IBM Mainframe Forum

if You look at the assembler listing ( the expansion of the instruction )
You will understand better what we are trying to tell You

I havent understood this completely.Is it that X'00' and XXXXBITS+1 point to the same storage location ?

NO! When assembled, the code that is to be executed for the TM 0(R5),X'00') is X'91005000' -- 91 is the op code, 00 is the test byte, 5 is the register and 000 is the offset from the register. The X'91' is stored at location XXXXBITS, the X'00' at XXXXBITS+1, and the X'5000' at XXXXBITS+2 and +3. So if you change XXXXBITS+1, then you change -- during execution of the program -- the TM from looking at X'00' to whatever you changed the byte to.

Thanks Robert..That clarified my doubts.

ramkumar1992sp wrote:Thanks Steve.

I see the instruction is modified using Register RB before the TM instruction is executed.This isn't a new program.It looks like it was written in 1972.

When the TM instruction is executed,aren't we checking the contents of RB register against the mask bits of zeroes.How is the value of XXXXBITS affecting it ?

Select all

LA    RC,XXXXBITS
---
---
NEWMEET  MVC   XXXXBITS+1(1),1(RB)  
-----
----
---
XXXXBITS TM    0(R5),X'00'          
         BC    5,SETCONFL          

 

Thanks,
Ram Kumar

No no, and no.

When the TM instruction is executed,aren't we checking the contents of RB register against the mask bits of zeroes.How is the value of XXXXBITS affecting it ?

Register 5 is an address. 0(5) directs the machine to compute an address by adding 0 to the contents of register 5. The contents of the mask in the instruction are then used to test the contents of the storage location specified by the address.

LA    RC,XXXXBITS

If register C is the program base register, as is often, though by no means always, true, the instruction is corrupting the register. The net effect is oops,

MVC   XXXXBITS+1(1),1(RB)

If this instruction is actually in your program and executed, the X'00' in the TM instruction is altered with the contents of a byte in storage whose address is computed by adding 1 to the contents of register B. This was poor programming in 1972, and is still poor programming.

You really need to hire someone who actually knows Assembler to correct your program. Trying to correct it using these forums will more likely than not foul it up worse. Me? I will not add to the confusion any more!