Re: Integer arithmetic, multiplication overflow

If the result register of an integer multiplication instruction is twice as wide as the factor registers then overflow will never occur.
I guess that in all processor architectures this is the case.

For instance, in IBM S/360 the integer multiplication instruction reads in assembler language format "M R1,D2(X2,B2)".
Here the first operand R1 (an even number) represents an even-odd pair of 32-bit registers.
The odd-numbered register R1+1 contains the first factor, a 32-bit signed integer.
The second operand D2(X2,B2) represents the second factor, a 32-bit signed integer in main storage.
The outcome of the multiplication is placed as a 63-bit signed integer in the two registers R1 and R1+1.
One checks easily that overflow never occurs, neither at the 2^63-1 end, nor at the -2^63 end.

BTW, You might get more and better response in a computer-oriented newsgroup.

Johan E. Mebius

Tomek wrote:

Hi,

I trying to solve the following thing:
Having two finite precision unsigned integer numbers A and B,
and knowing that maximal represented number is X,
calculate A*B - X, in case of A*B overflow. Use only
basic algebraic operations.

Any hints welcome.
Thanks,

Tomek



.

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