Dafny函数，while循环上的逻辑表达式无效 [英] Dafny function, invalid logical expression on while loop

问题描述

我是Dafny的新手，遇到了一些我无法弄清楚的错误。

I am new in Dafny and getting some errors that I can not figure it out.

• 在我的Dafny程序中，用于insertSort（< a href = https://rise4fun.com/Dafny/uj6 rel = nofollow noreferrer>代码在这里），我不明白为什么会得到无效的逻辑表达式上的While循环遍历变量 i 。 while（i< | input |）


• 在交换部分的同一代码中（ input [ j：= b]; input [j-1：= a]; ）我也得到预期方法调用，找到表达式。根据教程 input [j：= b] 将seq输入的索引j替换为b

• in my Dafny program for insertionSort (the code is here), I do not understand why I get an invalid logical expression on While loop over variable i. while (i < |input|)
• in the same code at the swapping part (input[j := b]; input[j-1 := a];) also I get expected method call, found expression. According to the tutorial input[j:=b] is replacing index j of seq input with the value of b

推荐答案

第一个错误是因为您被声明为函数而不是方法。在Dafny中，函数的主体应为表达式，而不是语句序列。因此，当解析器看到关键字 while时，它就意识到出了点问题（因为 while不能成为语句的一部分）并给出了错误消息。我不确定为什么错误消息引用的是逻辑表达式。

The first error is because you are declared a function rather than a method. In Dafny the body of a function is expected to be an expression, not a sequence of statements. So when the parser sees the keyword "while", it realizes something is wrong (since "while" can't be part of a statement) and gives an error message. I'm not sure why the error message refers to a "logical" expression.

无论如何，您可以通过声明方法而不是函数。

Anyway, you can fix this problem by declaring a method rather than a function.

您需要一种方法，因为您使用的是命令式算法而不是功能性算法。确实，您需要一个子例程，该子例程根据其输入来计算其输出而没有副作用。但是，在Dafny中，当您要使用的方法涉及诸如赋值和while循环之类的命令式构造时，您仍然会使用方法。

You need a method because you are using an imperative algorithm and not a functional algorithm. It's true that you want a subroutine that computes its output as a function of its input with no side effects. But, in Dafny, you still use a method for this when the way you want to do it involves imperative constructs like assignments and while loops.

第二个问题是 input [j：= b] 是一个表达式，而解析器期望一份声明。您可以通过重写 input [j：= b];来解决此问题。 input [j-1：= a]; 作为 input：= input [j：= b];输入：=输入[j-1]; 。

The second problem is that input[j := b] is an expression whereas the parser exepected a statement. You can fix this by rewriting input[j := b]; input[j-1 := a]; as input := input[j:=b]; input := input[j-1];.

不幸的是，这将导致另一个问题是，在Dafny中无法分配输入参数。因此，您最好创建另一个变量。

Unfortunately, that will lead to another problem, which is that, in Dafny, input parameters can't be assigned to. So you are better off making another variable. See below, for how I did that.

method insertionSort(input:seq<int>)
// The next line declares a variable you can assign to.
// It also declares that the final value of this variable is the result
// of the method.
returns( output : seq<int> )
    // No reads clause is needed.
    requires |input|>0
    // In the following I changed "input" to "output" a few places
    ensures perm(output,old(input))
    ensures sortedBetween(output, 0, |output|) // 0 to input.Length = whole input

{
    output := input ;
    // From here on I changed your "input" to "output" most places
    var i := 1;
    while (i < |output|) 
        invariant perm(output,old(input))
        invariant 1 <= i <= |output|
        invariant sortedBetween(output, 0, i)       
        decreases |output|-i
    {
        ...
            output := output[j := b];
            output := output[j-1 := a];
            j := j-1;
        ...
    }
}

由于无法更改输入参数，因此无论您在何处使用 old（input），都可以使用 input 。他们是同一回事。

By the way, since input parameters can't be changed, wherever you have old(input), you could just use input. They mean the same thing.

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