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Home/ Questions/Q 3441276
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T08:32:00+00:00 2026-05-18T08:32:00+00:00

I was having a look over this page: http://www.devbistro.com/tech-interview-questions/Cplusplus.jsp , and didn’t understand this

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I was having a look over this page: http://www.devbistro.com/tech-interview-questions/Cplusplus.jsp, and didn’t understand this question:

What’s potentially wrong with the following code?

long value;
//some stuff
value &= 0xFFFF;

Note: Hint to the candidate about the base platform they’re developing for. If the person still doesn’t find anything wrong with the code, they are not experienced with C++.

Can someone elaborate on it?

Thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-18T08:32:01+00:00Added an answer on May 18, 2026 at 8:32 am

    Several answers here state that if an int has a width of 16 bits, 0xFFFF is negative. This is not true. 0xFFFF is never negative.

    A hexadecimal literal is represented by the first of the following types that is large enough to contain it: int, unsigned int, long, and unsigned long.

    If int has a width of 16 bits, then 0xFFFF is larger than the maximum value representable by an int. Thus, 0xFFFF is of type unsigned int, which is guaranteed to be large enough to represent 0xFFFF.

    When the usual arithmetic conversions are performed for evaluation of the &, the unsigned int is converted to a long. The conversion of a 16-bit unsigned int to long is well-defined because every value representable by a 16-bit unsigned int is also representable by a 32-bit long.

    There’s no sign extension needed because the initial type is not signed, and the result of using 0xFFFF is the same as the result of using 0xFFFFL.

    Alternatively, if int is wider than 16 bits, then 0xFFFF is of type int. It is a signed, but positive, number. In this case both operands are signed, and long has the greater conversion rank, so the int is again promoted to long by the usual arithmetic conversions.


    As others have said, you should avoid performing bitwise operations on signed operands because the numeric result is dependent upon how signedness is represented.

    Aside from that, there’s nothing particularly wrong with this code. I would argue that it’s a style concern that value is not initialized when it is declared, but that’s probably a nit-pick level comment and depends upon the contents of the //some stuff section that was omitted.

    It’s probably also preferable to use a fixed-width integer type (like uint32_t) instead of long for greater portability, but really that too depends on the code you are writing and what your basic assumptions are.

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