CODEWITHSUNDEEP
cpointers
Toby
Toby

Reputation: 141

Is my logic correct for Double pointers/Pointers in C for this example

int age;

int *p;

int **q;

p=&age;

q=&p;

For these set of variables, I'm trying to figure out the logic of how the pointers work w.r.t each other. I have coded it but I want to mentally be able to work out different scenarios using pointer logic which I am struggling with.

Case 1: age == *p

So, a pointer points to a location in memory and stores the address of a variable. So, my thinking is that since p is a variable and *p stores it address and we also know that p=&age (p has address of age). So, *p ==age because they essentially have the same address.

Case 2: age == **q

For this I drew a diagram.

I had *q stores address of **q and **q stores address of variable q. So, since variable q = &p and p = & age, so q=& age and thus making the statement age==*q true

Case 3: age == *q

This is false but I'm not sure why. *q stores address of q which is the same as &p and since age is same as p which is same as q, it should be equal? I compiled it and its different but I dont see why though.

Case 4: p ==*q

This again is True. q=&p so q* stores q's address which is basically the address of p.

This is how I see things for this question. Any guidance/correction on how to approach this and any flaws in my logic would help a lot.

Upvotes: 0

Views: 41

Answers (2)

John Bollinger
John Bollinger

Reputation: 180113

Case 1: age == *p

So, a pointer points to a location in memory and stores the address of a variable. So, my thinking is that since p is a variable and *p stores it address and we also know that p=&age (p has address of age). So, *p ==age because they essentially have the same address.

True, and your reasoning is essentially correct. I would phrase the last bit differently, however: since p contains the address of age, it is necessarily the case that *p and age designate the same object. That's what unary * does. You could also say that for any identifier x that designates an object, *&x == x.

Case 2: age == **q

For this I drew a diagram.

I had *q stores address of **q and **q stores address of variable q. So, since variable q = &p and p = & age, so q=& age and thus making the statement age==*q true

True, and again your logic is essentially correct. You might also argue that, by the same reasoning as in Case 1, it is true that *q == p. From case 1, you know that *p == age, so we can combine those to say that *(*q) == age, which is the same as **q == age.

Case 3: age == *q

This is false but I'm not sure why. *q stores address of [p] which is the same as &p [...]

So far so good.

[...] and since age is same as p

STOP! Flag on the play! age is not the same as p. It is the same as *p, which is quite different.

Suppose I write the street address of my house on a piece of paper. Is the paper the same thing as the house? Of course not. In the same way, if I store the address of an object in a pointer variable, neither the pointer variable nor the pointer value stored in it is the same thing as the object to which the value points.

Case 4: p ==*q

True. This the same as case 1.

Upvotes: 2

Koori
Koori

Reputation: 61

For case 4, you are trying to compare the value of age against the address of age.

Should be &age == *q or age==**q

You can try to print the value of the pointer against the address of age and you might understand further.

Upvotes: 0

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