C char array seems to not pass full array-CodePudding

I would like to preface this with I am a junior but relatively experienced developer, but with very little C experience.

I have this test method that is supposed to pass an array of characters to find the "O".

TEST(MyTest, ReturnIndexOfO)
{
    Widget unitUnderTest;
    char x = 'X';
    char o = 'O';
    char *row[4] = {&o, &x, &x, &x};
    char *row2[4] = {&x, &o, &x, &x};

    EXPECT_EQ(unitUnderTest.findEmptySpace(*row, 4), 0);
    EXPECT_EQ(unitUnderTest.findEmptySpace(*row2,4 ), 1);
}

And this has properly invoked my findEmptySpace method, which is:

#define WHITE_SPACE 'O'
int Widget::findEmptySpace(char row[], int size)
{
    cout << "The row under test is:\n";
    for(int i = 0; i < size; i  ) {
        cout << row[i];
    }
    cout << "\n";
    for(int i = 0; i < size; i  ) {
        if(row[i] == WHITE_SPACE) {
            return i;
        }
    }
    return -1;
}

But unfortunately, my output seems indicate that not all of the characters are being read by my findEmptySpace method:

The row under test is:

OX

The row under test is:

X

So even with the truncated data, my first test case passes, but the second test case fails. Any idea why I am not seeing the data correctly?

CodePudding user response：

This expression

Test.findEmptySpace(*row, 4)

is equivalent to the expression

Test.findEmptySpace(row[0], 4)

and the element row[0] is a pointer with the value &o that points to the o of the type char

char o = 'O';.

So the function call (the inner loops within the function) with such parameters does not make a sense

It seems what you mean is the following

EXPECT_EQ(unitUnderTest.findEmptySpace(row, 4), 0);

and

#define WHITE_SPACE 'O'
int Widget::findEmptySpace(char * row[], int size)
{
    cout << "The row under test is:\n";
    for(int i = 0; i < size; i  ) {
        cout << *row[i];
    }
    cout << "\n";
    for(int i = 0; i < size; i  ) {
        if( *row[i] == WHITE_SPACE) {
            return i;
        }
    }
    return -1;
}

CodePudding user response：

   char *row[4] = {&o, &x, &x, &x};

This is not an "array of characters". This is an array of four pointers to characters.

unitUnderTest.findEmptySpace(*row, 4)

This passes the first one of these pointers, to findEmptySpace() in its first parameter. The first pointer is a pointer to o.

The second parameter to findEmptySpace() is 4. That's what the above statement does in C .

The function findEmptySpace() does the following:

    for(int i = 0; i < size; i  ) {
        cout << row[i];
    }

So it ends up printing the first four characters from the pointer that was passed to it.

The problem is that the passed in pointer is a pointer to just one character:

&o

That's the first pointer. And this function ends up trying to read the first four character from a pointer that's pointing to only one character.

This results in undefined behavior.