Finding like-minded numbers

[arnuld]

Again, my friend, who does not much access to internet, has given me a
working program and I need your views on improvement and design. I put
some error checking but my experience with C has made it a messy error
checking, as compared to C++ standards. I will welcome any constructive
criticism:
/* This program has actually emerged from a real world requirement. In India, vehicle
* registration numbers are alloted from 0 to 9999, and many people want to have their
* vehicle number based on their single lucky number e.g. like my friend has 1 as his
* lucky number. Vehicle numbers are always alloted in four digits, padded by zeroes if
* necessary, like 0001, 0567, hence the output needs to be int he same way.
*
* VERSION 1.0
*
*/
#include<iostream>
int get_lucky_num();
void print_lucky_nums( int& );
const int final_num = 10;

int main()
{
int lucky_num;
const int no_num = -1;
lucky_num = get_lucky_num();

if( lucky_num != no_num )
{
print_lucky_nums( lucky_num );
}
return 0;
}


int get_lucky_num()
{
std :: cout <<"Please Enter a single digit to get the collection :- ";

int num = final_num;
std::cin.clear();
std::cin >num;

/* This very carefully checks for anythigng greater than "final_num" but
is this the right check to know user entered the non-dgit character like
@ or even F , some user's are stupid anyway ;)
*/
while( (num >= final_num) )
{
std::cout << "INPUT = "
<< num
<< "\n";
std::cout << "\nAre you drunk? \nPlease enter a number less than "
<< final_num
<< ": ";

std::cin.clear();
// I get th enext line from somehwere but have no idea why it is here.
// I am only using it because if I don't use it ans uder enters something
// non-digit like F then program falls into infinite loop
std::cin.ignore(std::numeric_limits<std::streamsiz e>::max(), '\n');
std::cin >num;
}

return num;
}

// can't I have a way to pass the limit of 10,000 as some constant integer ?
void print_lucky_nums( int& num )
{
int k;

for(int i =0 ; i < final_num; ++i)
{
for(int f=0; f < final_num ; ++f)
{
for(int g=0; g < final_num; ++g)
{
for(int h=0; h < final_num; ++h)
{
k=i+f+g+h;

int div,modu,sum;
div = k / 10;
modu = k % 10;
sum = div + modu;

if( num == sum)
{
std :: cout << i
<< f
<< g
<< h
<< std::endl;
}
}
}
}
}
}

--
www.lispmachine.wordpress.com
my email is @ the above blog.
Google Groups is Blocked. Reason: Excessive Spamming

[arnuld]

On Wed, 24 Sep 2008 16:26:57 +0500, arnuld wrote:
Just changed the subject to reflect the content. I shopuld have done it
earlier. apologies.
--
www.lispmachine.wordpress.com
my email is @ the above blog.
Google Groups is Blocked. Reason: Excessive Spamming

[Barry]

On Sep 24, 7:26*pm, arnuld <[e-mail]:

Again, my friend, who does not much access to internet, has given me a
working program and I need your views on improvement and design. I put
some error checking but my experience with C has made it a messy error
checking, as compared to C++ standards. I will welcome any constructive
criticism:

/* This program has actually emerged from a real world requirement. In India, vehicle
** registration numbers are alloted from 0 to 9999, and many people want to have their
** vehicle number based on their single lucky number e.g. like my friend has 1 as his
** lucky number. Vehicle numbers are always alloted in four digits, padded by zeroes if
** necessary, like 0001, 0567, hence the output needs to be int he sameway.
**
** VERSION 1.0
**
**/

#include<iostream>

int get_lucky_num();
void print_lucky_nums( int& );

const int final_num = 10;

int main()
{
* int lucky_num;
* const int no_num = -1;

* lucky_num = get_lucky_num();

* if( lucky_num != no_num )
* * {
* * * print_lucky_nums( lucky_num );
* * }

* return 0;

}

int get_lucky_num()
{
* std :: cout <<"Please Enter a single digit to get the collection :- ";

* int num = final_num; *
* std::cin.clear();
* std::cin >num;

* /* This very carefully checks for anythigng greater than "final_num" but
* * *is this the right check to know user entered the non-dgit character like
* * *@ or even F , some user's are stupid anyway ;) *
* */
* while( (num >= final_num) )
* * {
* * * std::cout << "INPUT = "
* * * * * * * * << num
* * * * * * * * << "\n";

* * * std::cout << "\nAre you drunk? \nPlease enter a number less than "
* * * * * * * * << final_num
* * * * * * * * << ": "; *

* * * std::cin.clear();
* * * // I get th enext line from somehwere but have no idea why itis here.
* * * // I am only using it because if I don't use it ans uder enters something
* * * // non-digit like F then program falls into infinite loop
* * * std::cin.ignore(std::numeric_limits<std::streamsiz e>::max(), '\n');
* * * std::cin >num;
* * }

I think this while statement has to be something like this,
taking state of std::cin into account.

delete std::cin >num; in before "while"

while (std::cin >num && num >= final_num)
{
...
}

>
* return num;

}

// can't I have a way to pass the limit of 10,000 as some constant integer ?
void print_lucky_nums( int& num )
{
* int k;

* for(int i =0 ; i < final_num; ++i)
* * { *
* * * for(int f=0; f < final_num ; ++f)
* * * * {
* * * * * for(int g=0; g < final_num; ++g)
* * * * * * {
* * * * * * * for(int h=0; h < final_num; ++h)
* * * * * * * * {
* * * * * * * * * k=i+f+g+h;

* * * * * * * * * int div,modu,sum;
* * * * * * * * * div *= k / 10;
* * * * * * * * * modu = k % 10;
* * * * * * * * * sum *= div + modu;

* * * * * * * * * if( num == sum)
* * * * * * * * * * * {
* * * * * * * * * * * * std :: cout << i
* * * * * * * * * * * * * * * * * * << f
* * * * * * * * * * * * * * * * * * << g
* * * * * * * * * * * * * * * * * * << h
* * * * * * * * * * * * * * * * * * << std::endl;
* * * * * * * * * * * }
* * * * * * * * } * * * * * * * *
* * * * * * } * * * * *
* * * * }
* * }

}

My Firefox scans that "num" is not changed in "print_lucky_nums",
but why "int&" as param type?

--
Best Regards
Barry