Thanks Debanjan for intimating the bug in my code and Sorry i forget to add
48 in second putchar statement.
now rite code is
#include
void print(unsigned long int n)
{
if(n>9)
print(n/10);
putchar('0'+(n%10));
}
int main()
{
Logic is okay, but the second putchar() call should be offset-ed by '0'
character, not just 0-9 ...
I'd also avoid using hardcoded 48, not every compiler operates on ASCII
literal (e.g, EBCDIC on some IBM mainframes).
Best,
-Lego
On Thu, Oct 15, 2009 at 11:38 AM, Debanjan wrote:
>
> On Oct 15,
On Oct 15, 10:45 am, umesh kewat wrote:
> Hi,
> Here is the code for problem...
>
> void print(unsigned long int n)
> {
> if(n<10)
>
> putchar(n+48);
> else
> {
> print(n/10);
> putchar(n%10);
> }
>
> }
>
> int main()
> {
> unsigned long int n;
void print(unsigned long x) {
if (x >= 10) print(x / 10);
putchar('0' + (x % 10));
}
On Wed, Oct 14, 2009 at 10:45 PM, umesh kewat wrote:
> Hi,
> Here is the code for problem...
>
> void print(unsigned long int n)
> {
> if(n<10)
>
> putchar(n+48);
> else
> {
>
Hi,
Here is the code for problem...
void print(unsigned long int n)
{
if(n<10)
putchar(n+48);
else
{
print(n/10);
putchar(n%10);
}
}
int main()
{
unsigned long int n;
scanf("%ld",&n);
print(n);
return 0;
}
On Thu, Oct 15, 2009 at 9
On Oct 15, 9:03 am, ankur aggarwal wrote:
> 1. Given only putchar (no sprintf, itoa, etc.) write a routine putlong
> that prints out an unsigned long in decimal.
Well,this is rather very easy :
void putlong(unsigned long long num)
{
if(num > 9)
putlong(num/10);
putchar(num%1
no need to take care of negative numbers because given number is unsigned
long integer ..
apply simple algo :)
On Thu, Oct 15, 2009 at 1:07 PM, Siddharth S wrote:
>
> void putlong(long n)
> {
>if(n == 0) // special case
>{
> putchar('0');
> return;
>}
>else
>i
void putlong(long n)
{
if(n == 0) // special case
{
putchar('0');
return;
}
else
if(n < 0) // if number is negative, print "-" sign and continue as if it
is a positive number
putchar('-'), n = -n;
int arr[30];
int arr_cnt = 0;
while(n)
{
arr[arr_cnt+
