Question

正如标题所述，如何使用Hoare Triple验证以下功能？我读了很多关于它的讲座，但我不知道该怎么做。

int uguaglianza_insiemi(elem_lista_t *insieme_A,
                        elem_lista_t *insieme_B)
{
    int esito;  

    if ((insieme_A == NULL) && 
        (insieme_B == NULL))
        esito = 1;

    else if (insieme_A == NULL ||
            insieme_B == NULL)
            esito = 0;

    else if (insieme_A->valore != insieme_B->valore) 
            esito = 0;

    else esito = uguaglianza_insiemi(insieme_A->succ_p,
                                     insieme_B->succ_p);

    return (esito);
}

Answer 1

为避免在评论中进行冗长的讨论，我将尝试编写一些前置条件和后置条件。

由于无法在函数内部测试是否使用指向有效列表对象的指针来调用，该指针属于父/调用者：

// The following function must be called with pointers that are either null
// or point to valid list elements. The lists must be correct (no malloc bugs etc).
// The compiler must have checked that it is called with pointers to the proper types,
// as C has no typeof operator.
//
int uguaglianza_insiemi(elem_lista_t *insieme_A,
                        elem_lista_t *insieme_B)
{
    int esito;  

    if ((insieme_A == NULL) && 
        (insieme_B == NULL))
        esito = 1;    // both pointers are null: equal

    // not both pointes are null
    else if (insieme_A == NULL ||
            insieme_B == NULL)
            esito = 0;    // not both pointers are null, but one is: not equal

    // neither pointer is null and so they may be dereferenced
    else if (insieme_A->valore != insieme_B->valore) 
            esito = 0;    // neither pointer is null, but their element values aer not equal: not equal


    // the function can be called recursively now because its precondition has been met,
    // that both successor pointers are null or point to valid list elements (induction).
    else esito = uguaglianza_insiemi(insieme_A->succ_p,
                                     insieme_B->succ_p);
    // the post condition is that esito reflects equality of both (partial) lists                    
    return (esito);
}

我希望这是您和您的教授可以合作的东西。

{P}：必须使用为空或指向有效列表元素的指针来调用该函数。

C：uguaglianza_insiemi( *A, *B)

{Q}：函数结果反映了列表的相等性

在函数内部，此操作继续使用组成规则的if语句。

如何使用Hoare Triple验证该功能？

1 个答案: