Why won't this loop end??

I'm sorry, I'm green as grass to C++. I do have experience with Basic4Gl and some regular and visual Basic. I have a small test program set up to test out the things i've learned from the 1st 3 tutorials. These cover topics from variables up until loops. My program has a Do...While loops in it, the loop works, but I cannot get it to stop can somone please point out my problem?

Code:

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#include <iostream>

using namespace std;

int main()
{
    int age, number, yob;
    char retry;
    do {
    cout<<"How old are you?\n";
    cin>>age;
    cin.ignore();
    cout<<"What is your favorite number?\n";
    cin>>number;
    cin.ignore();
    yob = 2008 - age;
    if (age == 15) {
            cout<<"So you're "<< age<<"? Awsome! You're the same age as me!\n";
            }
    else {
        cout<<"So you're "<< age<<", were born in "<< yob<<", and you favorite number is "<< number<<".\n";
    }
    cout<<"Repeat, Y/N \n";
    cin>>retry;
    cin.ignore();
   
    }
    while ( retry = "Y"||"y" );
}

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> while ( retry = "Y"||"y" )
1. = is not the same as ==

Also, it's not the same as
while ( retry == "Y" || retry == "y" )

Quote:

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Originally Posted by Salem

> while ( retry = "Y"||"y" )
1. = is not the same as ==

Also, it's not the same as
while ( retry == "Y" || retry == "y" )

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I'm confused, I know that = and == are not the same, but if I do:

Code:

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while ( retry == "Y"||"y" );

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Then I get an error:

ISO C++ forbids comparison between pointer and interger

Quote:

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Originally Posted by The7thCrest

I'm confused, I know that = and == are not the same, but if I do:

Code:

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while ( retry == "Y"||"y" );

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Then I get an error

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Character literals look like 'y' or 'Y'.

Quote:

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Originally Posted by NeonBlack

Character literals look like 'y' or 'Y'.

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Nice ty, It works now, very helpful information. Thanks

I'm surprised. You should do as Salem says and

Code:

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while( retry=='y' || retry=='Y' ) ...

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as the other way won't work. You can't nest the || comparison operator like that.

It is still valid what you do, it just won't do what you think!
The retry == 'Y' || 'y' meanst retry EQUALS 'Y' OR 'y', which can be (retry == 'Y') || 'y' or it can be retry == ('Y' || 'y'). In any case 'Y' || 'y' should always be true since 'Y' != 0 and 'y' != 0 (0 is false, anything else true)
So either you get always true or always false or just have undefined behaviour...

Less usefully,

Code:

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(A == B||C) == (A == B || A == C)

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is seldom true for general types (which is what has been said).

Luckily, in most languages (including English), "is" distributes over "or," giving us a meaningful equality. Suppose A, B and C are all of the same type, then A == B || A == C compares objects of the same type. A == B||C compares typeof(A) to a bool. You probably got this long ago, and I'm still talking.