IBM Mainframe Forum

Hi Can we implement the equation like A = X * ((1 + Y)**Z) in mainframe assembler
Where both Y and Z can be a floating point number like 2.35 or 3.256

If yes can you please let me how to do it


Thanks

What have you tried so far, and what results did you get ?
What research have you done - as your question asks if it is possible or not - surely you have googled at least ?

I'm not so confident Mr. Sorichetti's link will be of direct assistance. The key is Y**Z where Y and Z are floating point. I don't think there is a single instruction to do Y**Z. If I ever knew how to do Y**Z, it has certainly flown away from my memory. Other than Y**Z the remainder is parsing out the operands and doing the arithmetic.

It's not so hard for an unregistered person to download the z/Architecture Principles of Operation. I just did that 11 days ago! It just takes some patience and Google.

I've had that book for some time (April 2017). I found it, originally, through http://www.cbttape.org. Awfully long winded, though. As a text book, I'd estimate it has enough material for at least 4 semesters

Actually, when I said I got the z/Arch POP 11 days ago, it turned out I was mistaken. That was the last time I looked at it. The file was allocated August 23.

Raghunaik3 wrote:Hi Can we implement the equation like A = X * ((1 + Y)**Z) in mainframe assembler
Where both Y and Z can be a floating point number like 2.35 or 3.256

If yes can you please let me how to do it


Thanks

Going back to the original question, I wrote up a little Fortran program -
I ran the program through the OS/360 Fortran G compiler in the hope its Assembler listing might provide some insight. It didn't. If I read it right, the code called a library routine to perform Y ** Z.

The output of the WRITE statement was

This seems to be right.

The Windows 7 calculator gives Y ^ Z = 8298.38 (rounded to 2 decimals)

I got essentially the same result from Calc in LibreOffice.