Texas Instruments MSP50C614 ±4. Data Memory Example, Which uses the absolute word memory address

Models: MSP50C614

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Bit, Byte, Word and String Addressing

Figure 4±4. Data Memory Example

Absolute Word

Data Memory Location (even) = 2 *

MS Byte

LS Byte

Data Memory

Memory Location

(Absolute word memory location)

Location (odd)

 

 

 

 

 

 

 

 

0x0000

0x0000

 

0x12

0x34

0x0001

 

 

 

 

 

 

0x0001

0x0002

 

0x56

0x78

0x0003

 

 

 

 

 

 

0x0002

0x0004

 

0x9a

0xbc

0x0005

 

 

 

 

 

 

0x0003

0x0006

 

0xde

0xf0

0x0007

 

 

 

 

 

 

0x0004

0x0008

 

0x11

0x22

0x0009

 

 

 

 

 

 

0x0005

0x000a

 

0x33

0x44

0x000b

 

 

 

 

 

 

Example 4.5.4

MOV STR, 4±2

 

 

 

 

MOV AP0, 2

 

 

 

MOVBS A0, *0x0003

Refer to Figure 4±4 for this example. The byte-string length is 4. It is loaded to the string register (STR) in the first instruction. AP0 is 2 and it points to AC2. Third instruction loads the value of the string at byte address, 0x0003, and subsequently stores its contents into four consecutive accumulators starting from AC2. The result is, AC2 = 0x0078, AC3 = 0x009A, AC4 = 0x00BC, AC5 = 0x00DE.

Example 4.5.5

MOV

STR,

4±2

 

MOV

AP0,

2

 

MOVS A0,

*0x0003

Refer to Figure 4±4 for this example. The byte-string length is 4. AP0 is loaded with 2 and points to AC2. The third instruction loads the value of the string at address 0x0002 (LSB bit is assumed 0) and stored into four consecutive accumulators starting from AC2. The result is, AC2 = 0x5678, AC3 = 0x9ABC, AC4 = 0xDEF0, AC5 = 0x1122. Same result can be obtained by replacing the third instruction by,

MOVS A0, *0x0001 * 2

which uses the absolute word memory address.

Example 4.5.6

MOV STR, 4±2

 

OV AP0,

2

 

MOV R0,

0x0005

MOVBS A0, *R0++

Refer to Figure 4±4 for this example. The byte string length is 4. AP0 points to AC2. R0 is loaded with 0x0005. The fourth instruction loads the value of the byte-string at the address in R0 (i.e, 0x0005 in byte mode). R0 auto±incre- ments by 1 after every fetch and stores the RAM contents into four consecutive accumulators starting from AC2. The result is, AC2 = 0x00BC, AC3 = 0x00DE, AC4 = 0x00F0, AC5 = 0x0011. There were four byte fetches and the new value of R0 = 0x0009.

Assembly Language Instructions

4-47

Page 139
Image 139
Texas Instruments MSP50C614 manual ±4. Data Memory Example, Which uses the absolute word memory address