d\n,i+++i,i,i---i);
printf(%d\n,i---i);
a=i---i;
printf(%d \n%d,a,i);
usigned long long x = 0x12345678;
int a = 0x09;
int b = 0x10;
printf(a=%x, b=%llx,a,b,c);
the result is: a=9,b=123456780010
i wonder why~~
can anyone explain it?
thanks.
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Check out what u have written...
On Sun, Sep 18, 2011 at 7:17 PM, wujin chen wujinchen...@gmail.com wrote:
usigned long long x = 0x12345678;
int a = 0x09;
int b = 0x10;
printf(a=%x, b=%llx,a,b,c);
the result is: a=9,b=123456780010
i wonder why~~
can anyone explain it?
thanks.
sorry , it should be :
usigned long long c = 0x12345678;
int a = 0x09;
int b = 0x10;
printf(a=%x, b=%llx,a,b,c);
2011/9/19 sagar pareek sagarpar...@gmail.com
Check out what u have written...
On Sun, Sep 18, 2011 at 7:17 PM, wujin chen wujinchen...@gmail.comwrote:
usigned long long x =
+1
On Wed, Aug 31, 2011 at 9:27 PM, SANDEEP CHUGH sandeep.aa...@gmail.comwrote:
its 4 3 i think
On Wed, Aug 31, 2011 at 9:25 PM, aditi garg aditi.garg.6...@gmail.comwrote:
8 3
On Wed, Aug 31, 2011 at 9:22 PM, rohit raman.u...@gmail.com wrote:
output??
int main()
{
char *d =
for 16 bit compiler ans is 2 3
for 32 bit compiler ans is 4 3
pls correct me if i m wrong
On Wed, Sep 14, 2011 at 11:36 AM, tech coder techcoderonw...@gmail.comwrote:
+1
On Wed, Aug 31, 2011 at 9:27 PM, SANDEEP CHUGH sandeep.aa...@gmail.comwrote:
its 4 3 i think
On Wed, Aug 31, 2011
for a 32 bit it is 4 ,3 and for 16 bit it is 2,3 and 64 bit it is 8,3..
also strlen(d) gives 3 bcz the coding of fn strlen() is such that as it
encounters null it will terminate..so irrespective of machine it prints 3 as
null at 4th place
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9 and 3
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@ prateek absolutely wrong , dear u need to brushup ur basics d is a pointer
,
it's can be 2 or 4 depebding on compiler
On Wed, Sep 14, 2011 at 6:26 AM, PRATEEK VERMA prateek...@gmail.com wrote:
9 and 3
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Algorithm
main()
{
char *p=persons;
clrscr();
if(p==persons)
printf(technical %s,p);
else
printf(true %s,p);
return 0;
}
..op : technical persons ..plz explain .. how come it works like an strcmp
operation???
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String literals are saved into a separate table in compiler. So second time
when you define persons it wont be created or allocated memory as it
already exists in table.
On Tue, Sep 6, 2011 at 9:04 PM, sivaviknesh s sivavikne...@gmail.comwrote:
main()
{
char *p=persons;
clrscr();
It is basically comparing the addresses of the two and since p contains the
memory address of persons it gives the output as technical persons...
infact ull be surprised to see dis
#includestdio.h
main(){char p[]=persons;char q[]=persons;if(p==q)printf
in case of constant strings..they all return the same pointer..where they
are defined..
but in case of non constant strings..like a[]=sumthing and
b[]=sumthing...they will have separate memory allocations..
On Tue, Sep 6, 2011 at 9:12 PM, aditi garg aditi.garg.6...@gmail.comwrote:
It is
addresses are compared here i think
On Tue, Sep 6, 2011 at 9:04 PM, sivaviknesh s sivavikne...@gmail.comwrote:
main()
{
char *p=persons;
clrscr();
if(p==persons)
printf(technical %s,p);
else
printf(true %s,p);
return 0;
}
..op : technical persons ..plz explain .. how come it works
String constants (literals) are saved into the .data section of the program,
Here is the sample program to show that. if() is essentially comparing the
addresses of two pointers which is same.
int main()
{
char *p=persons;
char *q=persons;
char *r=persons;
char *s=persons;
printf(%x %x %x
printf(%d,3.14*6.25*6.25);
...ans : 0 ..how and why?? why not type conversion take place??
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it wil not truncate the floating point to integer remember...
On Tue, Sep 6, 2011 at 9:55 PM, sivaviknesh s sivavikne...@gmail.comwrote:
printf(%d,3.14*6.25*6.25);
...ans : 0 ..how and why?? why not type conversion take place??
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coz both floating point are stored in ieee format which is different from
integers
On Tue, Sep 6, 2011 at 10:18 PM, sukran dhawan sukrandha...@gmail.comwrote:
it wil not truncate the floating point to integer remember...
On Tue, Sep 6, 2011 at 9:55 PM, sivaviknesh s
9 and 3
On Wed, Aug 31, 2011 at 9:22 PM, rohit raman.u...@gmail.com wrote:
output??
int main()
{
char *d = abc\0def\0;
printf(%d %d,sizeof(d),strlen(d));
getch();
}
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sizeof considers everything between double quotes + /0 character
strlen reads until first \0
On Thu, Sep 1, 2011 at 5:49 PM, sukran dhawan sukrandha...@gmail.comwrote:
9 and 3
On Wed, Aug 31, 2011 at 9:22 PM, rohit raman.u...@gmail.com wrote:
output??
int main()
{
char *d = abc\0def\0;
output is machine dependent.
On 9/1/11, sukran dhawan sukrandha...@gmail.com wrote:
sizeof considers everything between double quotes + /0 character
strlen reads until first \0
On Thu, Sep 1, 2011 at 5:49 PM, sukran dhawan sukrandha...@gmail.comwrote:
9 and 3
On Wed, Aug 31, 2011 at 9:22
its 4 3 i think
On Wed, Aug 31, 2011 at 9:25 PM, aditi garg aditi.garg.6...@gmail.comwrote:
8 3
On Wed, Aug 31, 2011 at 9:22 PM, rohit raman.u...@gmail.com wrote:
output??
int main()
{
char *d = abc\0def\0;
printf(%d %d,sizeof(d),strlen(d));
getch();
}
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You received
2 3 because d is pointer avriable
On Wed, Aug 31, 2011 at 9:25 PM, aditi garg aditi.garg.6...@gmail.comwrote:
8 3
On Wed, Aug 31, 2011 at 9:22 PM, rohit raman.u...@gmail.com wrote:
output??
int main()
{
char *d = abc\0def\0;
printf(%d %d,sizeof(d),strlen(d));
getch();
}
--
i said 4 3 according to 32 bit machine.. if its 16 bit thn it wud be 2 3
On Wed, Aug 31, 2011 at 9:30 PM, sachin goyal monugoya...@gmail.com wrote:
2 3 because d is pointer avriable
On Wed, Aug 31, 2011 at 9:25 PM, aditi garg aditi.garg.6...@gmail.comwrote:
8 3
On Wed, Aug 31, 2011 at
sizeof(d) is the sz of pointer, so depending on arch it will be 4 or 8 ...
On Wed, Aug 31, 2011 at 11:57 AM, SANDEEP CHUGH sandeep.aa...@gmail.comwrote:
its 4 3 i think
On Wed, Aug 31, 2011 at 9:25 PM, aditi garg aditi.garg.6...@gmail.comwrote:
8 3
On Wed, Aug 31, 2011 at 9:22 PM,
sorry my mistake...
it wud be 4 3
On Wed, Aug 31, 2011 at 9:30 PM, sachin goyal monugoya...@gmail.com wrote:
2 3 because d is pointer avriable
On Wed, Aug 31, 2011 at 9:25 PM, aditi garg aditi.garg.6...@gmail.comwrote:
8 3
On Wed, Aug 31, 2011 at 9:22 PM, rohit raman.u...@gmail.com
8 3 is correct..winshuttle mein aaya tha ;)
On Wed, Aug 31, 2011 at 9:30 PM, aditi garg aditi.garg.6...@gmail.comwrote:
No sizeof ignores \0
On Wed, Aug 31, 2011 at 9:27 PM, SANDEEP CHUGH sandeep.aa...@gmail.comwrote:
its 4 3 i think
On Wed, Aug 31, 2011 at 9:25 PM, aditi garg
yes it must be 4 3
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@dheeraj yea true,,,bt ut ws an array...its a pointer here...so output will
be 4 3...
infact it wud be 9 3 in the case u r considering
On Wed, Aug 31, 2011 at 9:32 PM, Dheeraj Sharma dheerajsharma1...@gmail.com
wrote:
8 3 is correct..winshuttle mein aaya tha ;)
On Wed, Aug 31, 2011 at 9:30
8 3 wud the answer if you use the following syntax
int main()
{
char d[] = abc\0def\0;
printf(%d %d,sizeof(d),strlen(d));
getch();
}
as you are using char *d
so sizeof will print only size of pointer variable
On Wed, Aug 31, 2011 at 9:32 PM, Dheeraj Sharma dheerajsharma1...@gmail.com
+1 aditi
On Wed, Aug 31, 2011 at 9:34 PM, aditi garg aditi.garg.6...@gmail.comwrote:
@dheeraj yea true,,,bt ut ws an array...its a pointer here...so output will
be 4 3...
infact it wud be 9 3 in the case u r considering
On Wed, Aug 31, 2011 at 9:32 PM, Dheeraj Sharma
for this
char *d = abc\ldef\l;
the output is 4 8 not 4 9 \l treated as one character
On Wed, Aug 31, 2011 at 9:34 PM, aditi garg aditi.garg.6...@gmail.comwrote:
@dheeraj yea true,,,bt ut ws an array...its a pointer here...so output will
be 4 3...
infact it wud be 9 3 in the case u r
8 3
On Wed, Aug 31, 2011 at 9:22 PM, rohit raman.u...@gmail.com wrote:
output??
int main()
{
char *d = abc\0def\0;
printf(%d %d,sizeof(d),strlen(d));
getch();
}
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Depends on the 32-bit or 64-bit...
sizeof(d) - returns the 4 in case of 32-bit, 8 incase of 64-bit
strlen(d) - 3
On Wed, Aug 31, 2011 at 9:32 PM, Dheeraj Sharma dheerajsharma1...@gmail.com
wrote:
8 3 is correct..winshuttle mein aaya tha ;)
On Wed, Aug 31, 2011 at 9:30 PM, aditi garg
#includestdio.h
#define max(a,b) (ab?a:b)
int main()
{
int j=max(3+2,2+8);
printf(%d,j);
return 0;
}
why this program show output as 9 ? please help me
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its output is
10
On Tue, Aug 23, 2011 at 12:03 AM, rohit rajuljain...@gmail.com wrote:
#includestdio.h
#define max(a,b) (ab?a:b)
int main()
{
int j=max(3+2,2+8);
printf(%d,j);
return 0;
}
why this program show output as 9 ? please help me
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Yeah its o/p is 10 :)
On Tue, Aug 23, 2011 at 12:45 AM, Deepak Garg deepakgarg...@gmail.comwrote:
its output is
10
On Tue, Aug 23, 2011 at 12:03 AM, rohit rajuljain...@gmail.com wrote:
#includestdio.h
#define max(a,b) (ab?a:b)
int main()
{
int j=max(3+2,2+8);
printf(%d,j);
output is 10 using gcc 4.5.2
Yanan Cao
On Mon, Aug 22, 2011 at 2:18 PM, sagar pareek sagarpar...@gmail.com wrote:
Yeah its o/p is 10 :)
On Tue, Aug 23, 2011 at 12:45 AM, Deepak Garg deepakgarg...@gmail.comwrote:
its output is
10
On Tue, Aug 23, 2011 at 12:03 AM, rohit
10 as 3+22+8?3+2:2+8 510?5:10 ' + ' has more precedence then ?: as well
as s o/p is 10..
correct me if i am wrong.
On Tue, Aug 23, 2011 at 12:50 AM, gmagog...@gmail.com
gmagog...@gmail.comwrote:
output is 10 using gcc 4.5.2
Yanan Cao
On Mon, Aug 22, 2011 at 2:18 PM, sagar pareek
its 10
On Tue, Aug 23, 2011 at 12:03 AM, rohit rajuljain...@gmail.com wrote:
#includestdio.h
#define max(a,b) (ab?a:b)
int main()
{
int j=max(3+2,2+8);
printf(%d,j);
return 0;
}
why this program show output as 9 ? please help me
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#includestdio.hconst char *fun();
int main()
{
char *ptr = fun();
return 0;
}const char *fun()
{
return Hello;
}
Why doesn't this code give error??
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This is becuase Hello is a constant string and constant strings get stored
in *Data Area, not in stack for the function you called. *Thats why pointer
to constant string will be returned and program will not produce any error.
Sanjay Kumar
B.Tech Final Year
Department of Computer Engineering
i think this must have produced a warning..conversion from const char *
to char*...
On Tue, Aug 16, 2011 at 8:52 PM, Sanjay Rajpal sanjay.raj...@live.inwrote:
This is becuase Hello is a constant string and constant strings get
stored in *Data Area, not in stack for the function you
int main ()
{
printf(%d,1+2+5);
getch();
return 0;
}
what should it return and how..??
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i got a warning in gcc and the output was garbage
On Sun, Aug 14, 2011 at 11:44 AM, Brijesh Upadhyay
brijeshupadhyay...@gmail.com wrote:
int main ()
{
printf(%d,1+2+5);
getch();
return 0;
}
what should it return and how..??
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5 will represent address of 5 in heap. so adding 3 to 5 increment tht
address by 3 and print it on the screen
On Sun, Aug 14, 2011 at 11:44 AM, Brijesh Upadhyay
brijeshupadhyay...@gmail.com wrote:
int main ()
{
printf(%d,1+2+5);
getch();
return 0;
}
what should it return and
I think the address of string 5 located will be added with 1 and 2...
On Sun, Aug 14, 2011 at 11:44 AM, Brijesh Upadhyay
brijeshupadhyay...@gmail.com wrote:
int main ()
{
printf(%d,1+2+5);
getch();
return 0;
}
what should it return and how..??
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int main()
{
int a[5]={1,2,3,4,5};
printf(%d,a[4]-a[0])
}
why it show 4 not 16?
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Pointer incrementation and subtraction are done in terms of memory blocks
and not addresses of memory.
For example,
int *p;
p++;
The pointer here jumps to the next integer location and not the next address
in memory.
Similarly,pointer subtraction will give the difference in indexes and not
the
i didn't tried it .. but it might be internal conversion only , like
whenever we do +1 to the address of int it automatically convert it
into +4(i.e int size)
On Fri, Aug 12, 2011 at 11:34 AM, rohit rajuljain...@gmail.com wrote:
int main()
{
int a[5]={1,2,3,4,5};
printf(%d,a[4]-a[0])
}
4*(sizeof(int *))
Thank you,
Siddharam
On Fri, Aug 12, 2011 at 11:56 AM, Varun Jakhoria varunjakho...@gmail.comwrote:
i didn't tried it .. but it might be internal conversion only , like
whenever we do +1 to the address of int it automatically convert it
into +4(i.e int size)
On Fri, Aug
On Fri, Aug 12, 2011 at 11:55 AM, Avinash Dharan avinashdha...@gmail.comwrote:
Pointer incrementation and subtraction are done in terms of memory blocks
and not addresses of memory.
For example,
int *p;
p++;
The pointer here jumps to the next integer location and not the next
address in
main()
{
int m,n;
m=3+max(2,3);
n=2*max(3,2);
printf(“%d,%d”,m,n);
}
ans:-m=2,n=3
why output is this???
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