JUnit 5测试

Question

如何使用java和@test注释使用junit测试用例实现该类中的3种方法的完整分支覆盖。

public class StringStack {
private int capacity = 10;
private int pointer = 0;
private String[] objects = new String[capacity];

public void push(String o) {
    if (pointer >= capacity)
        throw new IllegalArgumentException("Stack exceeded capacity!");
    objects[pointer++] = o;

}
public String pop() {
    if (pointer <= 0)
        throw new IllegalArgumentException("Stack empty");
    return objects[--pointer];

}

public boolean isEmpty() {
    return pointer <= 0;

}

我已经为isEmpty()方法编写了以下测试cases，虽然我很难为另外两种方法编写测试用例，因为它们都返回对象指针，而且我不知道如何在测试文件中初始化它们。

class squareTest {

    //1.
    @Test 
    public void push() {

        StringStack push1 = new StringStack();
        String e2 = push1.pop();
        try {
            Assert.fail( "Should have thrown an exception" );
        assertEquals(IllegalArgumentException("Stack empty"), e2);
        //java.lang.IllegalArgumentException: Stack empty

        }catch (Exception e) {
            String expmessage = "I should get this message";


        }




    }

    @Test
    public void testTC3()
    {
        try {
            StringStack.push(o);
            fail(); // if we got here, no exception was thrown, which is bad
        } 
        catch (Exception e) {
            final String expected = "Legal Values: Package Type must be P or R";
            assertEquals( expected, e.getMessage());
        }        
    }

    //3.EMPTY TEST CASES
    @Test
    public void empty()
    {
        StringStack test2 = new StringStack();
        boolean e1 = test2.isEmpty();
        assertEquals(true, e1);
    } 
    @Test
    public void notEmpty()
    {
        StringStack test3 = new StringStack();
        boolean ne1 = test3.equals("im not empty");
        assertEquals(false, ne1);
    }
}

Answer 1

也许下面的测试，同时帮助获得全面的覆盖。

推法试验

• 指针>=容量

@Test(exception = IllegalArgumentException.class)公共IllegalArgumentException.class WhiteBox.setInternalState(yourObject，“指针”，11)；String inputString = "Hello"；yourObject.push(inputString)；}

• 指针<容量

@Test public void push_PointerSmallerThanCapacity_ExceptionThrow(){ inputString = "Hello"；yourObject.push(inputString)；int指针= WhiteBox.getInternalState(yourObject，“指针”)；String[] objects = WhiteBox.getInternalState(yourObject，"objects")；assertEquals(inputString，objectspin-1)；}

pop法试验

• 指针<0

@Test(exception = IllegalArgumentException.class)公共pop_PointerNegative_ExceptionThrow(){ WhiteBox.setInternalState(yourObject，“指针”，-1)；字符串inputString = "Hello"；yourObject.push(inputString)；}

• 指针>0

@Test void pop_pointerGreaterThenZero_PopValue(){ //set指针WhiteBox.setInternalState(yourObject，“指针”，2)；String[] stringList = {"String0“、"String1”、"String2"}；//对象数组WhiteBox.setInternalState(yourObject，"objects"，stringList)；String actualOutput = yourObject.pop()；assertEquals(actualOutput，stringList1)；}

在这里，yourObject是您测试的类的对象。

Answer 2

我将给出第一个函数的例子。

public void push(String o) {
    if (pointer >= capacity)
        throw new IllegalArgumentException("Stack exceeded capacity!");
    objects[pointer++] = o;
}

您需要3次统一测试才能完全涵盖这一项。

指针

1. ==
2. 指针>容量
3. 和指针<容量

的最后一个

对于分支覆盖，您只需要1和3或2和3，虽然我建议这三种功能的关键部分。