可以在 prolog 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 的简单模型是状态机.

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

当我查看 prolog 时，它似乎是树搜索(或图形搜索)组合，同时在运行的约束条件下停止，直到找到其目标.

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

我听说你可以在 prolog 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 prolog 中将状态机模型表示为一个简单的 CPU?

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

推荐答案

Prolog 是一种图灵完备语言，因此您可以在其中表达任意计算，包括模拟 CPU.您可以将单条指令的执行表示为 CPU 的两个状态之间的关系，一个在指令执行之前，一个在指令执行之后(接下来要执行的指令也是状态的一部分，因为它例如可用在或通过 CPU 的寄存器之一):

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) :-
      ....

程序的完整执行可以声明性地描述为 CPU 的一系列状态转换.在程序上，您转换给定状态直到它达到某个定义的最终状态，例如，直到某个寄存器包含或指向特定值:

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)这可能是一个特定的状态，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，CPU总是执行IP 指向机器 RAM 中的指令.所以我们还需要在状态表示中包含机器的 RAM.让我们将机器的状态表示为一对 RAM-CPU，其中 RAM 是机器 RAM 内容的合适表示，而 CPU 是 CPU 寄存器的表示:

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，它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.

您可以通过此谓词的附加子句来描述其他可用指令的效果.

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

这篇关于可以在 prolog 中模拟一个简单的 CPU 吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！