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  1. #1
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    ignore this post - me dumb!

    Cheers

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

  2. #2
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    I am storring x (ranges from 0 - 9) into the integer array array[10].

  3. #3
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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.

  4. #4
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    Thanks.

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

  5. #5
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    Yeh..there probably is..i dunno tho..

    to be honest..my code isnt complex..

    neways

    Hope I helped..

    cheers
    Alex

  6. #6
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    I know it was't complex...

    I just need to get back into programming.

    Thanks.

  7. #7
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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};

  8. #8
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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

  9. #9
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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.

  10. #10
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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

  11. #11
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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.

  12. #12
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    Quote Originally Posted by Kramer55
    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?
    It doesn't "return" the size. A size_t is a type, an unsigned integral type. The result of the sizeof operator has type size_t.

    http://www.lysator.liu.se/c/rat/c3.html#3-3-3-4:
    The type of sizeof, whatever it is, is published (in the library header <stddef.h>) as size_t, since it is useful for the programmer to be able to refer to this type. This requirement implicitly restricts size_t to be a synonym for an existing unsigned integer type, thus quashing any notion that the largest declarable object might be too big to span even with an unsigned long. This also restricts the maximum number of elements that may be declared in an array, since for any array a of N elements,

    N == sizeof(a)/sizeof(a[0])

    Thus size_t is also a convenient type for array sizes, and is so used in several library functions. (See §4.9.8.1, §4.9.8.2, §4.10.3.1, etc.)
    It is suited to an example more like this.
    Code:
    #include <iostream>
    #include <cstddef>
    
    using namespace std;
    
    int main()
    {
       int array[] = {1,2,3,4,5,6,7,8,9,10};
       for ( size_t i = 0; i < sizeof array / sizeof array[0] ; ++i )
       {
          cout << array[i] << endl;
       }
       return 0;
    }
    Then you don't need to hard-code the array size in both the definition and the loop.

    [edit]
    Quote Originally Posted by Tronic
    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;
    }
    What a train wreck of an example.
    Last edited by Dave_Sinkula; 04-20-2005 at 02:11 PM.
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

  13. #13
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    >What a train wreck of an example.
    I doubt this one will make the FAQ.

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