related to pointer

[ashwini1680]

Hi,
I have a very basic problem related to strings using pointers in c++
//////////////////
i wrote code as,
int *i_ptr;
char *c_ptr;

int i=5;char c='A';
i_ptr=&i;
c_ptr=&c;

cout<<i_ptr;
cout<<c_ptr;

As an o/p i get an address of i_ptr but c_ptr doesn't store any addr WHY?

How sting are stored in memory(1 dim &2dim both) &how pointers access them?

As when we write a[1] we get addr if a is an array of pointers but in case of strings it returns string n not the address ,how?

[mschenkelberg]

I think it is because a char * is always interpreted as a "C String" , so it will try to output values until it reaches a null character. I think if you do something like &c_ptr, it will show the address

[weaknessforcats]

I think if you do something like &c_ptr, it will show the address

Nope. That will get the address of c_ptr but not the address inside c_ptr.

You are correct though, in that the operator<< for char* assumes a C-style string.

What you have to do is create an int out of the pointer.
[code=cpp]

unsigned int result(c_ptr);

[/code]

then you can display result and see the address.

This syntrax shown is the constructor form for creating a variable. It is equivalent to doing:
[code=cpp]
cout << static_cast<uns igned int> (c_ptr);
[/code]

[ashwini1680]

I think it is because a char * is always interpreted as a "C String" , so it will try to output values until it reaches a null character. I think if you do something like &c_ptr, it will show the address

Hi,
Thank you.I changed my code.now it's working.Still i am confused with 2d array of char using pointers.

[weaknessforcats]

Still i am confused with 2d array of char using pointers.

Start a new thread. This is a new question.

[ashwini1680]

Hi,
I have a very basic problem related to strings using pointers in c++
//////////////////
i wrote code as,
int *i_ptr;
char *c_ptr;

int i=5;char c='A';
i_ptr=&i;
c_ptr=&c;

cout<<i_ptr;
cout<<c_ptr;

As an o/p i get an address of i_ptr but c_ptr doesn't store any addr WHY?

How sting are stored in memory(1 dim &2dim both) &how pointers access them?

As when we write a[1] we get addr if a is an array of pointers but in case of strings it returns string n not the address ,how?

As the replies I got for the same problem i changed my code as
cout<<&c_ptr
but it gives the addr of pointer & not the addr of te var to which it points.
Where the addr is stored?
as we initialises pointer with the addr.
Same if i have

char *a[]={"Pointer","Co mputer","Scienc e"};
then
printing
a[0],a[1],a[2] gives me strings themseleves.How ?

because to swap the strings 2 & 3,we use
char *t;
t=a[1];
a[1]=a[2];
a[2]=t;
here it takes addresses & if we print a[0],a[1] we get string .I don't understand this whole concept.I am very Congused.Help me.

[shabinesh]

Hi,
*a[],**a,a[][] all means the same in their concept.
*a and a[] means the same ,so you needn't use any pointer for printing the values in "a",
char *t;
t =a[1];
a[1]=a[2];
a[2]=t;
i suppose its a mistake ,it should
char *t;
* t=a[1];
a[1]=a[2];
a[2]=*t;
for better refrence, check out 'The C Programming language' by K&R.

[ashwini1680]

Hi,
*a[],**a,a[][] all means the same in their concept.
*a and a[] means the same ,so you needn't use any pointer for printing the values in "a",
char *t;
t =a[1];
a[1]=a[2];
a[2]=t;
i suppose its a mistake ,it should
char *t;
* t=a[1];
a[1]=a[2];
a[2]=*t;
for better refrence, check out 'The C Programming language' by K&R.

hi,
I checked my code.It's correct,I have taken it from a book.
the point of confusion for me is that,
pointer var always stores addresses.
so when we write,
int *ptr;
int i=10;ptr=&i;
then if we write,
cout<<ptr
we get address of int i stored in ptr.
but in case of char
like

char *ptr;
char c='a'
ptr=&c
or
char *ptr
char c="pointer"
ptr=&c
Now if we print value of ptr we dont get address.Why?

if c is a char variable we get some strange characters & if c is a string then we get the string itself. How???

[Laharl]

In C, arrays of characters (also referred to syntactically with char*), are the only available implementations of strings. These strings end with a special character, '\0', and when a char* is passed to <<, it expects one of these C-style strings, even in C++, so it prints until it sees \0. Since in this case, it doesn't, you get weird results as the "array" goes out of bounds.

[shabinesh]

char *ptr;
char c='a'------------------(1)bug here
ptr=&c
or
char *ptr
char c="pointer"---------------(2) bug here
ptr=&c

(1).The char "a" is not stored in the var c the integer value is stored since you use a single quotes('a').
(2).The char c cannot hold a string it can hold only a character.

[weaknessforcats]

Almost nothing of what I have read here about pointers and arrays is correct.

First an int** does not point to a 2D array and an int***does not point to a 3D array. There are various reasons but the main one is that ther are no multi-dimensional arrays in C or C++. There are only one-dimensional arrays.

Read on :

First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:
[code=c]
int array[5];
[/code]

That is an array of 5 elements each of which is an int.

[code=c]
int array[];
[/code]

won't compile. You need to declare the number of elements.

Second, this array:
[code=c]
int array[5][10];
[/code]

is still an array of 5 elements. Each element is an array of 10 int.

[code=c]
int array[5][10][15];
[/code]

is still an array of 5 elements. Each element is an array of 10 elements where each element is an array of 15 int.


[code=c]
int array[][10];
[/code]

won't compile. You need to declare the number of elements.

Third, the name of an array is the address of element 0
[code=c]
int array[5];
[/code]

Here array is the address of array[0]. Since array[0] is an int, array is the address of an int. You can assign the name array to an int*.

[code=c]
int array[5][10];
[/code]

Here array is the address of array[0]. Since array[0] is an array of 10 int, array is the address of an array of 10 int. You can assign the name array to a pointer to an array of 10 int:
[code=c]
int array[5][10];

int (*ptr)[10] = array;
[/code]

Fourth, when the number of elements is not known at compile time, you create the array dynamically:

[code=c]
int* array = new int[value][5];
int (*ptr)[10] = new int[value][10];
int (*ptr)[10][15] = new int[value][10][15];
[/code]

In each case value is the number of elements. Any other brackets only describe the elements.

Using an int** for an array of arrays is incorrect and produces wrong answers using pointer arithmetic. The compiler knows this so it won't compile this code:

[code=c]
int** ptr = new int[value][10]; //ERROR
[/code]

new returns the address of an array of 10 int and that isn't the same as an int**.

Likewise:
[code=c]
int*** ptr = new int[value][10][15]; //ERROR
[/code]

new returns the address of an array of 10 elements where each element is an array of 15 int and that isn't the same as an int***.