C++ bool function

[thando7]

i am a nb on this portal and i require assistance with the foll:
I have to write a C++ boolean function, that determines whether an integer parameter is a square or not, and returns the answer to main( ).

could you plz assist me with that. I have tried using the square root function but it does not seem to work

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[JosAH]

i am a nb on this portal and i require assistance with the foll:
I have to write a C++ boolean function, that determines whether an integer parameter is a square or not, and returns the answer to main( ).

could you plz assist me with that. I have tried using the square root function but it does not seem to work

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Show us what you've tried and maybe we can help you.

kind regards,

Jos

[thando7]

bool IsSquare(int Num)
{
    bool Flag = true;
    for (int I = 2; I < (sqrt(I) / 2); I++)
        if ((Num % I) == 0)
        {
            Flag = false;
            break;
        }
    return Flag;
}    

i have tried something of this sort, but the only problem is that
the square root function is a type double and what confuses me is how to find
out whether a number is a square root or not

[Savage]

Hi

your for loop will never execute because sqrt(2)=1.4142.../2=0.705706...

and that is way more smaller then 2.

(And please don't double post your question.Just post in same thread again to
give it a push up.)

Savage

[Ganon11]

As Savage said, please don't double post your question. The threads have been merged, and your repeat post has been deleted.

I think you should be using sqrt(Num) (the number you are testing for square-ness) instead of sqrt(l) (the index variable).

[nmadct]

You could use the sqrt function to compute the square root of a given number, but since it uses a double-precision floating point value, your results may be slightly off from the true answer. You will probably want to test whether the result is *almost* an integer, i.e. less than 0.0000001 off from an integer value or something like that. If the square root is an integer (or "close enough"), then the number you started with is a perfect square.

Another way to do it, more accurate but probably slower, would be to try squaring a sequence of integers, counting up monotonically until the result either hits the target number (it is a perfect square) or passes it up (it's not a perfect square). I hope that makes sense.

The advantage of the second approach is that you do everything with integers so you can be sure of the correctness of your results.