Pointer outputs

Question

I'm just trying to understand the difference in how you output pointers.

let's say I have:

int x = 100; int*p = &x;

what would each of the following do?

cout << p << endl; cout << *p << endl; cout << &p << endl;

@TiagoRodrigues, just compiling and running can tell you a lot, especially using the sizeof function. — StephenH
– StephenH, Commented Apr 3, 2014 at 19:14
If it is a hardware type of company, I would be surprised if this kind of question is asked. The C++ questions may not even occur. Expect MATLAB or LabVIEW oriented questions though. If you know Verilog, VHDL, and/or SystemVerilog, I can see questions about them occurring too. — CPlusPlus OOA and D
– CPlusPlus OOA and D, Commented Apr 3, 2014 at 19:17
They do some programming too. Mostly scripting though. This is just in case a pointer question comes up. — TRod
– TRod, Commented Apr 3, 2014 at 19:18

4pie0 · Accepted Answer · 2014-04-03 19:09:52Z

3

cout << p << endl; // prints the adress p points to, // that is the address of x in memory cout << *p << endl; // prints the value of object pointed to by p, // that is the value of x cout << &p << endl; // prints the address of the pointer itself

answered Apr 3, 2014 at 19:09

4pie0

29.9k10 gold badges86 silver badges125 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

4pie0 Over a year ago

I didn't say how this will be printed but what it does

Deduplicator Over a year ago

@lizusek: Was not a criticism, just wanted to add that info.

slugonamission · Accepted Answer · 2014-04-03 19:34:49Z

int*p = &x; creates a pointer, p, which points to the variable x. A pointer is actually implemented as a variable which holds the memory address which it is pointing to. In this case, you will get the following. For the purposes of this example, assume that x is stored at 0x1000 and p is stored at 0x1004:

cout << p << endl; will print the address of x (0x1000).
cout << *p << endl; will dereference the pointer, and hence will print the value of x (100).
cout << &p << endl; takes the address of p. Remember that pointers are simply variables in their own right. For this reason, it will print 0x1004.

Note that the address of x is not always 0x1000, the size and value of the pointer varies by system. Sames goes for &p, size and value varies.
+1 if you fix the implication that the addresses are somehow predictable

Collectives™ on Stack Overflow

Pointer outputs

2 Answers 2

2 Comments

3 Comments

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

3 Comments

Related