CODEWITHSUNDEEP
cbinarylogicbit-manipulation
seg_fault
seg_fault

Reputation: 21

How to multiply an int with a fraction

I need to multiply an int by a fraction using bitwise operators without loops and such.

For example, I need to multiply by x by 3/8.

I thought you would:

int value = (x << 1) + x;  // Multiply by 3
value = (value >> 3);          // Divide by 8

But that doesnt work. I tried googling binary times fraction but that give floating point examples. I dont know exactly if this homework is for floating point but my hunch is no but getting me prepared for it. So any suggestions?

I need to round toward zero so any suggestions? This doesnt work for the number -268435457.

Upvotes: 2

Views: 4775

Answers (3)

Peter Lawrey
Peter Lawrey

Reputation: 533432

To avoid an overflow you can cast to (long long) and back to (int) for the final result. To cause the >> 3 to round toward zero you need to add 7 (8-1) for a negative number. There are a few ways of doing this. Using these approaches gives you -100663296.

Upvotes: 0

Foo Bah
Foo Bah

Reputation: 26251

You probably want

int value = (x << 1) + x;
value = (value >> 3);

note that:

(x << 1) + 1 = 2*x + 1; // ignoring issues about overflow

To adjust for negative values, you can explicitly check for sign:

int value = (x << 1) + x;
value = value >> 3;
value = value + ((x >> 31) & 1); // for 32 bit; for 64 bit you have to use x >> 63

Upvotes: 4

grdvnl
grdvnl

Reputation: 636

Did you try:

int value = (x << 1) + x;  // Multiply by 3
value = (value >> 3);      // Divide by 8

i.e. in your second statement replace 'x' with 'value'. Also, value will loose the decimal points.

Upvotes: 1

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