pointer-arithmetic

This question already has an answer here: C/C++: Pointer Arithmetic 6 answers Why is the difference between the two addresses wrong? http://codepad.org/NGDqFWjJ #include<stdio.h> int main() { int i = 10, j = 20; int *p = &i; int *q = &j; int c = p - q; printf("%d\n", p); printf("%d\n", q); printf("%d", c); return 0; } Output: -1083846364 -1083846368 1 First, pointer arithmetic isn't defined when performed on unrelated pointers. Second, it makes sense. When subtracting pointers you get the number of elements between those addresses, not the number of bytes. If you were to try that with char *p1

I've been reading K & R's book on C, and found that pointer arithmetic in C allows access to one element beyond the end of an array. I know C allows to do almost anything with memory but I just don't understand, what is the purpose of this peculiarity? C doesn't allow access to memory beyond the end of the array. It does, however, allow a pointer to point at one element beyond the end of the array. The distinction is important. Thus, this is OK: char array[N]; char *p; char *end; for (p = array, end = array + N; p < end; ++p) do_something(p); (Doing *end would be an error.) And that shows the

The C++ standard [sec 5.7] says: 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. So, am I correct in assuming that pointers one-past-the-end of other types than arrays are undefined? For example: int a = 0; vector<int> v(&a, (&a)+1); The above snippet compiles and works just fine (with g++), but is it valid? No, it is legal. 5.7(4) - one paragraph before your quote - says: "For the purposes of these operators, a pointer to

Consider the following code fragment: int (*p)[3]; int (*q)[3]; q = p; q++; printf("%d, %d\n", q, p); printf("%d\n", q-p); I know that pointer arithmetic is intelligent, meaning that the operation q++ advances q enough bytes ahead to point to a next 3-integers-array, so it does not surprises me that the first print is ' 12, 0 ' which means that incrementing q made it larger in 12. But the second print does surprises me. It prints 1! So why would it print 1 instead of 12? it just puzzles me. Like the ++ increment operator, the - subtraction operator with pointers also takes into account the