C/C++: Pointer Arithmetic

Question

I was reading a bit in Pointer Arithmetic, and I came upon 2 things I couldn't understand neither know it's use

address_expression - address_expression

and also

address_expression > address_expression

Can someone please explain them to me, how do they work and when they are used.

Edit:

What I meant to say is what do they produce if I just take two addresses and subtract them

And If I take two addresses and compare them what is the result or comparing based upon

Edit: I now understand the result of subtracting addresses, but comparing addresses I still don't get it.

I understand that 1<2, but how is an address greater than another one and what are they compared upon

Answer 1

Several answers here have stated that pointers are numbers. This is not an accurate description of pointers as specified by the C standard.

In large part, you can think of pointers as numbers, and as addresses in memory, provided (a) you understand that pointer subtraction converts the difference from bytes to elements (of the type of the pointers being subtracted), and (b) you understand the limits where this model breaks.

The following uses the 1999 C standard (ISO/IEC 9899, Second edition, 1999-12-01). I expect the following is more detailed than the asker requested, but, given some of the misstatements here, I judge that precise and accurate information should be given.

Per 6.5.6 paragraph 9, you may subtract two pointers that point to elements of the same array or to one past the last element of the array. So, if you have int a[8], b[4];, you may subtract a pointer to a[5] from a pointer to a[2], because a[5] and a[2] are elements in the same array. You may also subtract a pointer to a[5] from a pointer to a[8], because a[8] is one past the last element of the array. (a[8] is not in the array; a[7] is the last element.) You may not subtract a pointer to a[5] from a pointer to b[2], because a[5] is not in the same array as b[2]. Or, more accurately, if you do such a subtraction, the behavior is undefined. Note that it is not merely the result that is unspecified; you cannot expect that you will get some possibly nonsensical number as a result: The behavior is undefined. According to the C standard, this means that the C standard does not say anything about what occurs as a consequence. Your program could give you a reasonable answer, or it could abort, or it could delete files, and all those consequences would be in conformance to the C standard.

If you do an allowed subtraction, then the result is the number of elements from the second pointed-to element to the first pointed-to element. Thus, a[5]-a[2] is 3, and a[2]-a[5] is −3. This is true regardless of what type a is. The C implementation is required to convert the distance from bytes (or whatever units it uses) into elements of the appropriate type. If a is an array of double of eight bytes each, then a[5]-a[2] is 3, for 3 elements. If a is an array of char of one byte each, then a[5]-a[2] is 3, for 3 elements.

Why would pointers ever not be just numbers? On some computers, especially older computers, addressing memory was more complicated. Early computers had small address spaces. When the manufacturers wanted to make bigger addresses spaces, they also wanted to maintain some compatibility with old software. They also had to implement various schemes for addressing memory, due to hardware limitations, and those schemes may have involved moving data between memory and disk or changing special registers in the processor that controlled how addresses were converted to physical memory locations. For pointers to work on machines like that, they have to contain more information than just a simple address. Because of this, the C standard does not just define pointers as addresses and let you do arithmetic on the addresses. Only a reasonable amount of pointer arithmetic is defined, and the C implementation is required to provide the necessary operations to make that arithmetic work, but no more.

Even on modern machines, there can be complications. On Digital’s Alpha processors, a pointer to a function does not contain the address of the function. It is the address of a descriptor of the function. That descriptor contains the address of the function, and it contains some additional information that is necessary to call the function correctly.

With regard to relational operators, such as >, the C standard says, in 6.5.8 paragraph 5, that you may compare the same pointers you may subtract, as described above, and you may also compare pointers to members of an aggregate object (a struct or union). Pointers to members of an array (or its end address) compare in the expected way: Pointers to higher-indexed elements are greater than pointers to lower-indexed elements. Pointers to two members of the same union compare equal. For pointers to two members of a struct, the pointer to the member declared later is greater than the pointer to the member declared earlier.

As long as you stay within the constraints above, then you can think of pointers as numbers which are memory addresses.

Usually, it is easy for a C implementation to provide the behavior required by the C standard. Even if a computer has a compound pointer scheme, such as a base address and offset, usually all elements of an array will use the same base address as each other, and all elements of a struct will use the same base address as each other. So the compiler can simply subtract or compare the offset parts of the pointer to get the desired difference or comparison.

However, if you subtract pointers to different arrays on such a computer, you can get strange results. It is possible for the bit pattern formed by a base address and offset to appear greater (when interpreted as a single integer) than another pointer even though it points to a lower address in memory. This is one reason you must stay within the rules set by the C standard.

Answer 2

Pointer subtraction yields the number of array elements between two pointers of the same type.

For example,

int buf[10] = /* initializer here */;

&buf[10] - &buf[0];  // yields 10, the difference is 10 elements

Pointer comparison. For example, for the > relational operator: the > operation yields 1 if the pointed array element or structure member on the left hand side is after the pointed array element or structure member on the right hand side and it yields 0 otherwise. Remember arrays and structures are ordered sequences.

 &buf[10] > &buf[0];  // 1, &buf[10] element is after &buf[0] element

Answer 3

Subtracting two pointer addresses returns the number of elements of that type.

So if you have an array of integers and two pointers into it, subtracting those pointers will return the number of int values between, not the number of bytes. Same with char types. So you need to be careful with this, especially if you are working with a byte buffer or wide characters, that your expression is calculating the right value. If you need byte-based buffer offsets for something that does not use a single byte for storage (int, short, etc) you need to cast your pointers to char* first.