ST20-C1 datasheet(184/205 Pages) STMICROELECTRONICS

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The justification of this is as follows. If the depth of both expressions is less than the

number of registers available (

r), then there is no need to store the result of the first

evaluation in a temporary variable. The deeper expression is evaluated first to ensure

that the operation evaluates in the least number of stack registers. If this were not

done then the total depth requirement would have to be incremented due to the extra

location for storing the result of the first e valuation. If both expression depths are as

great as

r, then a local variable (temp) must be used to store the result of the first

expression. It is again better to evaluate the expression with the larger depth if

possible, because this minimizes the number of local variables required. If only one of

the expressions is as great as

r, then provided that expression is evaluated first, there

is no need to store its result in a local variable.

In the cases where a temporary variable

temp is required to hold the value of the first

expression in the evaluation of

e1 op e2, then that variable can be used as a

temporary variable in the evaluation of the first e xpression. Also a temporary variable

used in the evaluation of the first expression and not used to hold its result can be

used in the evaluation of second expression.

The code sequence

(e2; e1; rev; op)

at (*) in the above algorithm, can be optimized further to

(e1; e2; op)

removing the execution of the

rev instruction. But be aware that the latter uses an

extra stack register, and so there is trade-off here between evaluation depth and code

size.

C.4

Loading sequence

The three registers of the evaluation stack are used to hold operands of instructions.

Evaluation of an operand or parameter may involve the use of more than one register.

Care is needed when evaluating such operands to ensure that the first oper and to be

loaded is not pushed off the bottom of the evaluation stack by the evaluation of later

operands. The processor does not detect evaluation stack overflow.

Three registers are available for loading the first oper and, two registers for the second

and one for the third. Consequently, the instructions are designed so that Creg holds

the operand which, on average, is the most complex, and Areg the operand which is

the least complex.

In some cases, it is necessar y to evaluate the Areg and Breg operands in advance,

and to store the results in temporary variables. This can sometimes be avoided using

the reverse instruction. Any of the following sequences may be used to load the

operands

A, B and C into Areg, Breg and Creg respectively.

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