How does unary addition on C pointers work?

Question

I know that the unary operator ++ adds one to a number. However, I find that if I do it on an int pointer, it increments by 4 (the sizeof an int on my system). Why does it do this? For example, the following code:

int main(void)
{
  int *a = malloc(5 * sizeof(int));
  a[0] = 42;
  a[1] = 42;
  a[2] = 42;
  a[3] = 42;
  a[4] = 42;
  printf("%p\n", a);
  printf("%p\n", ++a);
  printf("%p\n", ++a);
  return 0;
}

will return three numbers with a difference of 4 between each.

that's not unary addition that's a pre-increment. Unary addition is a single +. — Scott Logan, Feb 15 '11 at 00:54
Do note that what you did just created a memory leak because you can no longer `free(a);` as it no longer points to the start of allocated memory. Attempting to free it at this point would be Undefined Behavior and may result in a segmentation fault (if you're lucky). — SiegeX, Feb 15 '11 at 01:21
@Bunnit "Unary addition is a single +" -- No, no more so than -x is unary substraction. Addition and substraction are binary operations. A single - is negation; a single + is a no-op. — Jim Balter, Feb 15 '11 at 12:17
@Jim Balter. Ok maybe "+x" is more commonly referred to as unary plus but it is an operator that can be overloaded therefore it is not always a no-op. It is pretty clear that this was not what the OP meant though, pre-increment is never called unary addition. — Scott Logan, Feb 15 '11 at 14:10
@Bunnit Actually, calling ++x "unary addition" is quite reasonable; it's a unary operator that does addition. In a functional language with currying, one can write something like fn add1 = add(1) or fn add1 = +(1) to define a unary function that adds 1 to its argument. OTOH, unary "+" does no addition and is not unary addition and is never called unary addition -- except occasionally on SO where doing so manages bizarrely to get 4 uprates. — Jim Balter, Feb 16 '11 at 00:02

Carl Norum · Accepted Answer · 2011-02-15T01:08:55.373

It's just the way C is - the full explanation is in the spec, Section 6.5.6 Additive operators, paragraph 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.

To relate that to your use of the prefix ++ operator, you need to also read Section 6.5.3.1 Prefix increment and decrement operators, paragraph 2:

The value of the operand of the prefix ++ operator is incremented. The result is the new value of the operand after incrementation. The expression ++E is equivalent to (E+=1).

And also Section 6.5.16.2 Compound assignment, paragraph 3:

A compound assignment of the form E1 op= E2 differs from the simple assignment expression E1 = E1 op (E2) only in that the lvalue E1 is evaluated only once.

score 3 · Answer 2 · answered Feb 15 '11 at 00:52

It's incrementing the pointer location by the size of int, the declared type of the pointer.

Remember, an int * is just a pointer to a location in memory, where you are saying an "int" is stored. When you ++ to the pointer, it shifts it one location (by the size of the type), in this case, it will make your value "4" higher, since sizeof(int)==4.

score 2 · Answer 3 · answered Feb 15 '11 at 00:53

2

The reason for this is to make the following statement true:

*(ptr + n) == ptr[n]

These can be used interchangeably.

answered Feb 15 '11 at 00:53

Erik

88,732
13
198
189

score 1 · Answer 4 · edited May 23 '17 at 11:44

1

In pointer arithmetic, adding one to a pointer will add the sizeof the type which it points to.

so for a given:

TYPE * p;

Adding to p will actually increment by sizeof(TYPE). In this case the size of the int is 4.

See this related question

edited May 23 '17 at 11:44

Community

1
1

answered Feb 15 '11 at 00:51

Gavin H

10,274
3
35
42

score 1 · Answer 5 · answered Feb 15 '11 at 00:52

1

Because in "C" pointer arithmetic is always scaled by the size of the object being pointed to. If you think about it a bit, it turns out to be "the right thing to do".

answered Feb 15 '11 at 00:52

andy mango

1,526
1
8
13

score 1 · Answer 6 · answered Feb 15 '11 at 00:52

1

It does this so that you don't start accessing an integer in the middle of it.

answered Feb 15 '11 at 00:52

Jeffrey Greenham

1,382
5
16
33

score 0 · Answer 7 · answered Feb 15 '11 at 00:56

Because a pointer is not a reference ;). It's not a value, it's just an address in memory. When you check the pointer's value, it will be a number, possibly big, and unrelated to the actual value that's stored at that memory position. Say, printf("%p\n", a); prints "2000000" - this means your pointer points to the 2000000th byte in your machine's memory. It's pretty much unaware of what value it's stored there.

Now, the pointer knows what type it points to. An integer, in your case. Since an integer is 4 bytes long, when you want to jump to the next "cell" the pointer points to, it needs to be 2000004. That's exatly 1 integer farther, so a++ makes perfect sense.

BTW, if you want to get 42 (from your example), print out the value pointed to: printf("%d\n", *a);

I hope this makes sense ;)

score 0 · Answer 8 · answered Feb 15 '11 at 06:43

Thats simple, cause when it comes down to pointer, in your case an integer pointer, a unary increment means INCREMENT THE MEMORY LOCATION BY ONE UNIT, where ONE UNIT = SIZE OF INTEGER .

This size of integer depends from compile to compiler, for a 32-bit and 16-bit it is 4bytes, while for a 64-bit compiler it is 8bytes.

Try doing the same program with character datatype, it will give difference of 1 byte as character takes 1 byte.

In Short, the difference of 4's that you've come across is the difference of SIZE OF ONE INTEGER in memory.

Hope this helped, if it didn't i'll be glad to help just let me know.

score 0 · Answer 9 · answered Feb 15 '11 at 12:21

0

"Why does it do this?" Why would you expect it to do anything else? Incrementing a point makes it point to the next item of the type that it's a pointer to.

answered Feb 15 '11 at 12:21

Jim Balter

16,163
3
43
66

How does unary addition on C pointers work?

9 Answers9

Linked