4 / 6 = .67

[MLambert]

I am sure there is an easy way to do this without string manipulation ... how ?

I want 2 decimal not .66666666667  when I divide 2 numbers.

Thks,

Mike

[krovit]

Code: QB64: [Select]

1. ? 4/6
2. ?
3.  
4. calc=4/6
5. ? calc
6. ?
7.  
8. calc#=4/6
9. ? calc#
10. ?
11.  
12. print using" #.##"; calc#
13. ?
14.  
15. ? VAL(mid$(str$(calc#),instr(str$(calc#),"."),3))
16. ?
17.  
18. ? val(mid$(str$(calc#+0.005),instr(str$(calc#+0.005),"."),3))
19. ?
20.  
21.  
22.  

Hello Mike,
it's a little rough and primitive but maybe it helps to understand (but maybe you already knew).

it's not a complication of QB64... is the typical system used by all programming languages.

Maybe I'm wrong but I don't think there is a direct command of the type: round-to-the-second-digit-this-number or the-result-of-this-calculation but you can do it yourself by creating a function.

Put line 18 code (appropriately modified) in a function and it all will become simple.

[SMcNeill]

X = INT(X * 100) / 100 to truncate at 2 decimal places.

X = INT(X * 100 + .5) / 100 to round at 2 decimal places.

[krovit]

Yes, it's the super classic system (which I also use, of course, like everyone else)

My contribution is a complication but made in order to illustrate. In fact, I'd better keep quiet ;)

[MLambert]

Many thks ... I haven't used masking for a long time and couldn't find/see INT.

Mike

[MLambert]

Hi,

tried the INT version ... 4/3 = 133.3333 ?? both INT solutions give me the same answer

i.e.  X = INT(4 * 100 + .5) / 3  or

X = INT(4 * 100) / 3

I would like 1.33 to 2 decimal places

and 4/6 = 1.67 and this rounded

I have tried the string manipulations and it seems to work fine.

Mike

[Qwerkey]

X = INT(X * 100 + .5) / 100 to round at 2 decimal places.

Seems unnecessarily complicated?  Why not the simple:
X = CINT(X * 100) / 100 to round at 2 decimal places?

[SMcNeill]

Seems unnecessarily complicated?  Why not the simple:
X = CINT(X * 100) / 100 to round at 2 decimal places?

Depends on if X * 100 is an integer value, or larger.  ;)

[MLambert]

Hi Krovit,

Using your method ... but it only returns the decimal places ... I need the whole number as well.

5/3 = .67  this is great .. but I want the 1.67.

Thks,

Mike

[STxAxTIC]

Code: QB64: [Select]

1. a = 5
2. b = 3
3.  
4. r = a / b
5. c = INT(r)
6. t = INT(100 * (r - c) + .5) / 100
7.  
8. PRINT c + t

[MWheatley]

X = INT(X * 100) / 100 to truncate at 2 decimal places.

X = INT(X * 100 + .5) / 100 to round at 2 decimal places.

This is how I've always done it, for 40+ years.  Most occasions, it's only once or twice at most in a program, ie for printing results to either screen or printer.  The idea of a function hadn't even occurred to me, because the usage is so rare.

I'd be intrigued to know if anybody has seen the value in rounding via a function, though.

Malcolm

[Petr]

This is way without string conversions. Function parameters are: decimal number, lenght

Code: QB64: [Select]

1. PRINT LNUM(0.0001, 4)
2. PRINT 22 / 7
3. PRINT LNUM(22 / 7, 1)
4. PRINT LNUM(22 / 7, 5)
5.  
6.  
7. FUNCTION LNUM! (num!, lenght)
8.     '1.2345, 2  =>  1.23
9.     integernum = num \ 1
10.     decimalnum = num / 1 - num! \ 1
11.     n = 10 ^ lenght
12.     newdecimalnum = decimalnum * n
13.     newdecimalnum = INT(newdecimalnum)  'or comment this row and rewrite it as  newdecimalnum = newdecimalnum \ 1 for rounding outputs.
14.     newdecimalnum = newdecimalnum / n
15.     LNUM = integernum + newdecimalnum
16. END FUNCTION
17.  
18.  

[SMcNeill]

Aren’t you guys just overcomplicating things?  A single line function is all anyone would ever really need...

Code: [Select]

FUNCTION RoundTo## (Num AS _FLOAT, Precision AS INTEGER)
    RoundTo = INT(Num * 10 ^ Precision + .5) / 10 ^ Precision
END FUNCTION

[Petr]

OK. This is the last thing I added to this forum.

[SMcNeill]

Hi,

tried the INT version ... 4/3 = 133.3333 ?? both INT solutions give me the same answer

i.e.  X = INT(4 * 100 + .5) / 3  or

X = INT(4 * 100) / 3

I would like 1.33 to 2 decimal places

and 4/6 = 1.67 and this rounded

I have tried the string manipulations and it seems to work fine.

Mike

 X = INT(X * 100 + .5) / 100 <— The original statement, which you misapplied here:  X = INT(4 * 100 + .5) / 3

The correct way would be either:
X = 4 / 3
X = INT( X * 100 + .5) / 100

Or:

X = INT( 4 / 3 * 100 + .5) / 100