in fy2 , the variable arr[10] has not been declared and hence it should
display error message while compiling
Thanks
Kumar Anurag
On Wed, Oct 24, 2012 at 11:06 PM, rahul sharma rahul23111...@gmail.comwrote:
int arr[10] // in fyl 1
now in fyl 2
extern int *arr
void foo()
{
arr[0]=10;
}
i think arr[10] and int *arr are two different declaration when ,when
compiler tried to link with the memory of int arr[10] it could nt find it ,
as u have declraed it to be integer type pointer , and in file 1 it could
find integer pointer .
On Wed, Oct 24, 2012 at 11:06 PM, rahul sharma
@all.can u provide me working code for use of extern..whenever i try to
use..linker error comes...like any simple codefor use of extern
On Thu, Oct 25, 2012 at 9:49 AM, Nishant Pandey
nishant.bits.me...@gmail.com wrote:
i think arr[10] and int *arr are two different declaration when ,when
int arr[10] // in fyl 1
now in fyl 2
extern int *arr
void foo()
{
arr[0]=10;
}
what kind of problem can be there?in what condition and y?
plz comment
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@dave sir: i understand your point, here my solution is only limited to
hold the divisor in range of integer..
On Sat, Sep 22, 2012 at 1:41 AM, Dave dave_and_da...@juno.com wrote:
@Rahul: What does this print for n = 193?
Dave
On Friday, September 21, 2012 12:14:18 AM UTC-5, Rahul Kumar
@Navin: It means that given a positive integer n whose decimal
representation ends in 3, find a multiple, m*n, which is written solely
with the digit 1. E.g., 3: 37 * 3 = 111; 13: 8547 * 13 = 111,111.
Dave
On Thursday, September 20, 2012 11:56:08 PM UTC-5, Navin Kumar wrote:
@all: Please
@Dave sir: Thanx for reply. your solution gives the exact multiple like 37
for 3, 8547 for 13. In the question i think we have to print the number
which is 13x8547 which will be very large (out of integer range).In
that case we have to store result in string.
On Fri, Sep 21, 2012 at 12:09
@Navin: No problem. Just print a 1 instead of a quotient digit. That makes
the code even simpler, like this:
int n=the number that ends with 3;
int divisor=1;
printf(1);
while( divisor != 0 )
{
printf(1);
divisor = 10 * (divisor % n) + 1;
}
printf(\n);
Dave
On Friday, September 21,
here is my code:
main()
{
int n=13;
int divisor=1;
int temp;
while( divisor n )
divisor = 10 * divisor + 1;
printf(%d\n , divisor);
temp = divisor;
while(n)
{
if(temp%n == 0)
break;
temp = 10 * (temp % n) + 1;
divisor = 10 * divisor + 1;
while( temp n )
{
I thought about it some more, and realize that my code wasn't correct. Try
this:
int n = the number that ends with 3; // e.g., int n = 13;
int d = 1;
while( d % n != 0 )
{
printf(1);
d = 10 * (d % n) + 1;
}
printf(1\n);
On Friday, September 21, 2012 11:07:29 AM UTC-5, Dave wrote:
@Rahul: What does this print for n = 193?
Dave
On Friday, September 21, 2012 12:14:18 AM UTC-5, Rahul Kumar Patle wrote:
here is my code:
main()
{
int n=13;
int divisor=1;
int temp;
while( divisor n )
divisor = 10 * divisor + 1;
printf(%d\n , divisor);
temp = divisor;
while(n)
what is the solution(not brute force) for 8th question ?
On Fri, Sep 14, 2012 at 5:19 PM, Bhupendra Dubey bhupendra@gmail.comwrote:
Which edition of barron?
On Wed, Sep 28, 2011 at 6:05 PM, VIHARRI viharri@gmail.com wrote:
1. Java uses stack for byte code in JVM - each instruction
@Bharat: Simulate long division, dividing a number ...1 by the number.
You can do this one digit at a time, printing the quotient digit by digit
until you bring down a zero. It could look something like this:
int n=the number that ends with 3;
int divisor=1;
while( divisor n )
divisor
@all: Please explain question number 8. I am not getting the question
exactly what it says ?
On Fri, Sep 21, 2012 at 9:30 AM, Dave dave_and_da...@juno.com wrote:
@Bharat: Simulate long division, dividing a number ...1 by the number.
You can do this one digit at a time, printing the
Which edition of barron?
On Wed, Sep 28, 2011 at 6:05 PM, VIHARRI viharri@gmail.com wrote:
1. Java uses stack for byte code in JVM - each instruction is of one
byte, so how many such instructions are possible in an operating
system.
2. Three processes p1, p2, p3, p4 - each have sizes
1. Java uses stack for byte code in JVM - each instruction is of one
byte, so how many such instructions are possible in an operating
system.
2. Three processes p1, p2, p3, p4 - each have sizes 1GB, 1.2GB, 2GB,
1GB. And each processes is executed as a time sharing fashion. Will
they be executed
Can any one pls post the questions that were asked in Adobe written test ???
And also pattern of the paper
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Regards:
---
K Pavan Kumar
Computer Science Engg.
2nd Mtech, IIT Madras.
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