I know that NULL == (void *)0
but it is mentioned that it can be represented as a value which doesn't contain all zeros. What bothers me is if those pieces of code are equivalent for all (any_type *)
:
any_type *val;
if (val) { ... };
and
if (val != NULL) { ... };
Yes. You can use the macro NULL
in this way. It is a null pointer constant which is equal to 0
.
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 asif((expr) != 0)
Substituting the trivial pointer expression
p
forexpr
, we haveif(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.
but it is mentioned that it can be represented as a value which doesn't contain all zeros.
Yes. True. But representation doesn't matter in this case:
Whenever a programmer requests a null pointer, either by writing
0
orNULL
, it is the compiler's responsibility to generate whatever bit pattern the machine uses for that null pointer.
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