Re: C++ Primer exercise 3.13

From:

=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?= <Erik-wikstrom@telia.com>

Newsgroups:

comp.lang.c++

Date:

Thu, 19 Jul 2007 12:01:55 GMT

Message-ID:

<ToIni.4399$ZA.1859@newsb.telia.net>

On 2007-07-19 13:41, arnuld wrote:

On Thu, 19 Jul 2007 09:23:21 +0000, Erik Wikstr??m wrote:

Sorry, but you have not solved the problem the author intended. First of
you should let the user enter the numbers. Then you should print the sum
of each pair of numbers, but if the user entered an odd number of
numbers you can't sum the last element with another so you you should
just print it:

ok, BTW, i found the statement ambiguous. so i created the programem on
that basis of what it could mean.

So an example run could look something like this: -- Please enter some
numbers:
1
2
3
4
5

You entered an odd number of numbers.

The sums of the pairs are:
3
7
5

sorry, after 1.5 hours of grueling work, my new programme still gives
"segmentatioon fault" on finding an odd list of numbers. for even amount
of numbers this programme works:

#include <iostream>
#include <vector>

int main()
{
  std::vector<int> ivec; /* empty vector */ int v_num;

  std::cout << "Please enter some numbers: " << std::endl; while(std::cin
  >> v_num)
    ivec.push_back(v_num);

  if((ivec.size() % 2) != 0)
    {
      std::cout << "oops!, you enetered an odd number of numbers" <<
      std::endl;
    }

  std::cout << "The sum of the pairs are: " << std::endl; /* actual
  programme */
  for(std::vector<int>::const_iterator iter = ivec.begin(); iter !=
  ivec.end(); iter += 2)

Replace the check to a less than (iter < ivec.end()), since you are
increasing the iterator with 2 each step you will step over ivec.end()
if the number of elements is odd..

    {
      if((iter + 1) == ivec.end())
    {
      std::cout << *iter << std::endl;

Or add break; here.

     }
      else
    {
      std::cout << *iter + *(iter + 1) << std::endl;
    }
    }

  return 0;
}

--
Erik Wikstr??m