bit test

Yulquen74 · 03 August 2014, 19:44

silly and stupid, but I need some help to complete a simple routine.
(I should really study 68k assembler more closely someday).

In the start of a program, I want to test the status of bit
15 of a memory address. If it is set, I want to jump to the exit label.

btst #15,$dd0800 ; (config register 1 of cpld)
beq Exit

looking at the address with q$dd0800 I can see that the bit is not set,
but the programs still exits without doing the rest.

thanks in advance for help.

thomas · 03 August 2014, 20:31

beq is wrong. It exits when the bit is not set. bne would be right

Yulquen74 · 03 August 2014, 20:38

thanks, it is.

Quote:

Originally Posted by thomas

beq is wrong. It exits when the bit is not set. bne would be right

mark_k · 03 August 2014, 20:57

Also, remember that BTST on memory is byte-sized; the bit number is modulo 8. So I suggest you have this in your code instead:
btst.b #15-8,$dd0800

Otherwise it's very easy (assuming there is a 16-bit register at $DD0800) to make the mistake of doing btst #6,$dd0800
with the intention of testing bit 6 of the word register. But that actually tests bit 14 of the register.

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03 August 2014, 19:44	#1
Yulquen74 Registered User Join Date: May 2013 Location: Kleppe / Norway Posts: 255	bit test silly and stupid, but I need some help to complete a simple routine. (I should really study 68k assembler more closely someday). In the start of a program, I want to test the status of bit 15 of a memory address. If it is set, I want to jump to the exit label. btst #15,$dd0800 ; (config register 1 of cpld) beq Exit looking at the address with q$dd0800 I can see that the bit is not set, but the programs still exits without doing the rest. thanks in advance for help.

03 August 2014, 20:31	#2
thomas Registered User Join Date: Jan 2002 Location: Germany Posts: 7,000	beq is wrong. It exits when the bit is not set. bne would be right

03 August 2014, 20:57	#4
mark_k Registered User Join Date: Aug 2004 Location: Posts: 3,343	Also, remember that BTST on memory is byte-sized; the bit number is modulo 8. So I suggest you have this in your code instead: btst.b #15-8,$dd0800 Otherwise it's very easy (assuming there is a 16-bit register at $DD0800) to make the mistake of doing btst #6,$dd0800 with the intention of testing bit 6 of the word register. But that actually tests bit 14 of the register.

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