simple question regarding pointers..

Hey guys.. I'm hoping for some clarification on this situation i created.

i was playing around with pointers and this situation came up.
(please forgive me if im just being absent minded and it's a syntax error.)

Code:

char* b_stringp;
b_stringp = "abcd";
cout << b_stringp << endl;

the above code does just as expected and outputs "abcd"

then i tried to do a similar thing with a int

Code:

int* b_int;
b_int = 234;
cout << b_int << endl;

im a little confused as to why this code didn't work?? it seems to have the same logic as the previous code.

then i thought to fix this problem would be to dynamically allocate memory.

Code:

int* a_int;
a_int = new int;
a_int = 234;
cout << a_int << endl;

and realized i had to put an astrix infront

this code tho did work

Code:

int* a_int = new int;
*a_int = 234;
cout << *a_int << endl;

from my understading this actualy makes a pointer that points to a area of memory labeled 234.

anyways i was hoping someone could clarify the above for me.. also show me how to properly make an int * and on another line dynamically allocate memory to it. and then insert data into it and "cout" it.. hope i can get some clarifying here

[EDIT as per Todd's post]I would have to say a pointer represents an address, and points to a location in memory.[/EDIT]

When you code something like a string literal (i.e. "I am awesome") your compiler will place that string literal inside your executable and interpret a pointer to that string literal as the address of that memory location.

Think of it like putting it in a filing cabinet and knowing its label. So when you are dealing with a string that is all well in good. But when you say something like int *x = 52 you are basically saying that the label inside the filing cabinet is 52, give me whatever is inside that folder. So instead of printing out "52" it goes to the memory address 0x34 and tries to read whatever int may reside there (which will likely cause an access voilation).

Originally Posted by master5001

Well a pointer points to an address in memory.

I would have to say a pointer represents an address, and points to a location in memory.

Yeah I think that is a better wording. I was trying to work in some genious analogy. Needless to say, Elysia's house address analogy was a little more coherently formed.

**Changed with due credit**

Code:

char* b_stringp;
b_stringp = "abcd";
cout << b_stringp << endl;

is different from the code where you use int * in the sense that char * is a special case of pointers - they are generally seen as strings, and cout will take a char * and print the letters point to by the pointer, rather than printing the value of the pointer.

If we really want to, we can set a pointer to a value that is not a valid memory location - but of course, that means that if we dereference the pointer, the code will crash. For example:

Code:

int *ptr = reinterpret_cast<int *>(234);
cout << ptr << endl;

That will print 234 (but in hex, so it will be something like 0xEA [0xF0 == 240, subtract 6 => 0xEA - I haven't verified with a calculator], possibly with a load of zeros at the front and no 0x).

Of course, if we do *ptr = 7 after that, it will give an "invalid memory access", unless it's DOS or some embedded OS that lacks the bigger OS's memory protection which prevent errant pointers from being used.

--
Mats

thanks for the reply guys..

did anyone have a way of
showing me how to properly make an int * and on another line dynamically allocate memory to it. and then insert data into it and "cout" it

Originally Posted by MegaManZZ

thanks for the reply guys..

did anyone have a way of
showing me how to properly make an int * and on another line dynamically allocate memory to it. and then insert data into it and "cout" it

Yes. You did, in fact. Copying and pasting from your post:

Code:

int* a_int;
a_int = new int;
*a_int = 234;
cout << *a_int << endl;

This may help some:
Memory Functions - Still Learning
Remember to use new/delete instead of malloc/free.