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Mnemonic Symbol= ic Fla= gs (bit 7-0) Instruction &= nbsp; Number Number = of Number of Comments
&nb= sp; = operation = S Z . H . V N C opcode  = ; of bytes &n= bsp; M cycles T states
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BIT b,r Fz <- NOT r= b X ? X 1 X X 0 * &= nbsp; 11 001 011 (CBh) 2 &nb= sp; 2 &nbs= p; 8  = ; r:
&nbs= p; &= nbsp; &nbs= p; &= nbsp; 01 <b> <r>  = ; &n= bsp;  = ; &n= bsp; 000 C
&nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; 001 BBIT b,(HL) Fz <- NOT (HL)b&n= bsp; X ? X 1 X X 0 * 11 001 011 (CBh) &n= bsp; 2 3&= nbsp; 12 &= nbsp; 010 E&nbs= p; &= nbsp; &nbs= p; &= nbsp; 01 <b> 110  = ; &n= bsp;  = ; &n= bsp; 011 D&nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; 100 LBIT b,(IX+d) Fz <- NOT (IX+d)b &nbs= p; X ? X 1 X X 0 * 11 011 101 (DDh) 4 &n= bsp; 5 &nb= sp; 20 &nb= sp; 101 HBIT b,(IY+d) Fz <- NOT (IY+d)b &nbs= p; X ? X 1 X X 0 * 11 111 101 (FDh) 4 &n= bsp; 5 &nb= sp; 20 &nb= sp; 111 A&nbs= p; &= nbsp; &nbs= p; &= nbsp; 11 001 011 (CBh)&nbs= p; &= nbsp; &nbs= p; &= nbsp; -- <d> ---&nbs= p; &= nbsp; &nbs= p; &= nbsp; 01 <b> 110  = ; &n= bsp;  = ; &n= bsp; b:&nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; 000 0&nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; 001 1SET b,r rb &l= t;- 1 &nbs= p; * * X * X * * * 11 001 011 (CBh) 2 &n= bsp; 2 &nb= sp; 8 &nbs= p; 010 2&nbs= p; &= nbsp; &nbs= p; = 11 <b> <r> &nb= sp; = &nb= sp; = 011 3&nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; 100 4SET b,(HL) (HL)b <- 1 &= nbsp; * * X * X * * * = 11 001 011 (CBh) 2 &n= bsp; 4 &nb= sp; 15 101 = ; 5&nbs= p; &= nbsp; &nbs= p; &= nbsp; 11 <b> 110 &nb= sp; = &nb= sp; = 110 6&nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; &nbs= p; &= nbsp; 111 7SET b,(IX+d) (IX+d)b <- 1 &nb= sp; * * X * X * * * 11 011 101 (DDh)&nb= sp; 4 &nbs= p; 6 23SET b,(IY+d) (IY+d)b <- 1 &nb= sp; * * X * X * * * 11 111 101 (FDh)&nb= sp; 4 &nbs= p; 6 23&nbs= p; &= nbsp; &nbs= p; &= nbsp; 11 001 011 (CBh)&nbs= p; &= nbsp; &nbs= p; &= nbsp; -- <d> ---&nbs= p; &= nbsp; &nbs= p; &= nbsp; 11 <b> 110RES b,r rb &l= t;- 0 &nbs= p; * * X * X * * * 11 001 011 (CBh) 2 &n= bsp; 2 &nb= sp; 8&nbs= p; &= nbsp; &nbs= p; = 10 <b> <r>RES b,(HL) (HL)b <- 0 &= nbsp; * * X * X * * * = 11 001 011 (CBh) 2 &n= bsp; 4 &nb= sp; 15&nbs= p; &= nbsp; &nbs= p; = 10 <b> 110RES b,(IX+d) (IX+d)b <- 0 &nb= sp; * * X * X * * * 11 011 101 (DDh)&nb= sp; 4 &nbs= p; 6 23RES b,(IY+d) (IY+d)b <- 0 &nb= sp; * * X * X * * * 11 111 101 (FDh)&nb= sp; 4 &nbs= p; 6 23&nbs= p; &= nbsp; &nbs= p; &= nbsp; 11 001 011 (CBh)&nbs= p; &= nbsp; &nbs= p; &= nbsp; -- <d> ---&nbs= p; &= nbsp; &nbs= p; = 10 <b> 110
Please note:
The bit to be tested by the BIT instruction is first inverted befo= re copied to Fz, ie. Fz =3D 1 if the corresponding bit is zero. Fz =3D 0 if= the corresponding bit is 1.