可以在序言中模拟一个简单的CPU吗? [英] Possible to simulate a simple CPU in prolog?

问题描述

我的理解是，CPU的简单模型就是状态机.

My understanding is that a simple model of a CPU is a state machine.

当我看序言时，它似乎是在进行树搜索(或图搜索)组合，同时停止运行约束直到找到目标.

When I look at prolog, it appears to tree-search (or graph search) combinations, whilst stopping at constraints running until its goals are found.

有人告诉我，您可以在序言中模拟一个简单的CPU.

I've been told that you can simulate a simple CPU in prolog.

是否可以在序言中表示像简单CPU这样的状态机模型?

Is it possible to represent a state machine model like a simple CPU in prolog?

推荐答案

Prolog是图灵完备的语言，因此您可以在其中表达任意计算，包括CPU仿真.您可以将一条指令的执行表示为CPU的两种状态之间的关系，一种在指令执行之前，另一种在指令执行之后(接下来要执行的指令也是状态的一部分，因为例如可以使用)或通过其中一个CPU寄存器):

Prolog is a Turing-complete language, so you can express arbitrary computations in it, including the simulation of a CPU. You can express the execution of a single instruction as a relation between two states of the CPU, one before the execution of the instruction and one after it (the instruction to be executed next is also part of the state, since it is for example available in or via one of the CPU's registers):

cpustate0_cpustate(State0, State) :-
      ....

一个程序的完整执行可以声明性地描述为CPU的状态转换序列.在程序上，您转换给定状态，直到达到某个定义的最终状态为止，例如，直到某个寄存器包含或指向特定值为止:

A complete execution of a program can be described declaratively as a sequence of state transitions of the CPU. Procedurally, you transform a given state until it reaches some defined final state, for example, until some register contains or points to a specific value:

cpu_before_after(State0, State) :-
    (    final_state(State0) -> State = State0
    ;    cpustate0_cpustate(State0, State1),
         cpu_before_after(State1, State)
    ).

编辑:例如，CPU的状态可能类似于Prolog术语cpu(10, 20, 0) 这可能是特定状态，其中第一个寄存器包含值10，第二个寄存器包含值20，第三个寄存器包含值0.让我们假设第三个寄存器是指令指针IP，并且CPU始终执行IP指向机器RAM中的指令.因此，我们还需要在状态表示中包括机器的RAM.让我们将机器的状态表示为一对RAM-CPU，其中RAM代表机器的RAM内容，而CPU代表CPU的寄存器:

EDIT: For example, a state of the CPU could look like the Prolog term cpu(10, 20, 0) which could be a specific state where the first register contains the value 10, the second register contains the value 20, and a third register contains the value 0. Let us assume that the third register is the instruction pointer IP, and the CPU always executes the instruction that IP points to in the machine's RAM. So we also need to include the machine's RAM in the state representation. Let us represent the machine's state as a pair RAM-CPU, with RAM being a suitable representation of the machine's RAM contents, and CPU a representation of the CPU's registers:

cpustate0_cpustate(State0, State) :-
     State0 = RAM0-cpu(_,_,IP0),   
     ram_at_value(RAM0, IP0, Instruction),
     instruction_state0_state(Instruction, State0, State).

假设现在有一条指令add，该指令将前两个寄存器的值相加并将结果存储在第一个寄存器中，而RAM保持不变:

Suppose now there is an instruction add, which adds the values of the first two registers and stores the result in the first register, leaving the RAM unmodified:

instruction_state0_state(add, RAM-cpu(A0,B,IP), RAM-cpu(A1,B,IP)) :-
    A1 #= A0 + B.

您可以通过该谓词的附加子句来描述其他可用指令的作用.

You can describe the effects of other available instructions by additional clauses of this predicate.

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