Question

我如何仅列出来自我的不满核心的特定约束结果？我有很多条件和打印整个核心不打印所有。我读到可以使用assert_and_track和unsat_core命令完成它。我找到了一些例子，但它们不起作用。有这方面的经验吗？

#include <sstream>
#include <iostream>
#include <fstream>
#include "Date.h"
#include "Address.h"
#include "student.h"


using namespace std;

int main(){

    string line;
    ifstream inputFile("studentdata.txt");
    bool keepGoing = true;
    Student *student = new Student[50];
    int i = 0;
    while(!inputFile.eof()){
        getline(inputFile, line);
        student[i].setInfo(line);
        i++;
    }
    int choice;
    cout << "A Heap of Students";
    while(keepGoing){ //This while loop creates the menu and asks for user input
        cout << "What would you like to do?" << endl;
        cout << "1. Print full report." << endl;
        cout << "2. Print simple report." << endl;
        cout << "3. Sort records." << endl;
        cout << "4. Quit" << endl;
        cin >> choice;

        //prints full student report
        if(choice == 1){
            for(int i = 0; i < 50; i++){ //Full print loop
                cout << student[i] << endl;
            }
            cout << endl;
            keepGoing = true;
        }
        //Just first and last name
        else if(choice == 2){
            cout << "First Last" << endl;
            for(int i = 0; i < 50; i++){ //Simple print loop
            cout << student[i].getFirstName() << " " << student[i].getLastName() << endl;
            }
            cout << endl; //formatting
            keepGoing = true;
        }
        //sort function
        else if(choice == 3){
            for(int j = 1; j < 50; j++){
                stringstream key = student[j].getLastName;
                int i = j - 1;
                while(i > 0 && student[i].getLastName > key){
                    student[i+1] = student[i];
                    i = i - 1;
                }
                Student[i + 1] = key;
            }

            for j = 1 to A.length
            key = A[j]
            i = j - 1
            while i > 0 and A[i] > key
            A[i + 1] = A[i]
            i = i - 1
            A[i + 1] = key

            keepGoing = true;
        }
        //quit
        else if(choice == 4){
    cout << "Goodbye!" << endl;
            keepGoing = false;
        }
    }
        return 0;
}

那么如何只打印p1或p2（只是真/假结果）？例如p2 = false，或者它可以按照打印（c）模式显示的方式显示 - 但仅适用于p1。

Answer 1

实现这一目标的最简单方法是简单地将标签贴图保存到约束中，我们可以做这个例子

M = {}
M['p1'] = (A > B)
M['p2'] = (B > C)
M['p3'] = (C > A)

以便稍后我们可以根据这些约束打印核心，例如如下

core = s.unsat_core()
for e in core:
    print(M[str(e)])

当然，如果您只想打印一些条目而不是全部条目，这也会有效。

如何显示特定的不满足约束，而不是整个核心（Z3，Python）

1 个答案: