Pascal- How to convert Real to Integer variable

Question

I'm writing a task in pascal.

Everything is ok, just my result is not right.

I'm summing some numbers

Example: 2.3 + 3.4+ 3.3 = 9

But output shows: 9.000000 + EEE or something like that.

So- how to convert, to be only 9, not this REAL variable.

Answer 1

To actually convert:

var
  i: integer;
...
  i := round(floatVar);

To output only the integer part:

writeln(floatVar:9:0);

Answer 2

For purpose of showing pretty output on the screen you can use something like this:

Writeln(result:0:2);

Result on screen would be this:

9.00

What this means someone would ask? Well first number 0 means how wide filed is. if you say it's 0 then Pascal writes it at the very left side of screen. If you said writeln(result:5:2) result would be:

   9.00

In other words i would print form the right side and leave 5 chars to do so.

Second number 2, in this example means you want that result printed with 2 decimal places. You can place it only if you want to print on screen value that is real, single, double, extended and so on.You can round to any number of decimals, and if you do writeln(result:0:0) you would get ouput:

9

If you are printing integer and want to have some length of field, lets sat 5 you would do: writeln(int:5). If you added :2 to the end you would get compile time error.

This all also works for something like this: writeln(5/3.5+sqrt(3):0:3),

You should know that this does not round variable itself but just formats output. This is also legal:

program test;
var
    a:real;
    n,m:integer;
begin
    readln(a,m,n);
    writeln(a:m:n);
end.

What i did here is i asked user if on how many decimals and with what length of field he wants to write entered number a. This can be useful so i'm pointing it out. Thank you for reading. I hope i helped

Answer 3

Let's consider this quite simpler equation:

3.5 + 2.5

What do you expect? 6, right? Let's try this code

write(3.5 + 2.5);

Unfortunately, it's a floating-point number, so it would produce a number represented in a scientific way:

6.00000000000E+00

or 6.0000000000 x 10⁰, or 6 x 10^o. Whatever, you only care about 6, who need this weird useless long number? So the idea is to cut off the decimal part and output to the console only the integer part, which can be done with this line of code:

write(3.5 + 2.5 : 0 : 0);

Ok, now it outputs a beautiful number as expected

6

Seems like the problem is solved, but you say that:

I'm summing some numbers

Example: 2.3 + 3.4+ 3.3 = 9

Ohh so that the evenly, beautiful integer is just randomly appeared? Here the problem comes, how do you expect this equation would output?

3.6 + 2.5

It should be 6.1, right? Let's try it with the worked line of code:

write(3.6 + 2.5 : 0 : 0);

And the output is...

6

Unexpected, right? So how about rounding to some decimal places, like 1?

write(3.5 + 2.5 : 0 : 1);
write(3.6 + 2.5 : 0 : 1);

Then, 3.5 + 2.5 = 6.0 and 3.6 + 2.5 = 6.1. But 6.0 may look quite long, so how to make it output 6 for 6.0 and 6.1 for 6.1?

Actually, you can't make the program auto-detect if a real variable contains an integer value because the way a real var is stored is completely different from an integer var (how different they are, please contact Google; but you can do it manually by making a function to do the job).

So my solution is, to be easy, making the output rounded to some decimal places, and that's it.

Answer 4

You can convert to string, get the int part, e convert to int number!

Or Float to Str than Str to Int:

nPage := StrToInt(FloatToStr(Int(nReg / nTPages))) + 1;