C: bdata unsigned BUFR16=0; sbit BIT8=BUFR16^8; the assemble language which is compiled by Keil uVision3: C:0x1172 9210 MOV BIT8(0x22.0),C ...... ;BUFR16<<=1; C:0x117D E522 MOV A,0x22 C:0x117F 25E0 ADD A,ACC(0xE0) C:0x1181 F522 MOV 0x22,A C:0x1183 E521 MOV A,BUFR16(0x21) C:0x1185 33 RLC A C:0x1186 F521 MOV BUFR16(0x21),A ......
At a guess, the question has something to do with the bit ordering within a U16 declared in bdata. See the note at the end of the Bit-Addressable Objects section in Chapter 3 of the compiler manual. The sbit data type uses the specified variable as a base address and adds the bit position to obtain a physical bit address. Physical bit addresses are not equivalent to logical bit positions for certain data types. Physical bit position 0 refers to bit position 0 of the first byte. Physical bit position 8 refers to bit position 0 of the second byte. Because int variables are stored high-byte first, bit 0 of the integer is located in bit position 0 of the second byte. This is physical bit position 8 when accessed using an sbit data type.