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Started by chrisparton1991 Aug 8, 2006 at 12:48 AM 7 replies 1.3k views

Original Post

chrisparton1991

Aug 8, 2006 at 12:48 AM

Hi ppl. I was mucking around with functions and I realised that you can't access an integer/floating point/etc outside of its function :( I know this isn't big news, to you or myself, but is there any workaround to solve my problem? This is my function:


void getTriangleArea(float base, float height, bool output)
{
    float area = (0.5 * base) * height;//formula
    if(output == true)//if you want it outputted
    {
        cout << area << endl;//outputs triangle area
    }
    else//or not
    {}
}

I was trying something like this(assuming my function is in a header file):


#include <iostream>
#include <stdio.h>
#include "myHeader.h"

using namespace std;

int main()
{
getTriangleArea(30, 10, false)
cout << area << endl;

system("pause");
return 0;
}

Of course, that doesn't work and even if it did, it would be useless. Just an example. Can my problem be solved while keeping my function? Or will I have to use some sort of class instead? [Edited by - chrisparton1991 on December 4, 2009 7:18:58 PM]

jeroenb

Aug 8, 2006 at 12:58 AM

If you want to output the area in the main program make your program this way:

float getTriangleArea(float base, float height){    return area = (0.5 * base) * height;//formula}

What I did was making it a real function: having a return type (a function without a return type (void) is theoretically not a function but a procedure. Then you can use this function like this:

void main(){   float area = getTriangleArea(30, 10);   cout << area << endl;}

chrisparton1991

Author

chrisparton1991

Aug 8, 2006 at 4:50 AM

Thanks for your help, but I can't get it to work how I want it.

I was trying to get it to work in something like this:

#include <iostream>#include <stdio>#include "myHeader.h"getTriangleArea(50, 100, false);float whatever = (area * 2.5);system("pause");return 0;

Once again, just an example ;)

Am I completely missing the point here?

Wan

154

Wan

Aug 8, 2006 at 5:01 AM

The variable area only exists in the scope of the function getTriangleArea.
If you want to be able to access it, you'll have to declare it in a global scope.
This would work for example:

#include <iostream>#include <stdio>#include "myHeader.h"float area = 0.0f;  // declare areavoid getTriangleArea(float base, float height, bool output){    area = (0.5 * base) * height; // don't redeclare    // etc..}int main(){    getTriangleArea(30, 10, false)    cout << area << endl;    // etc..}getTriangleArea(50, 100, false);float whatever = (area * 2.5);

But it would be extremely bad practice. Like previous posters already pointed out, you want the function getTriangleArea to return the area.

jflanglois

Aug 8, 2006 at 5:11 AM

Quote:
Original post by chrisparton1991
Thanks for your help, but I can't get it to work how I want it.

I was trying to get it to work in something like this:

*** Source Snippet Removed ***

Once again, just an example ;)

Am I completely missing the point here?

You kind of are. Take a look at Scope and Lifespan. The above should be:

float area = getTriangleArea( 50, 100, false );float whatever = (area * 2.5);

That said, it is possible to do it kind of like you want (edit: WanMaster shows something closer with globals):

struct TriangleArea {  operator()( float base, float height ) {    area = (0.5 * base) * height;  }  float area;};int main() {  TriangleArea ta;  ta( 30, 10 );  cout << ta.area << endl;}

I don't quite see the point of it, though.

jfl.

Zahlman

257

Zahlman

Aug 8, 2006 at 12:05 PM

Quote:
Original post by Anonymous Poster
Did you know threre's a "for beginners" section in these forums?

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