Question

I am having trouble with a C++ object-orientated script. When I create an object, I wish to calculate an AttributeQ based on its attributes MyAValue, MyBValue, and MyCValue.

While using the Visual 2010 debugger, I noticed that TempAttribueQ seems to always be 0 (except before it is initialized of course). Assuming Delta != 0, BVal == Maximum, and DeltaA == DeltaC, then TempAttribueQ should be 1/3 not 0.

At first I thought it was a scope problem, but the variable is defined outside the if-else statements. I have tried initializing TempAttribueQ as some outrageous number, which it keeps up until the if-else statements when it becomes 0 when it shouldn’t.

This is my code…

void SetMyAttribueQ(){

    double TempAVal = MyAValue;
    double TempBVal = MyBValue;
    double TempCVal = MyCValue;

    double Minimum = min(min(TempAVal, TempBVal), TempCVal);
    double Maximum = max(max(TempAVal, TempBVal), TempCVal);
    double Delta = Maximum - Minimum;

    double DeltaA = 0;
    double DeltaB = 0;
    double DeltaC = 0;

    double TempAttribueQ = 0;

    if(Delta == 0) {
        MyAttribueQ = TempAttribueQ; // this->SetMyAttribueQ(TempAttribueQ);
    }
    else {
        DeltaA = /* (a removed equation goes here... */
        DeltaB = /* (a removed equation goes here... */
        DeltaC = /* (a removed equation goes here... */

        if(AVal == Maximum)
            TempAttribueQ = (DeltaC - DeltaB);
        else if(BVal == Maximum)
            TempAttribueQ = (1/3) + (DeltaA - DeltaC);
        else
            TempAttribueQ = (2/3) + (DeltaB - DeltaA);

        MyAttribueQ = TempAttribueQ;
    }
}

What is preventing TempAttribueQ from getting a value of 1/3 or 2/3? Or, what is causing it to be set to be set to 0?

Answer 1

When you divide one integer by another, you get an integer result. Change either or both the constants to non-integer to fix it – C++ rules will result in the other being converted to floating point before the division takes place. All of the following will work:

• 1.0 / 3.0
• 1 / 3.0
• 1.0 / 3

An integer will get converted back to a double invisibly, which is why you weren’t seeing any errors in your code.