Outputting a list of squared Doubles

This is a discussion on Outputting a list of squared Doubles within the C++ Programming forums, part of the General Programming Boards category; Some of you may have seen the program in which I outputted a list of squared numbers starting from 0 ...

  1. #1
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    Outputting a list of squared Doubles

    Some of you may have seen the program in which I outputted a list of squared numbers starting from 0 up until that given number. This time I am trying to do the same thing with a list of doubles. I believed the concept was about the same but I was proven wrong when my program sucked up every last bit of my memory. I think my problem may be where I am using my loops and how it deals with the doubles. Thanks for the help!

    main
    Code:
    #include "squareddoubleheader.h"
    #include <conio.h>
    #include <iostream>
    #include <vector>
    
    using std::cin;
    using std::cout;
    using std::vector;
    
    int main(){
        
        cout << "What is the double you wish to square? ";
        double doub;
        cin >> doub;
        
        vector<double> doubled;
        squareddouble(doub,  doubled);
        
        for(int i = 0; i <= doub; i++){
                cout << i << "       " << doubled[i];
                }
                
        cout << "Press enter to continue..";
        _getch();
        
        return 0;
        
    }
    source
    Code:
    #include "squareddoubleheader.h"
    #include <vector>
    
    using std::vector;
    
    void squareddouble(double doub, vector<double>& squaredoub){
         for(double d = 0.0; d <= doub; doub++){
                    squaredoub.push_back(d*d);
                    }
         }

  2. #2
    The larch
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    Your error is here
    Code:
    void squareddouble(double doub, vector<double>& squaredoub){
         for(double d = 0.0; d <= doub; doub++){
    However, you probably should keep the values that are used as indices integral. Only cast the integral type to double when calculating the square:

    Code:
    void squareddouble(int count, vector<double>& squaredoub){
         for(int d = 0; d <= count; d++){
                    squaredoub.push_back(static_cast<double>(d)*d);
                    }
         }
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  3. #3
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    OK I did the fix and it seems to still be eating my memory away. Nothing gets outputted either.
    Last edited by dnguyen1022; 01-19-2009 at 11:17 AM.

  4. #4
    The larch
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    Post new code and tell us the value that you input.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  5. #5
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    Yes sir.

    source
    Code:
    #include "squareddoubleheader.h"
    #include <vector>
    
    using std::vector;
    
    void squareddouble(double doub, vector<double>& squaredoub){
         for(int d = 0; d <= doub; doub++){
                    squaredoub.push_back(static_cast<double>(d)*d);
                    }
         }
    main
    Code:
    #include "squareddoubleheader.h"
    #include <conio.h>
    #include <iostream>
    #include <vector>
    
    using std::cin;
    using std::cout;
    using std::vector;
    
    int main(){
        
        cout << "What is the double you wish to square? ";
        double doub;
        cin >> doub;
        
        vector<double> doubled;
        squareddouble(doub,  doubled);
        
        for(int i = 0; i <= doub; i++){
                cout << i << "       " << doubled[i];
                }
                
        cout << "Press enter to continue..";
        _getch();
        
        return 0;
        
    }
    I've tried 1, 1.11, 1.0, 10, 3.211...all produce the same result of nothingness.

  6. #6
    The larch
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    Look at my first reply. Have you fixed that issue?
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  7. #7
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    oh...sorry I overlooked that. Got it now, now I have to split up the numbers into columns and rows

  8. #8
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    So given a double, how would I determine its width without the decimals? For example 100.1111, should be a width of 3 for 100. or 1.1 should just be 1.

  9. #9
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    Quote Originally Posted by dnguyen1022 View Post
    So given a double, how would I determine its width without the decimals? For example 100.1111, should be a width of 3 for 100. or 1.1 should just be 1.
    The log10 method we described before works for (positive) integers. So as long as you have a set number of decimals, your width of the whole number can be calculated from that.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  10. #10
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    Actually how would I use this function? Do I just simply replace arr with doub?

    Code:
    for (unsigned i = 0; i != sizeof(arr)/sizeof(arr[0]); ++i) {
            std::cout << arr[i] << ' ' << static_cast<unsigned>(std::log10(arr[i])) + 1 << '\n';
        }
    And how do I declare the array?

    Code:
    for (unsigned i = 0; i != sizeof(doub)/sizeof(arr[0]); ++i) {
            cout << arr[i] << ' ' << static_cast<unsigned>(log10(arr[i])) + 1 << '\n';
        }
    That is what I did but its giving me an arr not declared error.
    Last edited by dnguyen1022; 01-19-2009 at 12:00 PM.

  11. #11
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    Quote Originally Posted by dnguyen1022 View Post
    Actually how would I use this function? Do I just simply replace arr with doub?

    Code:
    for (unsigned i = 0; i != sizeof(arr)/sizeof(arr[0]); ++i) {
            std::cout << arr[i] << ' ' << static_cast<unsigned>(std::log10(arr[i])) + 1 << '\n';
        }
    And how do I declare the array?
    That code is simply to show the lenght in digits of each number in arr[]. You have a vector with your squared numbers, so you need to figre the length of those. The part that does static_cast<unsigned>(std::log10(...) is what you need - the rest is just to walk through the testing array - which is not what you want to do.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  12. #12
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    Oh, okay so I was able to get the length, but its giving me 1 less than the actualy length which I believe is why there is a + 1 in the example code. Is this correct?

  13. #13
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    Quote Originally Posted by dnguyen1022 View Post
    Oh, okay so I was able to get the length, but its giving me 1 less than the actualy length which I believe is why there is a + 1 in the example code. Is this correct?
    Yes - and if you want to have space for "." between the integer and fraction part, you may need to add another 1.

    Note also that log() is invalid for negative numbers.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  14. #14
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    Alright, I know how to deal with it if its a negative number. Thanks for your help!

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