Thread: Question array loop

  1. #16
    Registered User Noir's Avatar
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    You're spoiling the fun
    Heheh, what fun? It's not like this is an NP complete problem.

  2. #17
    Frequently Quite Prolix dwks's Avatar
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    Nope, C++ compilers have to support all of the C headers as well as the C++ version of those headers, but the C headers are deprecated.
    Is that right? Hmm . . . *googles it*

    You're right; if you don't care about namespaces you can use either:
    The C International Standard specifies 18 headers which must be provided by a conforming hosted implementation. The name of each of these headers is of the form name.h. The C++ Standard Library includes the C Standard Library and, hence, includes these 18 headers. Additionally, for each of the 18 headers specified by the C International Standard, the C++ standard specifies a corresponding header that is functionally equivalent to its C library counterpart, but which locates all of the declarations that it contains within the std namespace. The name of each of these C++ headers is of the form cname, where name is the string that results when the ".h" extension is removed from the name of the equivalent C Standard Library header. For example, the header files <stdlib.h> and <cstdlib> are both provided by the C++ Standard Library and are equivalent in function, with the exception that all declarations in <cstdlib> are located within the std namespace.
    From http://publib.boulder.ibm.com/infoce.../stdcpplib.htm
    dwk

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  3. #18
    Registered User Noir's Avatar
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    You're right; if you don't care about namespaces you can use eithe
    Well now, miracles do happen. I was talking out my butt based on stuff I'd heard before, but actually being right is a nice change.

  4. #19
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    I am still bit confussed someone else on another forum said to do it this way as I am new to c++ and not yet heard or the rand function.
    Code:
    int myArray[10];
    int i;
    for ( i = 0; i < 5; i++) // it is offset so 0-9
    {
    //can add a cout here//
    myArray[i] = rand() % 100;
    }
    Surley what that is doing is saying i = o if 5 is less than 0 add one so that gonna got like count up 1,2,3,4,5, but what that doing how is that adding up myArray what is 10,20,30,40,50,60,70,80,90,100 ?

    Or is i setting up the 10 differnt myArrays then the next bit saying

    myArray[i](what will have 9 there = a randomd number modual 100 ?
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  5. #20
    Deathray Engineer MacGyver's Avatar
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    rand() produces a pseudo-random number. When you % it with a number, it's putting a limit on how big the number can be. This means you'll receive a number in between 0 and 99 (inclusive).

    Your index of i inside the for loop will not fill the array. It will only cycle 5 times. This means that array[0] to array[4] will be filled with random numbers from 0 to 99 while array[5] to array[9] will be filled with any arbitrary values that you cannot guarentee.

  6. #21
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    so basically what happen is it start at 0 and then increments it each time applying a random number so the loop goes 0 random number, 1 random number, so on so on untill 9 as I want an Array with 10 values

    then how would i work out the average once the loop has run throught all 9 and gave each a random value?

  7. #22
    The larch
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    Add them up (for example in the same loop) and divide by how many elements your array happens to have.

  8. #23
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    so if that add a random value to each array how do I then add them and divide by 10?

  9. #24
    Deathray Engineer MacGyver's Avatar
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    Create a variable... call it total and set it to 0. Now write a loop that goes through every element of your array and adds it to total and stores the result inside total.

    Hint:

    Code:
    ...
    total += array[i];
    ...
    Now when you're done with that, divide total by the size of the array, and that will be the average of all numbers inside your array.

  10. #25
    Lean Mean Coding Machine KONI's Avatar
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    Quote Originally Posted by MILLWALLL
    so if that add a random value to each array how do I then add them and divide by 10?
    That question has been answered in post #2, #4 and #13.

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