Datasheet

Chapter 2. ColdFire Core 2-41
Instruction Timing
certain hardware resources within the processor are marked as “busy” for two clock
cycles after the nal DSOC cycle of the store instruction. If a subsequent store
instruction is encountered within this two-cycle window, it is stalled until the
resource again becomes available. Thus, the maximum pipeline stall involving
consecutive store operations is two cycles.
The OEP can complete all memory accesses without memory causing any stall
conditions. Thus, timing details in this section assume an innite zero-wait state
memory attached to the core.
All operand data accesses are assumed to be aligned on the same byte boundary as
the operand size:
16-bit operands aligned on 0-modulo-2 addresses
32-bit operands aligned on 0-modulo-4 addresses
Operands that do not meet these guidelines are misaligned. Table 2-9 shows how the
core decomposes a misaligned operand reference into a series of aligned accesses.
2.7.1 MOVE Instruction Execution Times
The execution times for the MOVE.{B,W,L} instructions are shown in the next tables.
Table 2-12 shows the timing for the other generic move operations.
NOTE:
For all tables in this chapter, the execution time of any
instruction using the PC-relative effective addressing modes is
equivalent to the time using comparable An-relative mode.
ET with {<ea> = (d16,PC)} equals ET with {<ea> = (d16,An)}
ET with {<ea> = (d8,PC,Xi*SF)} equals ET with {<ea> = (d8,An,Xi*SF)}
The nomenclature “(xxx).wl” refers to both forms of absolute
addressing, (xxx).w and (xxx).l.
Table 2-9. Misaligned Operand References
A[1:0] Size Bus Operations Additional C(R/W)
1
1
Each timing entry is presented as C(r/w), described as follows:
C is the number of processor clock cycles, including all applicable operand fetches and writes, as
well as all internal core cycles required to complete the instruction execution.
r/w is the number of operand reads (r) and writes (w) required by the instruction. An operation
performing a read-modify write function is denoted as (1/1).
x1 Word Byte, Byte 2(1/0) if read
1(0/1) if write
x1 Long Byte, Word, Byte 3(2/0) if read
2(0/2) if write
10 Long Word, Word 2(1/0) if read
1(0/1) if write
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