What does SETB 53h mean in 8051 assembly language?

I understand that SETB in 8051 sets a bit in a particular register. So if it is addressed as:
SETB P1.5  or SETB ACC.2, it sets the 5th bit of port1 or 2nd bit of ACC.
But what does SETB 53h mean? Which bit of which register does it set? There are other similar instructions like CPL 64h or JB 53h.
SETB 53h
CPL 64h
JB 76h

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przachAsked:
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t0t0Commented:
Bits 00h to 7Fh are the user's program's variable bit memory area. SETB 53h basically sets the corresponding memory bit 55h to '1'.

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t0t0Commented:
Sorry....typo:

Bits 00h to 7Fh are the user's program's variable bit memory area. SETB 53h basically sets the corresponding memory bit 53h to '1'.
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t0t0Commented:
Forgot to mention. You should look up the 8051's memory map for a better understanding. See:

ftp://ftp.hte.com/uconline/library/8051info/51memmap.pdf
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t0t0Commented:
Take a look at the second page of:

http://www.nxp.com/acrobat_download/various/80C51_FAM_PROG_GUIDE_1.pdf

you'll notice the bits 00-7F are laid out in 8-bit widths (don't ask me why...). You'll notice 53h falls in the region covered by the 11th byte, more specifically, the 4th bit (counting from the left) of byte 11.

I tend to think of it as a single linear address range and just count to the 83rd bit.

Just to recap then, 53h is the 83rd (or so) bit in a 128-bit wide (16 bytes) user memory area starting at 00 and ending at 7Fh.
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przachAuthor Commented:
Thank you very much. That makes it clear.
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t0t0Commented:
Thank you.
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