Setting pointer to an array c++

Question

I have this problem for CS. I'm using the code p3 = &s1[3]; however when I cout p3 it displays the value instead of the address. Is this to expected and the problem is right or is there something I am doing wrong?

Given the following declarations:
int i = 3, j, *p1 = &i, *p2;
char s1[20] = "Wow, more pointers!", *p3;
Write a statement that makes p3 equal to the address of the fourth element of s1.

my apologies I was typing fast I was already using the code p3 = &s1[3]; so it compiles fine the output is still the array of characters and I want it to be the address of that element alone — user4892930, Oct 11 '15 at 22:48
We don't care how fast you type, but we expect complete code that was tested before posting. We don't like to chase ghosts. — Karoly Horvath, Oct 11 '15 at 22:57

score 4 · Answer 1 · answered Oct 11 '15 at 22:52

The << operator for streams has several overloads. One of them takes a char*, and prints a string (a sequence of chars starting from that address, until the next 0). A different one takes a void* (a pointer to anything), and prints the address it points to.

If you cast the pointer to void* yourself, it will force the compiler to choose the void* overload that prints the address:

cout << (void*)p3;

Your pointer assignment is not wrong; the problem is when you print it.

score 2 · Answer 2 · edited May 23 '17 at 11:52

2

That's how std::cout treats (char*). This question contains comprehensive answer.

edited May 23 '17 at 11:52

Community

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answered Oct 11 '15 at 22:58

Gerald Agapov

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score 0 · Answer 3 · answered Oct 11 '15 at 22:17

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By assigning to *p3 you are not modifying the pointer p3, you are modifying the memory that it points to. To modify the pointer p3 itself, just assign to p3.

answered Oct 11 '15 at 22:17

Isabelle Newbie

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The original post has no assignment to p3. – cliffordheath Oct 11 '15 at 22:22
@cliffordheath: read the entire question in full – dreamlax Oct 11 '15 at 22:24
@dreamlax: The OP has no assignment to p3. It assigns to the uninitialised *p3 instead. – cliffordheath Oct 12 '15 at 00:42
@cliffordheath I don't see anybody claiming that the original post contained an assignment to `p3`, but whatever. I think the OP's question has been answered. – Isabelle Newbie Oct 12 '15 at 17:27

Emil Laine · Answer 4 · 2015-10-11T22:25:55.563

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The assignment asks you to make "p3 equal to the address of the fourth element of s1", but you're assigning the address to *p3, not to p3.

*p3 is what p3 points to. p3 is the address.

Btw your *p3 = &s1[3]; shouldn't even compile. If it does, enable your compiler warnings and errors.

edited Oct 11 '15 at 22:25

answered Oct 11 '15 at 22:17

Emil Laine

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The original post has no assignment to p3. – cliffordheath Oct 11 '15 at 22:21
@cliffordheath: The OP says "I'm using the code `*p3 = &s1[3];`" – dreamlax Oct 11 '15 at 22:22
"my apologies I was typing fast I was already using the code p3 = &s1[3];" - apparently `*p3` was a typo in the question, and this is not the problem. – user253751 Oct 11 '15 at 22:50
@dreamlax: I stand by my claim. The original version has no assignment to p3. – cliffordheath Oct 12 '15 at 00:33

score -1 · Answer 5 · answered Oct 11 '15 at 22:21

p3 = s1+3;

Because the elements are at s1+0, s1+1, s1+2, s1+3, ...

Note that this is exactly equivalent to saying:

p3 = &s1[3];

... and some people who are uncomfortable with pointer arithmetic might prefer that. Personally I prefer the first version I gave.

Setting pointer to an array c++

5 Answers5