Reading in a variable number of integers with cin

[H.]

I'm writing a driver for something I wrote, and can't remember how to
do this very simple thing. The code I have at this point in time:

int myArr[100];
int temp;
int ind = 0;
cout << "enter some integers: ";
while (cin >> temp)
{
myArr[ind++] = temp;
}

The problem is that it doesn't exit until the user enters a non-number.
It won't exit if the user simply hits enter.

What I'm trying to create is something that allows the user to separate
values by spaces, and press the Enter key when done entering them. A
variable number of items will be separated by spaces, e.g.:
"Enter the integers to be sorted with spaces between them and Enter to
finish"

[int2str]

H. wrote:

I'm writing a driver for something I wrote, and can't remember how to
do this very simple thing. The code I have at this point in time:
The code you posted doesn't compile.
Please make sure you post compilable code here if you want help.
int myArr[100];
How do you know 100 is a good value? What if I want to enter 101
numbers?
Use a std::vector instead to keep your options open.
int temp;
Always initialize your variables.
int ind = 0;
cout << "enter some integers: ";
while (cin >> temp)
{
myArr[ind++] = temp;
At index 101, this will blow ;)
}

The problem is that it doesn't exit until the user enters a non-number.
It won't exit if the user simply hits enter.

What I'm trying to create is something that allows the user to separate
values by spaces, and press the Enter key when done entering them. A
variable number of items will be separated by spaces, e.g.:
"Enter the integers to be sorted with spaces between them and Enter to
finish"

I'm probably doing your homework for you here, but I'll hope that the
sorting algorithm is the real challenge here.

Here's some code for you to work with:
#include <iostream> // std::cout, std::endl
#include <string> // std::string
#include <sstream> // std::stringstream
#include <vector> // std::vector
#include <algorithm> // std::copy
#include <iterator> // std::ostream_iterator

int main()
{
std::cout << "Enter integers separated by spaces:\n";

std::string line;
if ( std::getline( std::cin, line ))
{
std::stringstream ss( line );
std::vector<long> values;
long l = 0;

while( ss >> l )
values.push_back( l );

if ( values.size() )
std::copy( values.begin(), values.end(),
std::ostream_iterator<long>(std::cout, "\n"));
}
}

Cheers,
Andre

[H.]

Thanks for all that, it was very educational. (Unfortunately, It's an
overkill for what I'm trying to do, but again, thanks though.) Vectors
are great; they can't be used for what I'm doing, arrays are needed.)
And hard-coding the array number to 100 is fine for the testing driver
I'm writing; there will never be 101 numbers. Basically the goal is to
create something as simple as possible to get the job done, with no
extra classes besides iostream. It is being used as a driver for a
homework problem, but I already created a simpler driver that hard
codes the number of entries, which is good enough for the assignment at
hand, since the assignment is not about the numbers but manipulating
them in a given way with recursion. Actually, for the driver, I could
skip the user input altogether and just feed the recursive function all
of the test lists, and so in that sense, it's not a homework problem at
all, just a "for my edification" problem. I'll figure it out, because
I'm sure it can't really be that hard a thing to do.

in*****@gmail.com wrote:

H. wrote:

I'm writing a driver for something I wrote, and can't remember how to
do this very simple thing. The code I have at this point in time:

The code you posted doesn't compile.
Please make sure you post compilable code here if you want help.

int myArr[100];

How do you know 100 is a good value? What if I want to enter 101
numbers?
Use a std::vector instead to keep your options open.

int temp;

Always initialize your variables.

int ind = 0;
cout << "enter some integers: ";
while (cin >> temp)
{
myArr[ind++] = temp;

At index 101, this will blow ;)

}

The problem is that it doesn't exit until the user enters a non-number.
It won't exit if the user simply hits enter.

What I'm trying to create is something that allows the user to separate
values by spaces, and press the Enter key when done entering them. A
variable number of items will be separated by spaces, e.g.:
"Enter the integers to be sorted with spaces between them and Enter to
finish"

I'm probably doing your homework for you here, but I'll hope that the
sorting algorithm is the real challenge here.

Here's some code for you to work with:
#include <iostream> // std::cout, std::endl
#include <string> // std::string
#include <sstream> // std::stringstream
#include <vector> // std::vector
#include <algorithm> // std::copy
#include <iterator> // std::ostream_iterator

int main()
{
std::cout << "Enter integers separated by spaces:\n";

std::string line;
if ( std::getline( std::cin, line ))
{
std::stringstream ss( line );
std::vector<long> values;
long l = 0;

while( ss >> l )
values.push_back( l );

if ( values.size() )
std::copy( values.begin(), values.end(),
std::ostream_iterator<long>(std::cout, "\n"));
}
}

Cheers,
Andre

[dragoncoder]

I am not sure if I understand what you want but I think typing Ctrl-d
on the unix terminal should end the while loop.

Is that what you want?
H. wrote:

I'm writing a driver for something I wrote, and can't remember how to
do this very simple thing. The code I have at this point in time:

int myArr[100];
int temp;
int ind = 0;
cout << "enter some integers: ";
while (cin >> temp)
{
myArr[ind++] = temp;
}

The problem is that it doesn't exit until the user enters a non-number.
It won't exit if the user simply hits enter.

What I'm trying to create is something that allows the user to separate
values by spaces, and press the Enter key when done entering them. A
variable number of items will be separated by spaces, e.g.:
"Enter the integers to be sorted with spaces between them and Enter to
finish"

[wittempj]

If you want to keep the user input part you can do something like below
#include <string>
#include <sstream>
#include <iostream>

int main()
{
std::string inp;
int temp(0);

while (std::cin >> inp)
{
if ((inp == "q") ||(inp == "Q"))
{
break;
}
std::istringstream i(inp);
if (!(i >> temp))
{
std::cerr << "Invalid item entered: " << inp << std::endl;
}
// add number here to your array
}
return 0;
}

[Daniel T.]

In article <11*********************@f14g2000cwb.googlegroups. com>,
"H." <hb****@gmail.com> wrote:

Thanks for all that, it was very educational. (Unfortunately, It's an
overkill for what I'm trying to do, but again, thanks though.) Vectors
are great; they can't be used for what I'm doing, arrays are needed.)
Then it must be a homework assignment.
Basically the goal is to
create something as simple as possible to get the job done, with no
extra classes besides iostream.

What 'int2str' provided was pretty simple. What do you have against
using the tools the language provides?

Your requirements were to enter a series of numbers separated by white
space and ending with the return/enter key. The code he provided does
exactly that. If you want a simpler solution, I suggest you pose a
simpler problem.
--
Magic depends on tradition and belief. It does not welcome observation,
nor does it profit by experiment. On the other hand, science is based
on experience; it is open to correction by observation and experiment.

[Ivan Vecerina]

<in*****@gmail.com> wrote in message
news:11**********************@f14g2000cwb.googlegr oups.com...
: Here's some code for you to work with:
: #include <iostream> // std::cout, std::endl
: #include <string> // std::string
: #include <sstream> // std::stringstream
: #include <vector> // std::vector
: #include <algorithm> // std::copy
: #include <iterator> // std::ostream_iterator
:
: int main()
: {
: std::cout << "Enter integers separated by spaces:\n";
:
: std::string line;
: if ( std::getline( std::cin, line ))
: {
: std::stringstream ss( line );

Here:
: std::vector<long> values;
: long l = 0;
:
: while( ss >> l )
: values.push_back( l );

Since you already use ostream_iterator for output,
how about using the single statement:
std::vector<long> values
( std::istream_iterator<long>( ss )
, std::istream_iterator<long>() );

: if ( values.size() )
: std::copy( values.begin(), values.end(),
: std::ostream_iterator<long>(std::cout, "\n"));
: }
: }
:
: Cheers,
: Andre

.... only for the sake of consistency/completeness,
Ivan
--
http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form

[Daniel T.]

In article <11**********************@g43g2000cwa.googlegroups .com>,
"H." <hb****@gmail.com> wrote:

I'm writing a driver for something I wrote, and can't remember how to
do this very simple thing. The code I have at this point in time:

int myArr[100];
int temp;
int ind = 0;
cout << "enter some integers: ";
while (cin >> temp)
{
myArr[ind++] = temp;
}

The problem is that it doesn't exit until the user enters a non-number.
It won't exit if the user simply hits enter.

What I'm trying to create is something that allows the user to separate
values by spaces, and press the Enter key when done entering them. A
variable number of items will be separated by spaces, e.g.:
"Enter the integers to be sorted with spaces between them and Enter to
finish"

void fn( vector<int>& myArr )
{
cout << "enter some integers: ";
string line;
if ( getline( cin, line ) )
{
myArr.clear();
stringstream ss( line );
copy( istream_iterator<long>( ss ), istream_iterator<long>(),
back_inserter( myArr ) );
}
}

Note, with the above: if the user enters a non-numeric char then the
vector will be loaded up to that point but not after. If you want it to
skip incorrect input, then I suggest going through the line first
looking for non-digits...

bool is_digit( char c ) {
return isdigit( c );
}

// after "myArr.clear();"
replace_if( line.begin(), line.end(),
not1( ptr_fun( &is_digit ) ), ' ' );

I have a question... I was unable to put isdigit directly into the
ptr_fun (I had to wrap it.) Any help there?

--
Magic depends on tradition and belief. It does not welcome observation,
nor does it profit by experiment. On the other hand, science is based
on experience; it is open to correction by observation and experiment.

[Daniel T.]

In article <ds**********@news.hispeed.ch>,
"Ivan Vecerina" <IN*****************@ivan.vecerina.com> wrote:

<in*****@gmail.com> wrote in message
news:11**********************@f14g2000cwb.googlegr oups.com...
: Here's some code for you to work with:
: #include <iostream> // std::cout, std::endl
: #include <string> // std::string
: #include <sstream> // std::stringstream
: #include <vector> // std::vector
: #include <algorithm> // std::copy
: #include <iterator> // std::ostream_iterator
:
: int main()
: {
: std::cout << "Enter integers separated by spaces:\n";
:
: std::string line;
: if ( std::getline( std::cin, line ))
: {
: std::stringstream ss( line );

Here:
: std::vector<long> values;
: long l = 0;
:
: while( ss >> l )
: values.push_back( l );

Since you already use ostream_iterator for output,
how about using the single statement:
std::vector<long> values
( std::istream_iterator<long>( ss )
, std::istream_iterator<long>() );

For that matter, copy will work:

copy(istream_iterator<long>(ss), istream_iterator<long>(),
back_inserter(values));
: if ( values.size() )
The above 'if' is unnecessary. Should values.begin() and values.end() be
equal, copy will do the right thing.
: std::copy( values.begin(), values.end(),
: std::ostream_iterator<long>(std::cout, "\n"));
: }
: }
:
: Cheers,
: Andre

... only for the sake of consistency/completeness,
Ivan

--
Magic depends on tradition and belief. It does not welcome observation,
nor does it profit by experiment. On the other hand, science is based
on experience; it is open to correction by observation and experiment.

[int2str]

H. wrote:

And hard-coding the array number to 100 is fine for the testing driver
No, it's not fine. Not even for testing.
I'm writing; there will never be 101 numbers.

I don't believe you. Or to put it another way, there is no guarantee
that there will "never" be 101 numbers.
To look at your code snipper again:

while (cin >> temp)
{
myArr[ind++] = temp;

This simply contradicts you. With the while loop being what it is,
there in fact CAN be 101 numbers. If you want to GUARANTEE that there
will NEVER be 101 numbers, then have the code do the talking:

while( cin >> temp && ind < 100 )
....

And of course us a constant value instead of a magic number.

Whenever you declare an array of fixed size, you need to make 100% sure
that it cannot go out of bounds. Make it a habbit.

Cheers,
Andre

[H.]

Whenever you declare an array of fixed size, you need to make 100% sure
that it cannot go out of bounds. Make it a habbit.

Good point, thanks.

[int2str]

H. wrote:

the assignment is not about the numbers but manipulating
them in a given way with recursion.

You know, especially if the assignment isn't about array handling
itself, then you should go with a vector. You can access the elements
in a vector just like in an array (ex. my_vect[1]). So the "meat" of
your code (your recursive function) wouldn't look much different
whether you used arrays or a vector.

Cheers,
Andre