CODEWITHSUNDEEP
assemblyx86
Solomon
Solomon

Reputation: 189

x86 Assembly - whole number rounding issues

I am trying to round to the nearest whole number for my assignment in assembly language and I've been spinning my wheels trying to figure it out. If for example I perform a division function 137/6, how do I get the result to round to the nearest whole number?

Upvotes: 0

Views: 3320

Answers (2)

Weather Vane
Weather Vane

Reputation: 34585

For positive values add half the denominator to the numerator before the division. I leave the more complex situation of negative values to you.

mov ecx, [denominator]      ; divisor
mov eax, ecx                ; copy to numerator register
shr eax, 1                  ; half divisor
add eax, [numerator]        ; add to numerator
div ecx                     ; (numerator + denominator/2) / denominator

Upvotes: 3

rkhb
rkhb

Reputation: 14409

"Round half towards positive infinity" for a positive number (explanations in the comments):

xor edx, edx                ; Clear EDX for division
mov eax, [numerator]        ; Dividend stored in the data section (eg. dd 137)
mov ecx, [denominator]      ; Divisor stored in the data section (eg. dd 6)
div ecx                     ; EDX:EAX / ECX = EAX remainder EDX
shl edx, 1                  ; EDX *= 2
cmp edx, ecx                ; Fraction part < 0.5 (remainder*2 < divisor) ?
jb .done                    ; Yes: skip rounding
add eax, 1                  ; No: round half up (http://en.wikipedia.org/wiki/Rounding#Round_half_up)
.done:                      ; EAX = rounded result of division

Upvotes: 2

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