Question

위8원의 12 55 56 91 A1 B1 C1 B1 00 01 02 UXU303 Ox0210 Ox0214 Ox0218 Ox021C 40 Ox0302 Ox0301 Ox0300 30 20 D1 D2 D3 D4 D5 D6 D7 D8 10 The following MSP430 memory system consists of 2 consecutive 64-Byte RAM chips starting at location ox200, and 512 Byte of Flash ending at 0XFFFF. Fill in the blanks (4) with the correct starting or ending addresses. C Starting Address Starting Address Ending Address (Binary) (Hex) (Hex) 64B RAM1 0000 0010 0000 0000 Ox0200 64B RAM2 OXFFFF 512B FLASH (i0) 1) For each of the following program segments, execute the program one instruction at a time. At the end of each instruction execution, indicate any changes in registers, condition flags, and/or memory locations. Once you compile the program segments below and before starting the execution on one instruction at a time by using the CCS step-over function. you need to manually fill in the initial data in the registers and memory. Note that the result of any instruction may be an operand for succeeding instructions. New Memory Contents Memory Address N ZVC R5 R6 Contents Ox1100 0200 0 0 01 0x0208 0XA006 Instruction Ox2200 0202 O oOx020f Oxf 5 ADDC.W@R5+, R6 ooxo20A Ox3300 0x3300 0204 MOV &0204h, R6 0206 0x4400 MOV.W R6, 0(R5) 0x5500 Ox OO o OX20A 0208 BIC.W #0x3100, R6 Ox 200 0x6600 020A Ox 3300 SUBC 6(R6),-2(R5) New Memory Contents Memory Address NZVC R5 SP Contents 0x5800 0 1 0 1 0x0200 0X020A 0200 Instruction O oo Oxo200 Dx020A RRC.W -4(SP) 0202 0x5700 ooPXoloo ox0208 PUSH 4(R5) 0204 0x5600 802axo oots X 0208 0x5500 0206 XOR -2(SP), R5 o ox0200 0x0208 0x5400 bx660 AND SP, R5 O olo coo0x6600 ox020A 020A 0x5300 POP R5

Answer 1

Answer :- The RAM1 starts from 0x0200 and is of 64 byte. Thus the ending address will be 0x023F. As we need 63 more address locations since starting from zero and 63 = 0x3F. So 0x0200 + 0x003F = 0x023F.

Next, RAM2 starts from 0x0240 i.e. just after the RAM1 ending address. Adding 0x003F to 0x0240 we get the ending address is 0x027F.

Flash has size of 512 byte and ending address is 0xFFFF. 512 in hexadecimal is equal to 0x0200. Hece the starting address is = 0xFFFF - 0x0200 + 0x0001 = 0xFE00.