(Z3Py) 函数声明有什么限制吗? [英] (Z3Py) any limitations in functions declaring?

问题描述

函数声明有什么限制吗?

Is there any limitations in functions declaring?

比如这段代码返回unsat.

For example, this piece of code returning unsat.

from z3 import *

def one_op (op, arg1, arg2):
    if op==1:
        return arg1*arg2
    if op==2:
        return arg1-arg2
    if op==3:
        return arg1+arg2

    return arg1+arg2 # default

s=Solver()

arg1, arg2, result, unk_op=Ints ('arg1 arg2 result unk_op')

s.add (unk_op>=1, unk_op<=3)

s.add (arg1==1)
s.add (arg2==2)
s.add (result==3)
s.add (one_op(unk_op, arg1, arg2)==result)

print s.check()

Z3Py 如何解释声明的函数?它只是偶尔调用它还是一些隐藏的机器也在这里?

How Z3Py interpret declared function? Is it just calling it some times or some hidden machinery also here?

推荐答案

在函数调用one_op(unk_op, arg1, arg2)中，unk_op是一个Z3表达式.那么op==1和op==2(在one_op的定义中)等表达式也是Z3符号表达式.由于 op==1 不是 Python 布尔表达式 False.函数 one_op 将始终返回 Z3 表达式 arg1*arg2.我们可以通过执行 print one_op(unk_op, arg1, arg2) 来检查.注意one_op定义中的if语句是Python语句.

In the function call one_op(unk_op, arg1, arg2), unk_op is a Z3 expression. Then, expressions such as op==1 and op==2 (in the definition of one_op) are also Z3 symbolic expressions. Since op==1 is not the Python Boolean expression False. The function one_op will always return the Z3 expression arg1*arg2. We can check that by executing print one_op(unk_op, arg1, arg2). Note that the if statements in the definition of one_op are Python statements.

我相信您的真正意图是返回包含条件表达式的 Z3 表达式.您可以通过将 one_op 定义为:

I believe your true intention is to return a Z3 expression that contains conditional expressions. You can accomplish that by defining one_op as:

def one_op (op, arg1, arg2):
    return  If(op==1,
               arg1*arg2,
               If(op==2,
                  arg1-arg2,
                  If(op==3,
                     arg1+arg2,
                     arg1+arg2)))

现在，命令 If 构建 Z3 条件表达式.通过使用这个定义，我们可以找到一个令人满意的解决方案.这是完整的示例:

Now, the command If builds a Z3 conditional expression. By using, this definition, we can find a satisfying solution. Here is the complete example:

from z3 import *

def one_op (op, arg1, arg2):
    return  If(op==1,
               arg1*arg2,
               If(op==2,
                  arg1-arg2,
                  If(op==3,
                     arg1+arg2,
                     arg1+arg2)))

s=Solver()

arg1, arg2, result, unk_op=Ints ('arg1 arg2 result unk_op')

s.add (unk_op>=1, unk_op<=3)
s.add (arg1==1)
s.add (arg2==2)
s.add (result==3)
s.add (one_op(unk_op, arg1, arg2)==result)

print s.check()
print s.model()

结果是:

sat
[unk_op = 3, result = 3, arg2 = 2, arg1 = 1]

这篇关于(Z3Py) 函数声明有什么限制吗?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！