Function Returning Structures

This is a discussion on Function Returning Structures within the C++ Programming forums, part of the General Programming Boards category; I just want to know, how to use the function "sum" effectively, returned in the following code. distance is a ...

  1. #1
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    Function Returning Structures

    I just want to know, how to use the function "sum" effectively, returned in the following code.

    distance is a structure.

    Code:
    distance &sum(distance L1, distanceL2);
    distance L3; // global structure
    
    distance &sum(distance L1,distance L2)
    {
    L3.feet=L1.feet+L2.feet+(L1.inches+L2.inches)/12;
    L3.inches=(L1.inches+L2.inches)%12;
    return L3;
    }

  2. #2
    Frequently Quite Prolix dwks's Avatar
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    Well . . . sum() takes two distance structures, adds their contents and stores the result in the global variable L3, also returning a reference to L3. You could use it like so:
    Code:
    distance one = {1, 9}, two = {2, 11};
    distance both = sum(one, two);
    In my opinion it's not written very well. First of all, L1, L2, and L3 are very bad variable names. They could be anything. "result", "one" and "two" might be better. Secondly, using a global variable for just that is extremely wasteful. And why return a reference to the variable? It's already global . . . .
    dwk

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    #1 : SUPPOSE sum() returned a value, is it right to do

    cout << sum(L1,L2).feet;
    -------------

    #2 How to print values of L3 in the original case, where sum() returns a reference to L3 case?

  4. #4
    Frequently Quite Prolix dwks's Avatar
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    Quote Originally Posted by pritin View Post
    #1 : SUPPOSE sum() returned a value, is it right to do
    Code:
     cout << sum(L1,L2).feet;
    You mean, will that work? Sure. This would work too:
    Code:
    sum(L1, L2);
    cout << L3.feet;
    #2 How to print values of L3 in the original case, where sum() returns a reference to L3 case?
    You mean, how would you print the previous value of L3 after sum() has modified it? You can't. That's why I don't like using global variables.

    You'd have to save the value first:
    Code:
    distance prev = L3;
    sum(L1, L2);
    // L3 is the new
    // prev is the old L3
    dwk

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    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
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