Using hexadecimal values for all of your answers, determine the one X affected b

Question

Using hexadecimal values for all of your answers, determine the one X affected by each of the following instructions. MOVE.W 10002000 MOVE.W $1000 >1 MOVE.B 1000JD1 CLR.L SFFFFFC 4.42. Compare the operation of the following instructions when (Al) = 10000 and (1000) FFE016 before each execution. MOVE.W A1,D1 MOVE.W (A1),D1 MOVE.W 1000, D1 MOVEJW #1000JD1 Determine the contents of the destination in hexadecimal after each instruction MOVE.W #'AB',D1 MOVE.W #$C1,D1 MOVE.W #1000, D1 Using only the instructions and techniques discussed thus far. write the symbolic instructions to store the low-order word of D1 into memory locations 1001 and 100- is* after execution (1001) = (Dl) [15:8] and (1002) = (Dl) [7:0]. (Remember That word length operands must start at even locations in memory and that they occupy 2 bytes) MACHINE LANGUAGE FOR THE MC68000 The instructions that the processor executes are coded in machine language formal in memory. These instructions can be created by a program programmer sequences directly. More likely, an assembler program translates assembly language

Explanation / Answer

Just to be sure all the instructions have been tested with Easy68K simulator.

1) MOVE.W 1000,2000

the content of memory location 1000(decimal) =03e8(hexa) is moved into the memory location 2000(decimal)=07D0(hexa). The moved data has the length of 1 word =10 bits

MOVE.W $1000,D1

The content of the memory location 1000(hexa) is moved into data register D1. The moved data has the length of 1 word.

MOVE.B 1000, D1

The content of the memory location 1000(decimal) =03e8(hexa) is moved into (the lower byte) of D1 register. The moved data has the length of 1 byte (8 bits).

CLR.L $FFFFFC

The content of the memory location 00FFFFFC(hexa) is cleared. All bits of are set to zero. The length of the cleared data is 1 Long Word =2 Words =4bytes=32 bits.

3) MOVE.W #'AB', D1

the direct data 'AB' (text) =4142(hexa) is moved in data register D1. The content of D1 will be D1 =00004142(hexa) if initially D1 was cleared, or D1=xxxx4142(hexa) if initially D1 was containg a nonzero value.

MOVE.W #$C1, D1

the direct data C1(hex) is moved into data register D1. The content of D1 will be D1=000000C1(hexa) (if initially D1 was cleared) or D1 =xxxx00C1 (if initially D1 was containing a value)

MOVE.W   #1000,D1

the direct data 1000(decimal) =03e8(hexa) is moved into data register D1. The content of D1 will be D1=000003e8(hexa) (D1 initially cleared) or D1=xxxx03e8 (D1 initially nonzero).

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