Thread: Subtracting from memory address- why does it do this?

You should always use %p to print pointers, and cast the pointer to a void *

Code:

printf("&a = %p\n", (void *) &a);

%d may or may not be suitable for representing the value of a pointer.

It does it because that's what the C standard says it is supposed to do. Adding/subtracting 1 from any pointer moves it by the size of the thing pointed to. Thus, on your system, and int must be 4 bytes. This makes sense since if you increment a pointer, it should point to the next object, not the next byte. Think of using a pointer to step through an array. If it incremented by a single byte, the pointer could end up pointing to the middle of an object and you could have alignment issues (possibly something like a bus error that causes your program to crash). At the very least you would be accessing data in a nonsensical way. Furthermore, it would require different code for systems with different sized ints, longs, etc which would be a maintenance nightmare with no practical benefit. Note, the same type may be different sizes on different systems -- int was commonly only 16 bits back in the day, then 32 for 32 bit systems and now it's 64 on many (but not all) systems/implementations.

Relevant portion of the C standard, if you want to get technical

Originally Posted by C99 6.5.6 p8

8 When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. If the pointer operand points to an element of
an array object, and the array is large enough, the result points to an element offset from
the original element such that the difference of the subscripts of the resulting and original
array elements equals the integer expression. In other words, if the expression P points to
the i-th element of an array object, the expressions (P)+N(equivalently, N+(P)) and
(P)-N(where N has the value n) point to, respectively, the i+n-th and i−n-th elements of
the array object, provided they exist. Moreover, if the expression P points to the last
element of an array object, the expression(P)+1 points one past the last element of the
array object, and if the expression Q points one past the last element of an array object,
the expression (Q)-1 points to the last element of the array object. If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined. If the result points one past the last element of the array object, it
shall not be used as the operand of a unary*operator that is evaluated.

That actually seems incredibly useful. Thanks.