Re-Learning

This is a discussion on Re-Learning within the C++ Programming forums, part of the General Programming Boards category; Hi, I have a quick question on some old code I have that I am going through ( I haven't ...

  1. #1
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    Re-Learning

    Hi, I have a quick question on some old code I have that I am going through ( I haven't done any programming for about 10 months)

    Code:
    #include <iostream>
    
    using namespace std;
    
    int main()
    {
    	int array[10];
    	int x;
    
    	for(x = 0; x < 10; x++)
    	{
    		cout << array[x] << endl;
    	}
    
    	
    
    	return 0;
    }
    I cannot for the life of me figure out what is wrong with the program.

    I do not receive any errors, but the output is 10 values of
    -858993460.

    What am I doing that is completely wrong?

    Thanks, from a n00b re-learning C++.

    (I plan on looking back on my old books and the tutorials on this site, but I just couldn't see what was wrong.)

  2. #2
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    ignore this post - me dumb!

    Cheers

    Alex
    Last edited by macman; 04-17-2005 at 02:00 PM.

  3. #3
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    I am storring x (ranges from 0 - 9) into the integer array array[10].

  4. #4
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    KK..i think ive got it...

    u need summit like this:-

    Code:
    #include <iostream>
    using namespace std;
    
    int main () {
    	
    	int array[10], x;
    	
    	for(x=0; x < 10 ; x++) array[x] = x;
    	
    	//displays array
    	for(x=0; x < 10; x++)
    		cout  << array[x] << "\n";
    	
    
    	
    	
    	
    	
        return 0;
    }
    cheers

    Alex
    Last edited by macman; 04-17-2005 at 02:01 PM.

  5. #5
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    Thanks.

    Is there a simpler way? I thought I remembered shorter ways to make an array.

  6. #6
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    Yeh..there probably is..i dunno tho..

    to be honest..my code isnt complex..

    neways

    Hope I helped..

    cheers
    Alex

  7. #7
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    I know it was't complex...

    I just need to get back into programming.

    Thanks.

  8. #8
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    Quote Originally Posted by Kramer55
    Thanks.

    Is there a simpler way? I thought I remembered shorter ways to make an array.
    You can initialize when you declare.

    Code:
    int array[10] = {0,1,2,3,4,5,6,7,8,9};

  9. #9
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    yeah..that would work..

    Its just depends on the amount of intergers..
    mine works well for large numbers..but hey..theres a choice now

    Cheers

    Alex

  10. #10
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    Just one question, what is the reason for

    Code:
    	for(x=0; x < 10 ; x++)
    	{ 
    		array[x] = x;
    	}
    Does that give the first value of x (0) in the for loop and assign it to x in the array[x] = x where that x is then used in the next for loop?

    If so, that seems almost like a pointer.

  11. #11
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    Code:
     	Just one question, what is the reason for
    
    Code:
    
    	for(x=0; x < 10 ; x++)
    	{ 
    		array[x] = x;
    	}
    This effectively does this:

    array[0]=0
    array[1]=1
    array[2]=2
    .
    .
    .
    array[9]=9


    so if you were to test something like this:

    cout<<array[2];

    the output would be '2'.

    Go read up on the tutorial for arrays in this site me thinks.


  12. #12
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    Quote Originally Posted by Kramer55
    I am storring x (ranges from 0 - 9) into the integer array array[10].
    the code that I gave u..gives exactly what u wanted..

    array[0]=0
    array[1]=1
    array[2]=2


    so i agree with treenef.. Some revision is needed me thinks..but thats cool..we are all here to learn. I would suggest biting into a good c++ book..with projects n stuff..

    Neways

    Good luck with your revision..and on your path to mastering c++!

    Cheers

    Alex

  13. #13
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    Another quick question, what is the point of having size_t in this code?

    I know it returns the size, but what is the point of having the size?

    Code:
    #include <iostream>
    
    using namespace std;
    
    int main () {
        
        int array[10];
        for ( size_t i = 0; i < 10 ; ++i ) { 
            array[i] = i;
            cout << array[i] << endl;  // or just cout << i << endl;
        }
    }
    Thanks.

  14. #14
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    The reason it's doing that is because you did not define the contents of the array, so you can't expect it to print out 0-9 or something like that.

  15. #15
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    Is that a typedef? A class name?

    I'm going to assume a typedef, which is a really funky typedef unless it's going to be used alot.

    When you have this code:
    Code:
    for(int i = 0; i < i < 10; i++)
    {
    
    }
    It will iterate 10 times, or more specifically, every time i is less than the value 10, and greater than or equal to the value of zero. Sometimes people confuse the "byte size" of a variable, with the "max-value" of a variable.

    Say you have the code..
    Code:
    #include <iostream>
    #include <conio.h>
    
    int main()
    {
      char *name;
      std::cout << "Enter in name\n" 
                    << "> ";
        std::cin >> name;
      for(char *c = name; *c < sizeof(name); c+=sizeof(char))
      {
    	  std::cout << *c;
      }
      getch();
    return 0;
    }
    This is an example of iterating through a sequence of variables by way of the size of the variable. Most often you'll be dealing with the size of a variable when you're working with basic optimizations and pointers.

    An alternate way of doing it would be to just use ints, as in my first example.

    Typedef'ing size_t only serves to confuse someone that looks at your code, is size_t relating to the size of the variable in bytes? In indices? Or is it just some confusing naming convention?
    OS: Windows XP Pro CE
    IDE: VS .NET 2002
    Preferred Language: C++.

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