ACSL-- 无法证明功能

Question

我试图证明这个功能，但没有成功。我也在用另一个函数，但是我证明了。

谁能帮帮我？

        I'm using Frama-C Sodium version with Alt-ergo 3 as prover.

  Given a not-empty string x.  An integer number p such that
    0 < p ≤|x| is meant to be "a period of x" if

      x[i] = x[i + p]

    for i = 0, 1, ... , |x| − p − 1.
    Note that, for each not-empty string, the length of the string
    is a period of itself.  In this way, every not-empty string
    has got at least one period.  So is well formed the concept of minimum
    period of string x, denoted by per(x):

      per(x) = min { p | p is period of x }.

    Write a C function

       unsigned per(const char x[], unsigned l)

    that, given a string x of length l, returns per(x).

这是我目前编写的代码和规范：

/*@ 
    requires l > 0;
    requires p >= 0;

    behavior zero:
      assumes  p == l;
      ensures \result == 1;

    behavior one: 
      assumes l != p && (l%p) == 0;
      assumes \forall int i; 0 <= i < l-p-1 ==> x[i] == x[i+p];
      ensures \result == 1;

   behavior two:
      assumes l != p && (l%p) == 0;
      assumes !(\forall int i; 0 <= i < l-p-1 ==> x[i] == x[i+p]);
      ensures \result == 0;

   behavior three:
     assumes p != l && l%p != 0;
     ensures \result == 0;

   complete behaviors;
   disjoint behaviors;
*/

unsigned has_period(const char x[], unsigned int p, unsigned l) {
    if (p == l) return 1;
    if ((l % p) != 0) return 0;
        /*@
            loop assigns i;

            loop invariant \forall int j; 0 <= j < i ==> (x[j] == x[j+p]);
            loop invariant i <= l-p-1;
            loop invariant i >= 0;
        */

        for (int i = 0 ; i < l-p-1 ; ++i) {
            if (x[i] != x[i + p])
               return 0;
        }     

    return 1; 
}

/*
   predicate has_period(char* x, unsigned int p, unsigned l) =
      \forall int i; i < (l-p-1) ==>  x[i] == x[i+p]; 
*/

/*@
    requires l > 0;
    requires \valid(x+(0..l-1));

    ensures 1 <= \result <= l;
    ensures \forall unsigned int i; i < (l-\result-1) ==> x[i] == x[i+\result];
    ensures \forall unsigned int p; 1 <= p < \result ==> !(\forall int i; i < (l-p-1) ==> x[i] == x[i+p]);
*/

unsigned per(const char x[], unsigned l) {
     int p = 1;

    /*@
        loop assigns p;

        loop invariant 1 <= p <= l;
        loop invariant \forall unsigned j; 1 <= j < p ==> !(\forall int i; i < (l-j-1) ==> x[i] == x[i+j] || (l%p) == 0);
        loop invariant p >= 0;
    */

    while(p < l && !has_period(x,p,l)) 
        ++p;

    return p;
}

Answer 1

如果您告诉我们您的具体问题是什么，而不是笼统地告诉我们，那将会有所帮助 "it doesn't work"，但这里有一些要点：

您的合同缺少 assigns 条款。使用 WP 时，这些 not 是可选的，特别是对于在您的开发中调用的 has_period 等函数。 WP 需要这些子句来了解哪些位置在调用期间可能发生了变化，并且通过取补，哪些位置保持不变。
您很好地 semi-formal 定义了周期性字符串。您应该使用它来定义一个 ACSL 谓词 periodic（采用与您的 C 函数 has_period 相同的参数），并根据该谓词编写您的规范。这将大大简化您的注释。特别是 has_period 不需要 4 个行为：如果 periodic 成立，则必须 return 1，如果 periodic 不成立，则必须 return 0.
使用 -wp-rte 会导致一堆未经证实的义务。根据您的任务，您可能想要加强您的规范，尤其是关于 x.

ACSL-- 无法证明功能

ACSL - Can't prove function

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frama-c

acsl