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.

Answer 1

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.

Answer 2

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.