Hello

I need help understanding /interpreting (or putting into words the behaviours of pointers and arrays). Please see the included code.

I have included 4 cases/scenarios of using pointers and arrays in which I am trying to understand their behaviour. I think I understand what is happening in cases 1, 3 and 4. However I don’t know what is happening in case 2.

Here are my descriptions of what I think is happening in case 1, 3 and 4.
Case 1
char myName[] = "John Smith";
myName is an array so when you do
cout << myName << endl;
you are just selecting the ith item of the array and outputting it.

Case 3
char* myPointer2 = myName;
myPointer2 is a pointer that points to data type of type char.
If you have: int anArray[5];
anArray is actually a pointer that points to the first element of the array
Since by definition the name of any array is a pointer when you do
myPointer2, it is if you are performing the same action as in Case one that is
myPointer2 is equivalent to myName

Case 4
char* myPointer3 = myName;
When you do cout << *(myPointer3+i)
You are telling the computer to output the contents of the memory address myPointer3+i

Questions:
Why does Case 2 gives the output it gives? In other words why doesn’t it just print one letter at a time?

I also want to add that I noticed that if I do the following
Line A: cout << myName << endl;
Line B: int someNums[] = {1, 2, 3, 4, 5};
Line C: cout << someNums << endl;

Line A results in the following being output to the screen: John Smith
However Line C results in the following being output to the screen: 003FF854
003FF854 is a memory address. Why is that the output as opposed to 12345

My Code:

#include <iostream>
using namespace std;

int main()
{
    char myName[] = "John Smith";
    cout << "char myName[] = \"John Smith\";" << endl;
    cout << "strlen(myName)  = " << strlen(myName) << endl << endl;
    
    //Case One
    cout << "Case 1" << endl;
    cout << "using: cout << " << "myName" << endl;
    for (int i = 0; i < (int) strlen(myName); i++)
    {
        cout <<"i = " << i << ", " << myName << endl;
    }
    cout << endl << "===" << endl << endl;

    //=====================================================

    //Case Two
    cout << "Case 2" << endl;
    cout << "char* myPointer1 = myName;" << endl;
    cout << "using: cout << " << "myPointer1+i" << endl;
    char* myPointer1 = myName;
    for (int i = 0; i < (int) strlen(myName); i++)
    {
        cout <<"i = " << i << ", " << myPointer1+i << endl;
    }
    cout << endl << "===" << endl << endl;

    //=====================================================

    // Case Three
    cout << "Case 3" << endl;
    cout << "char* myPointer2 = myName;" << endl;
    cout << "using: cout << " << "myPointer2" << endl;
    char* myPointer2 = myName;
    cout << strlen(myName) << endl;
    for (int i = 0; i < (int) strlen(myName); i++)
    {
        cout <<"i = " << i << ", " << myPointer2 << endl;
    }
    cout << endl << "===" << endl << endl;

    //=====================================================

    //Case Four
    cout << "Case 4" << endl;
    cout << "char* myPointer3 = myName;" << endl;
    cout << "using: cout << " << "*(myPointer3+i)" << endl;

    char* myPointer3 = myName;
    cout << strlen(myName) << endl;
    for (int i = 0; i < (int) strlen(myName); i++)
    {
        cout <<"i = " << i << ", " << *(myPointer3+i) << endl;
    }


    return 0;
}

The Results:

char myName[] = "John Smith";
strlen(myName)  = 10

Case 1
using: cout << myName
i = 0, J
i = 1, o
i = 2, h
i = 3, n
i = 4,
i = 5, S
i = 6, m
i = 7, i
i = 8, t
i = 9, h

===

Case 2
char* myPointer1 = myName;
using: cout << myPointer1+i
i = 0, John Smith
i = 1, ohn Smith
i = 2, hn Smith
i = 3, n Smith
i = 4,  Smith
i = 5, Smith
i = 6, mith
i = 7, ith
i = 8, th
i = 9, h

===

Case 3
char* myPointer2 = myName;
using: cout << myPointer2
10
i = 0, J
i = 1, o
i = 2, h
i = 3, n
i = 4,
i = 5, S
i = 6, m
i = 7, i
i = 8, t
i = 9, h

===

Case 4
char* myPointer3 = myName;
using: cout << *(myPointer3+i)
10
i = 0, J
i = 1, o
i = 2, h
i = 3, n
i = 4,
i = 5, S
i = 6, m
i = 7, i
i = 8, t
i = 9, h
Press any key to continue . . .

 

The difference in case 2 is that you're not de-referencing the pointer. So it just prints the pointer, rather than the value being pointed to.

Actually, it increments the POINTER every iteration of the loop (similar to case 4), but it is not being de-referenced before it is output.

 

Thanks for your reply.

The difference in case 2 is that you're not de-referencing the pointer. So it just prints the pointer, rather than the value being pointed to.

Actually, it increments the POINTER every iteration of the loop (similar to case 4), but it is not being de-referenced before it is output.

I understand and agree with the part in blue.

What do you mean by it just prints the pointer? My understanding of pointers is that when you print a pointer you are printing the address that the pointer is holding.

When does the computer know to stop printing?

 

What do you mean by it just prints the pointer? My understanding of pointers is that when you print a pointer you are printing the address that the pointer is holding.

Yes you are correct.

When does the computer know to stop printing?

What does that means I have no idea........

 

std::cout is overloaded for various types. the 'cout << myPointer' is using the version of cout which takes a pointer to a char, it keeps printing til it hits the null at the end of your char array. someone can probably explain it better than me mind...