Char Array output - C++

Question

Apologies if I missed another thread, I looked through the recommended and none are quite like what I am experiencing. I am trying to display the alphabet in both upper-and-lower-case letters. Before I commit to the full thing I wanted to test with just a few:

int main()
{
    char *letters[79] = { "Aa ","Bb ","Cc ", "/0"} ;
    int pause;
    cout << "Letters are: " << letters << endl ;
    cout << "Continue (Y or N)" << endl ;
    cin >> pause;
    cout << endl;

    return 0;
}

When I attempt to compile this has been the result:

console output of above code

What I cannot understand is why I am getting what appears to be a hex output on those letters instead of the letters themself.

Why are you using arrays? Why are you using char* instead of std::string? To learn, or because you think that's the best approach? Knowing this helps others help you. :) — druckermanly
– druckermanly, Commented Oct 9, 2017 at 6:24
Learning, I am still in college for this and its pretty much my third week, this is the very edge of an assignment using arrays and pointers. — DeSa
– DeSa, Commented Oct 9, 2017 at 6:26
letters refer to the memory address of the first element of the array. If you want to print the elements , then use indexes like letters[0] for first element , letters[1] for second element and so on. — Bhawan
– Bhawan, Commented Oct 9, 2017 at 6:29
Side note: what't mthe purpose of "/0"? My crystall ball tells me it does not what you thinkt it does. — Jabberwocky
– Jabberwocky, Commented Oct 9, 2017 at 6:41
@MichaelWalz a secondary book I am using said to use it when I am using characters in an array to ... its considered a null character. — DeSa
– DeSa, Commented Oct 9, 2017 at 6:46

Farhad · Accepted Answer · 2017-10-09 06:37:49Z

1

You missed to mention array index letters[0]

Try this:

char *letters[79] = { "Aa ","Bb ","Cc ", "/0"};
cout  << "Letters are: " << letters[0] << endl;

output:

Letters are:  Aa

Or in loop like this :

char *letters[79] = { "Aa ","Bb ","Cc ", "/0"};
for(int i=0; i<sizeof(*letters); i++)
    cout  << "Letters are: " << letters[i] << endl;

output:

Letters are:  Aa  

Letters are:  Bb  

Letters are:  Cc  

Letters are:  /0

edited Oct 9, 2017 at 6:37

answered Oct 9, 2017 at 6:27

Farhad

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Serkan Pekçetin · Accepted Answer · 2017-10-09 06:45:31Z

0

Your console output shows the address of the letters, because you are telling it to print the pointer, which points to a memory address. You need a loop to iterate over it. The rest is pointer arithmetic.

Also, "Aa" is not a char type, but "A" and "a" are. So, you should also define them separately; otherwise, it is very likely that you get a compiler warning. Please note that the last character should be "\0", not "/0". If you could try this:

int main()
{
    char *letters[79] = { "A","a","B", "b", "C", "c", "\0"} ;
    int pause;
    cout << "Letters are: ";
    for(int i = 0; i < 79; i++){
        cout << letters[i];
    }
    cout << endl << "Continue (Y or N)" << endl ;
    cin >> pause;
    cout << endl;
    return 0;
}

Then, the output is:

Letters are: AaBbCc

answered Oct 9, 2017 at 6:45

Serkan Pekçetin

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1 Comment

DeSa Over a year ago

Thank, between you, aghilpro, and Micheal up above this has come along way ... now, to solve the "unhandled exception thrown write access violation _first was nullptr" error that has popped up.

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