Had C++ Interview - went not well :(

[CollyMcK]

Hi I just had a C++ Interview - it did not go well - I was hoping I could get some questions cleared up so I can perhaps learn if I get another one.

The question was something like this:

int a = 2;
int b = 4;

????? a | b;

????? a || b;

The question mark I cant exactly remember the syntax but it was a printf asking for the result - am i right in the first is a BITWISE OR and the second is LOGICAL OR - what would the results of these be?

Thanks.

[filipe]

The first one is a bitwise or. Here's a binary representation of what goes on:

2 == 00000010
4 == 00000100

2 | 4 == 00000110 == 6

As for the logical or, it will print 1 if the expression is true, and 0 is it's false. Actually it will print 2, because there's no parens, and because of short circuit evaluation. If there were, in C++, it would print true or false directly.

[moorecm]

Get Programming Interviews Exposed by John Mongan, Noah Suojanen, and Eric Giguere.

Don't even think about going to an interview until you have read it. This book helped me land my current position--and it's a good one.

[CollyMcK]

Thanks for your response Guys - filipe - yeah it actually just hit me when I came out that a BitWise Or is just a fancy addition - blanked in the interview.

Moorecm - Cheers I will look into this Book

[filipe]

yeah it actually just hit me when I came out that a BitWise Or is just a fancy addition

It's not really addition. Consider 2 | 6.

[CollyMcK]

Thank filipe - so the result in what you have said here

2 | 6 would == 6 Correct?

[filipe]

Yes.

[Duthomhas]

It will still print 1 or 0, because it is still a logical evaluation. Short-circuiting doesn't change the type of the expression.

[cppmatt]

It will still print 1 or 0, because it is still a logical evaluation. Short-circuiting doesn't change the type of the expression.

Not on my system it won't.

cout << (2|6) << endl;

6

[moorecm]

It depends on the context, which was left out of the original post... No need to get alarmed.

[Disch]

Duaos was referring to ||, not |

cout << (2|6) << endl;  // prints 6
cout << (2||6) << endl;  // prints 1 / true 

[filipe]

Hmm. After I wrote that post, I thought maybe I could have made a mixup with C. I was at work, with no compiler (I'm not a professional programmer), but I recently found a copepad.org that allows you to compile and get the output online. When I compiled it there, (2 || 6) printed as true, and 2 || 6 printed as 2.

Now at home, I compiled it with Visual C++ and GCC. Both output 1 for the first expression, and 2 for the second. Even for the expression 2 && 6 both compilers will output 2. So there's really no short circuit evaluation happening, but it doesn't evaluate to a bool.

[Duthomhas]

cout << (2||6) << endl;
prints 1 because it is true. Try this to see the difference:
cout << boolalpha << 7 << ", " << (2||6) << endl;

cout << 2 || 6 << endl;
will fail to compile on conformant implementations, because the << operator has higher precedence than the || operator, and operator << (bool, smanip) is not defined in the standard. If you only use:
cout << 2 || 6;
then you can trick yourself, but the precedence rules actually group it as:
(cout << 2) || 6;
Hence, the logical operation (which produces a boolean value) happens after the output operation...

While C lacks a strict bool data type, it implements all the machinery for one. Hence, the output will still be a "1" or a "0".

Hope this helps.

[filipe]

prints 1 because it is true.

Yes, I'm aware of that. I was pointing out that whatever compiler codepad.org uses, it outputs the word "true" instead of 1.

will fail to compile on conformant implementations, because the << operator has higher precedence than the || operator ... Hence, the logical operation (which produces a boolean value) happens after the output operation...

Thanks a lot.

[jenia]

I even decided to register to get it.

Let’s run this code:

cout << (2|6) << endl;  //  prints 6
6 || cout << 2||6;	     // no printing, it was ignored.
6 || (cout << 2)||6;       // no printing, it was ignored 

But if the << operator has higher precedence than the || operator, it should have printed 2 ...

After that printing -> 6||2 -> It gives 6 -> 6||6 ->6.

But in any case at first I should see output of the code, in this case 2.
Am I right?
Compiler (VS 2010) gives nothing.
Thanks

[firedraco]

The reason why those don't work is because of short circuiting, since 6 is true, it ignores the cout.

"Precendence" does not mean "do this first always", it means when you get to something like:
std::cout<<2||6;

It determines whether or not the code should be:

std::cout.operator<<(2) || 6;
//or
std::cout.operator<<(2||6);

[jenia]

Thanks firedraco :)