Print numbers

[andrew.smith.cpp]

Hello
how Can i get this kind of output?

1,2,3,5,8,13,21
i try it with loop
but could print it.


Thanks

 

[arnuld]

Hello
welcome

how Can i get this kind of output?

1,2,3,5,8,13,21

yes, I solved it. you can find it here:
http://lispmachine.wordpress.com/complangc-welcome-message/

i try it with loop
but could print it.

I thin you meant "but could not print it". BTW, AAMOF, IMVHO, you can use
std::cout to print something to standard output.

 

[Linlin Yan]

Hello
how Can i get this kind of output?

1,2,3,5,8,13,21
i try it with loop
but could print it.

Thanks

It's a Fibonacci series, so it is easy to print or get with loop.
But why did you post the question in this C++ newsgroup?

 

[James Kanze]

Hello
how Can i get this kind of output?

1,2,3,5,8,13,21
i try it with loop
but could print it.

If printing it is a problem:
std::cout << "1,2,3,5,8,13,21\n" ;
should do the trick. But that looks like the start of a
Fibonacci sequence. If so, and you want to output an arbitrary
number of elements, you'll need something like:

std::cout.setf( std::ios::fixed, std::ios::floatfield ) ;
std::cout.precision( 0 ) ;
double sqrt5( sqrt( 5.0 ) ) ;
double psi( (1.0 + sqrt5) / 2.0 ) ;
for ( int i = 1 ; i <= count ; ++ i ) {
if ( i != 1 ) {
std::cout << ',' ;
}
std::cout << (pow( psi, i ) - pow( -psi, -i )) / sqrt5 ;
}
std::cout << '\n' ;

 

[arnuld]

If printing it is a problem:
std::cout << "1,2,3,5,8,13,21\n" ;
should do the trick.

But that looks like the start of a
Fibonacci sequence. If so, and you want to output an arbitrary
number of elements, you'll need something like:

std::cout.setf( std::ios::fixed, std::ios::floatfield ) ;
std::cout.precision( 0 ) ;
double sqrt5( sqrt( 5.0 ) ) ;
double psi( (1.0 + sqrt5) / 2.0 ) ;
for ( int i = 1 ; i <= count ; ++ i ) {
if ( i != 1 ) {
std::cout << ',' ;
}
std::cout << (pow( psi, i ) - pow( -psi, -i )) / sqrt5 ;
}
std::cout << '\n' ;

I did not know that OP was looking for a Fibonacci sequence. I just did it
this way:


/* a program to print the sum of last 2 numbers.

OUTPUT should be: 0,1,1,2,3,5,8,13,21

0 + 1 = 1
1 + 1 = 2
1 + 2 = 3
2 + 3 = 5
3 + 5 = 8
5 + 8 = 13
.........
.........

*/


#include <iostream>


int main()
{
int x = 0;
int y = 1;
int sum = 0;
const int max_num = 10;


for( int i = 0; i != max_num; ++i )
{
sum = x + y;

/* if statement purely exists to print zero */
if( !x )
{
std::cout << x << ", "
<< sum << ", ";
}
else
{
std::cout << sum << ", ";
}

x = y;
y = sum;

}

std::cout << std::endl;


return 0;
}


and it prints fine, except that it puts a comma at the end and I have out
a limit of 10 numbers:

[arnuld@dune programs]$ g++ -ansi -pedantic -Wall -Wextra test.cpp
[arnuld@dune programs]$ ./a.out
0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,
[arnuld@dune programs]$

 

[James Kanze]

I did not know that OP was looking for a Fibonacci sequence. I
just did it this way:

/* a program to print the sum of last 2 numbers.

Which is the definition of a Fibonacci sequence.

The "classical" implementation is:

int
fib( int n )
{
return n <= 0 ? 1 : fib( n - 1 ) + fib( n - 2 ) ;
}

It's sometimes used as a good example of when not to use
recursion; if you just want a few specific values, and add a
cache, however, it's not that bad:

int
fib( int n )
{
static std::vector< int > cache( 2, 1 ) ;
if ( n >= static_cast< int >( cache.size() ) ) {
cache.push_back( fib( n - 1 ) + fib( n - 2 ) ) ;
}
return n < 0 ? 1 : cache[ n ] ;
}

Both such solutions suffer from the fact that int's overflow for
very small values of n, however. My solution above doesn't.

and it prints fine, except that it puts a comma at the end

Which is a separate (and general) problem: how to format
sequences of data.

and I have out a limit of 10 numbers:

Try outputting 100 values, and see what happens.

 

[red floyd]

I did not know that OP was looking for a Fibonacci sequence. I
just did it this way:

/* a program to print the sum of last 2 numbers.

Which is the definition of a Fibonacci sequence.

The "classical" implementation is:

int
fib( int n )
{
return n <= 0 ? 1 : fib( n - 1 ) + fib( n - 2 ) ;
}

It's sometimes used as a good example of when not to use
recursion; if you just want a few specific values, and add a
cache, however, it's not that bad:

int
fib( int n )
{
static std::vector< int > cache( 2, 1 ) ;
if ( n >= static_cast< int >( cache.size() ) ) {
cache.push_back( fib( n - 1 ) + fib( n - 2 ) ) ;
}
return n < 0 ? 1 : cache[ n ] ;
}

Both such solutions suffer from the fact that int's overflow for
very small values of n, however. My solution above doesn't.

and it prints fine, except that it puts a comma at the end

Which is a separate (and general) problem: how to format
sequences of data.

and I have out a limit of 10 numbers:

Try outputting 100 values, and see what happens.

I suspect it's homework, but you probably knew that, James.

I wish I could remember one of the IOCCC fibonacci programs. But I
can't. So here's one that will hopefully confuse the OP.

#include <iostream>
#include <ostream>
#include <algorithm>
#include <iterator>
template <int N> struct fib {
enum { value = fib<N-1>::value + fib<N-2>::value };
};

template<> struct fib<0> {
enum {value = 1 };
};

template<> struct fib<1> {
enum {value = 1};
};

int main()
{
int const fv[] = { fib<1>::value, fib<2>::value, fib<3>::value,
fib<4>::value, fib<5>::value, fib<6>::value,
fib<7>::value };

std::copy(fv, fv+7, std:utput_iterator<int>(std::cout,","));
}

 

[James Kanze]

[...]

I suspect it's homework, but you probably knew that, James.

Exactly. Thus, a few "exotic" suggestions. I wonder what his
prof would say if he turned in the one with the floating point.

Of course, as he originally stated the problem, my original
solution (std::cout << "1,1,2,3,..") is both the simplest and
the most efficient---and so, the correct solution. I rather
doubt, however, that that was what the prof was looking for.

I wish I could remember one of the IOCCC fibonacci programs.
But I can't. So here's one that will hopefully confuse the
OP.

#include <iostream>
#include <ostream>
#include <algorithm>
#include <iterator>
template <int N> struct fib {
enum { value = fib<N-1>::value + fib<N-2>::value };

};

template<> struct fib<0> {
enum {value = 1 };
};

template<> struct fib<1> {
enum {value = 1};
};

int main()
{
int const fv[] = { fib<1>::value, fib<2>::value, fib<3>::value,
fib<4>::value, fib<5>::value, fib<6>::value,
fib<7>::value };
std::copy(fv, fv+7, std:utput_iterator<int>(std::cout,","));
}

Yes. And if you want the number of values to be variable, you
ouput the code to a temporary file, looping over the
initialization of fv, and then use system to compile and run
that.

I like it!

 

[arnuld]

If printing it is a problem:
std::cout << "1,2,3,5,8,13,21\n" ;
should do the trick.

But that looks like the start of a
Fibonacci sequence. If so, and you want to output an arbitrary
number of elements, you'll need something like:

std::cout.setf( std::ios::fixed, std::ios::floatfield ) ;
std::cout.precision( 0 ) ;
double sqrt5( sqrt( 5.0 ) ) ;
double psi( (1.0 + sqrt5) / 2.0 ) ;
for ( int i = 1 ; i <= count ; ++ i ) {
if ( i != 1 ) {
std::cout << ',' ;
}
std::cout << (pow( psi, i ) - pow( -psi, -i )) / sqrt5 ;
}
std::cout << '\n' ;

I did not know that OP was looking for a Fibonacci sequence. I just did it
this way:


/* a program to print the sum of last 2 numbers.

OUTPUT should be: 0,1,1,2,3,5,8,13,21

0 + 1 = 1
1 + 1 = 2
1 + 2 = 3
2 + 3 = 5
3 + 5 = 8
5 + 8 = 13
.........
.........

*/


#include <iostream>


int main()
{
int x = 0;
int y = 1;
int sum = 0;
const int max_num = 10;


for( int i = 0; i != max_num; ++i )
{
sum = x + y;

/* if statement purely exists to print zero */
if( !x )
{
std::cout << x << ", "
<< sum << ", ";
}
else
{
std::cout << sum << ", ";
}

x = y;
y = sum;

}

std::cout << std::endl;


return 0;
}


and it prints fine, except that it puts a comma at the end and I have out
a limit of 10 numbers:

[arnuld@dune programs]$ g++ -ansi -pedantic -Wall -Wextra test.cpp
[arnuld@dune programs]$ ./a.out
0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,
[arnuld@dune programs]$