Can you figure this out?

[willakawill]

A single line in my code looks like this:

result = 0.57 * ((100 - 1) / (100 + 200));

I have replaced all variables except result with literals, as above, and get this same problem;
result is declared as float (same problem when declared as double) and defined as 0.0

result always stays 0.000 when it should be 0.1881
this calculation works in excel and vb and on a calculator.

Clearly I am missing something important. The app is MFC appwizard dialog. Everything else in this function works except for this one line.

Any ideas?

Thanks

[sivadhas2006]

Hi,

You may use like this.

 
result = 0.57 * (double(100 - 1) / double(100 + 200));
 
 

Regards,
M.Sivadhas.

[willakawill]

Hi,

You may use like this.

 
result = 0.57 * (double(100 - 1) / double(100 + 200));
 
 

Regards,
M.Sivadhas.

thank you. Casting is not the problem.
I just built the expression up from scratch to see where the error began and it is in the bracketing. This works:

result = 0.57 * (100 - 1) / (100 + 200);

and this does not work;

result = 0.57 * ((100 - 1) / (100 + 200));

Curious eh?

[sivadhas2006]

thank you. Casting is not the problem.
I just built the expression up from scratch to see where the error began and it is in the bracketing. This works:

result = 0.57 * (100 - 1) / (100 + 200);

and this does not work;

result = 0.57 * ((100 - 1) / (100 + 200));

Curious eh?

Yes,

In the first expression, the result will come like this.

result = 0.57 * (100 - 1) / (100 + 200)
result = 0.57 * 99 / 300
result = 56.43 /  300
result = 56.43 /  300
result = 0.1881
 

In the second expression, the result will come like this.

result = 0.57 * ((100 - 1) / (100 + 200))
result = 0.57 * (99 / 300)
result = 56.43 /  0
result = 0
 

The main reason is due to backet precedence.

Regards,
M.Sivadhas.

[willakawill]

result = 0.57 * (99 / 300)
result = 56.43 / 0

This does not make sense to me.
If we assume that bracketing overrides operator prescidence we get this;

result = 0.57 * (99/300)
result = 0.57 * (0.33)
result = 0.1881

this bracketing works in vb because I copied and pasted the expression and it worked out. vb also uses bracketing to override operator prescidence so something else must be going on with C++ here. I don't know what.

Perhaps the extra set of brackets has caused the (99/300) part to be evaluated as an integer division resulting in 0 thus ending the expression as

result = 0.57 * (0)

[horace1]

thank you. Casting is not the problem.
I just built the expression up from scratch to see where the error began and it is in the bracketing. This works:

result = 0.57 * (100 - 1) / (100 + 200);

and this does not work;

result = 0.57 * ((100 - 1) / (100 + 200));

Curious eh?

in the expression

result = 0.57 * ((100 - 1) / (100 + 200));
 

the () have the highest precedence therefore the subexpression ((100 - 1) / (100 + 200)) is evaluated first giving 99/300 which (as both operands are int) equals 0
hence result = 0.57 * 0 will be 0

in the expression

result = 0.57 * (100 - 1) / (100 + 200);
 

* and / have the same precedence and associate from left to right so
(1) (100 – 1) is evaluated giving 99 (an int)
(2) 0.57 * 99 is evaluated – as 0.57 is a double 99 is converted to a double, therefore result is 56.43
(3) (100 + 200) is evaluated giving 300 an int
(4) 56.43 / 300 is evaluated – as 56.43 is a double 300 is converted to a double, therefore result is 0.1881

[willakawill]

Excellent thanks.
So I see I can get the original to work by changing this;

result = 0.57 * ((100 - 1) / (100 + 200));

to this;

result = 0.57 * ((100 - 1) / (100 + 200 * 1.0));

and it works. or I could just follow the advice given above with the double cast

I did try out this one;

result = 0.57 * double((100 - 1) / (100 + 200));

which fooled me into thinking that it was not a casting issue

Thanks again both of you.

[horace1]

yes, in

double result = 0.57 * double((100 - 1) / (100 + 200));
 

the ((100 - 1) / (100 + 200)) is still evaluated as an int giving 0 and then converted to a double

you need to cast one of the ints to a double then evaluate, e.g.

result = 0.57 * ( double(100 - 1) / (100 + 200));
 

as soon as you have a double the rest of the expression will be converted to double