Is if(p), where p is a pointer, a valid and portable test?

Q

Is if(p), where p is a pointer, a valid and portable test?

✍: Guest

A

It is always valid.
When C requires the Boolean value of an expression, a false value is inferred when the expression compares equal to zero, and a true value otherwise. That is, whenever one writes
if(expr)
where ``expr'' is any expression at all, the compiler essentially acts as if it had been written as
if((expr) != 0)
Substituting the trivial pointer expression ``p'' for ``expr'', we have
if(p) is equivalent to if(p != 0)
and this is a comparison context, so the compiler can tell that the (implicit) 0 is actually a null pointer constant, and use the correct null pointer value. There is no trickery involved here; compilers do work this way, and generate identical code for both constructs. The internal representation of a null pointer does not matter.
The boolean negation operator, !, can be described as follows:
!expr is essentially equivalent to (expr)?0:1
or to ((expr) == 0)

which leads to the conclusion that
if(!p) is equivalent to if(p == 0)
``Abbreviations'' such as if(p), though perfectly legal, are considered by some to be bad style (and by others to be good style;

2016-02-29, 524👍, 0💬