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Storing data to memory

Storing the register contents uses the same addressing modes. The assembly instructions used for storing are STB , STH , and STW . Read and understand these instructions in the TI manual.

(Storing to memory): Write assembly instructions to store 32-bit constant 53fe 23e4h to memory address 0000 0123h .

Intentionally left blank.

Sometimes, it becomes necessary to access part of the data stored in memory. For example, if you store the 32-bit word 0x11223344 at memory location 0x8000 , the four bytes having addresses location 0x8000 , location 0x8001 , location 0x8002 , and location 0x8003 contain the value 0x11223344 . Then, if I read the byte data at memory location 0x8000 , what would be the byte value to be read?

The answer depends on the endian mode of the memory system. In the little endian mode , the lower memory addresses contain the LSB part of thedata. Thus, the bytes stored in the four byte addresses will be as shown in .

0x8000 0x44
0x8001 0x33
0x8002 0x22
0x8003 0x11

In the big endian mode , the lower memory addresses contain the MSB part of the data. Thus, we have

0x8000 0x11
0x8001 0x22
0x8002 0x33
0x8003 0x44

In this course, we use the little endian mode by default and all the lab programming must assume the little endian mode.

(Little endian mode): What will be the value in A0 after executing the following assembly instructions? (functional unit specificationswere omitted.)

  • MVKL 0x80000000, A10
  • MVKH 0x80000000, A10
  • MVKL 0x12345678, A9
  • MVKH 0x12345678, A9
  • STW A9, *A10
  • LDB *+A10[2],A0
What will be the value in A0 if the system uses the big endian mode?

Intentionally left blank.

In fact, the above addressing method describes the so-called linear addressing mode (default upon reset), where the offset or increment/decrement of pointers occurwithout bound. There is a circular addressing modes that can handle a finite size buffer efficiently. Youwill implement circular buffers for the FIR filtering algorithm in the FIR filtering experiments later.

In the C62x CPU, it takes exactly one CPU clock cycle to execute each instruction. However, the instructions such as LDW need to access the slow external memory and the results of the load are not availableimmediately at the end of the execution. This delay of the execution results is called delay slots .

For example, let's consider loading up the content of memory content at address pointed by A10 to A1 and then moving the loaded data to A2 . You might be tempted to write simple 2 line assembly codeas follows:

1 LDW .D1 *A10, A1 2 MV .D1 A1,A2

What is wrong with the above code? The result of the LDW instruction is not available immediately after LDW is executed. As a consequence, the MV instruction does not copy the desired value of A1 to A2 . To prevent this undesirable execution, we need to make the CPU wait until the resultof the LDW instruction is correctly loaded to A1 before executing the MV instruction. For load instructions, we need extra 4 clock cycles until the loadresults are valid. To make the CPU wait for 4 clock cycles, we need to insert 4 NOP (no operations) instructions between LDW and MV . Each NOP instruction makes the CPU idle for one clock cycle. The resulting code will be likethis:

1 LDW .D1 *A10, A1 2 NOP3 NOP 4 NOP5 NOP 6 MV .D1 A1,A2

or simply you can write

1 LDW .D1 *A10, A1 2 NOP 43 MV .D1 A1,A2

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Finite impulse response. OpenStax CNX. Feb 16, 2004 Download for free at http://cnx.org/content/col10226/1.1
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