Thread: gotta be a better way than this?

hi,

this code prints out numbers like so.

1 2 (there should be a 25 space gap)
3 4
5 6 etc.

but i am not happy with my method of doing it . Could anyone show me a better alternative.

code.

#include <iostream>
#include <iomanip> // for use of set(x)
using namespace std;

int main ()

{

int count=1;
int num=0;

for(count = 1; count <= 20;count++)
{

num +=1;
cout << setw(3) << num ;
num +=1;
cout << setw(25) << num <<endl;
}
cout << endl;

return 0;

}

thanks

#include <iostream>
#include <iomanip> // for use of set(x)
using namespace std;

int main ()

{

int count=1;
int num=0;

for(count = 1; count <= 20;count++)
{

cout << setw(3) << (num += 1);
cout << setw(25) << (num += 1) <<endl; ;

}
cout << endl;

return 0;

}

dang

Code:

#include <iostream> 
#include <iomanip> // for use of setw(x) 
using std:: cout; 
using std::endl;

int main () 
{ 
   int count=1, num = 0; 

   for(count = 1; count <= 20;count++) 
   { 
      cout << setw(3) << ++num << setw(25) << ++num << endl; 
   } 

cout << endl; 

return 0; 
}

I originally tried it that way but the results are as follows:

2 1
4 3
6 5
8 7
10 9
12 11
14 13
16 15
18 17
20 19
22 21
24 23
26 25
28 27
30 29
32 31
34 33
36 35
38 37
40 39

Press any key to continue

At least that's what I get with Visual C++ 6.0 service pack 3.

I'm a little confused as to why that is and would love an explanation if someone has one.

Originally posted by Dang
I originally tried it that way but the results are as follows:



At least that's what I get with Visual C++ 6.0 service pack 3.

I'm a little confused as to why that is and would love an explanation if someone has one.

I see what you mean, same thing happened to me. I shortened the code to:

Code:

#include <iostream> 
#include <iomanip> // for use of setw(x) 
using std:: cout; 
using std::endl;
using std::setw;

int main () 
{ 
  for( int count = 0; count <= 20; ) 
  { 
     cout << setw(3) << count++ << setw(25) << count++ << endl; 
   } 

  cout << endl; 

  return 0; 
}

and that gives the results:
1 0
3 2
5 4
.
.
.
etc. I don't know why, I've tried initializing count to 1, to 0, using count++ and ++count. Did the same thing with my original code in the earlier post, same results. Dunno why. Did forget using std:setw; first time.

I am glad it also freaks you guys out. Dont you hate it when you think something should be just a simple little program and it turns out to be a bit of a headache. Still i suppose that is what programming is about and when one of the smarter guys gives a good explanation i think i will be quite enlightened (for a short while )

(response to last couple of messages.)

I'm a little confused as to why that is and would love an explanation if someone has one.

I suspect it's because the expression -

Code:

cout << setw(3) << ++num << setw(25) << ++num << endl;

is being evaluated from right to left.

This should give you the intended output using the above method -

Code:

 #include <iostream> 
#include <iomanip> // for use of setw(x) 

using std:: cout; 
using std::endl;
using std::setw;

int main () 
{ 
   int count=1, num = 0; 

   for(count = 1; count <= 20;count++) 
   { 
      cout << setw(3) << ++num; 
      cout << setw(25) << ++num << endl; 
   } 

cout << endl; 

return 0; 
}

Thanks Zen!

Originally posted by Salem
> cout << setw(3) << --num2 << setw(25) << ( num2 = num1 += 2 ) << endl;
Evaluation order has nothing to do with it - if you modify the same object twice in the same expression, the result is undefined - period.
http://www.eskimo.com/~scs/C-faq/q3.2.html

Ok. But my limited understanding of multiple <<'s is that it's a cascading operator, so that:

Code:

cout << ++num << ++num.....;

is really a series of function calls:

Code:

cout << ++num;
cout << ++num;
.....

At least that's what I get from studying overloaded operators (just starting that). Or is the problem with the ++ operators? Thanks.

But my limited understanding of multiple <<'s is that it's a cascading operator

Yes, if you were calling it like a normal function it would look something like this -

(cout.operator <<(++i)).operator <<(++i);

However, it is undefined as to what or which will be evaluated first. Check this link

Ok, I see from the link why the problem is occurring. The evaluation of the expression is undefined.
I guess the next question is why is expression evaluation in a case like this undefined? Wouldn't it have been better for Bjarne or whoever designs/updates/creates C++ to have the behavior defined? The undefined behavior is counter intuitive, and I don't see a reason behind it.