Texas Instruments TMS320C67X/C67X+ DSP Programming Considerations, Single Assignment Programming

Programming Considerations

5.6 Programming Considerations

The interaction of the C6000 CPUs and sources of interrupts present program- ming issues for you to consider when you are developing your code.

5.6.1Single Assignment Programming

Using the same register to store different variables (called here: multiple assignment) can result in unpredictable operation when the code can be interrupted.

To avoid unpredictable operation, you must employ the single assignment method in code that can be interrupted. When an interrupt occurs, all instruc- tions entering E1 prior to the beginning of interrupt processing are allowed to complete execution (through E5). All other instructions are annulled and refetched upon return from interrupt. The instructions encountered after the return from the interrupt do not experience any delay slots from the instructions prior to processing the interrupt. Thus, instructions with delay slots prior to the interrupt can appear, to the instructions after the interrupt, to have fewer delay slots than they actually have.

Example 5−10 shows a code fragment which stores two variables into A1 using multiple assignment. Example 5−11 shows equivalent code using the single assignment programming method which stores the two variables into two different registers.

For example, suppose that register A1 contains 0 and register A0 points to a memory location containing a value of 10 before reaching the code in Example 5−10. The ADD instruction, which is in a delay slot of the LDW, sums A2 with the value in A1 (0) and the result in A3 is just a copy of A2. If an interrupt occurred between the LDW and ADD, the LDW would complete the update of A1 (10), the interrupt would be processed, and the ADD would sum A1 (10) with A2 and place the result in A3 (equal to A2 + 10). Obviously, this situation produces incorrect results.

In Example 5−11, the single assignment method is used. The register A1 is assigned only to the ADD input and not to the result of the LDW. Regardless of the value of A6 with or without an interrupt, A1 does not change before it is summed with A2. Result A3 is equal to A2.

Example 5−10. Code Without Single Assignment: Multiple Assignment of A1

LDW	.D1	*A0,A1
ADD	.L1	A1,A2,A3
NOP	3
MPY	.M1	A1,A4,A5	; uses new A1

5-22 Interrupts SPRU733