Question 5.3

Is the abbreviated pointer comparison ``if(p)'' to test for non-null pointers valid? What if the internal representation for null pointers is nonzero?

When C requires the Boolean value of an expression (in the if, while, for, and do statements, and with the &&, ||, !, and ?: operators), 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; see question 17.10).