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ROB 502: Programming for Robotics: Homework 3

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int val1 = 10;
int val2 = 20;
int *b = &val1; // b points to val1
int *c = b; // c copies the value of b, so it also points to val1
int **d = &b; // points to b

There is nothing all the special about taking a pointer to a pointer. Just like val1 is a local variable, b is also a local variable, and the pointers for them both work basically the same. The difference is that there is now one more level of indirection. So d points to b, which currently points to val1.

printf("%d\n", **d); // 1

Since d points to b which points to val1, the answer is 10. The important thing to notice is that the types do work. The type definition int **d means any one of the following:

  1. int** d: d is a pointer to a pointer to an integer, the value actually stored in d.
  2. int* *d: *d is a pointer to an integer, and that would be b, since d points to b.
  3. int **d: **d is an integer, and that would be 10, since d points to b which points to val1 which has the value 10.
*b = 1; // b points to val1, so now val1 = 1, and val1 does not equal 10 anymore!
b = &val2; // b NOW points to val2, and does not point to val1 anymore!

When we modify *b this doesn’t change b, it changes val1. When we modify b we are changing only the value of b and it will work as a pointer to something else now.

printf("%d\n", *b); // 2

This is a tricky one since we just did so many things with b. The important thing is that *b = 1 actually was a change to val1, and not to b. Then b = &val2 actually causes b to point to val2, which has a value of 20, and it doesn’t really matter what we did before.

printf("%d\n", *c); // 3

Since the beginning, c has been a pointer to val1, and it still is! Right now val1 has a value of 1, which was set when b pointed to val1.

printf("%d\n", **d); // 4

d is a pointer to b and b is a pointer to val2, so again we get a value of 20.

*d = &val1; // d points to b, so now b is a pointer to val1 again!

printf("%d\n", *b); // 5

Since we just used d to make b point to val1, we get the value of val1: 1.

printf("%d\n", *c); // 6

c really hasn’t changed at all! It is still a pointer to val1 and so again we get 1.

printf("%d\n", **d); // 7

d still points to b which still points to val1: 1.

If you have would like to see this all a bit more visually, I encourage you to run this example with CTutor.