Programming Forums > Application Development > C++

b1g4L · Feb 4th, 2006, 6:23 PM

I have an array of chars (1 bit).

char *array = new char [15];

If I want to set a pointer to the 6th index in that array, does this code do that?

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char *address = array + 5

DaWei · Feb 4th, 2006, 6:34 PM

Yes, but a char is not one bit, it's one byte (8 bits). While pointers to arrays and arrays appear to be the same thing in some contexts, it isn't strictly true. If this seems confusing to you, you might want to refer to the "Pointer Basics" link in my signature, in particular the section "What is not a pointer."

b1g4L · Feb 4th, 2006, 7:30 PM

Yea, 1 byte (sorry). Does the +1 move to the next spot in memory (in the array) then?

DaWei · Feb 4th, 2006, 7:33 PM

It's called 'array arithmetic.' Again, see my tutorial. For a char array it moves one byte. It moves one element. If the elements are integers, it moves the size of one integer, and so forth. That's one reason the pointer has a type.

b1g4L · Feb 4th, 2006, 10:48 PM

How can I get the memory address of a specific index in an array? Say I wanted to know the memory address of array [5]? Does array [5] need to contain data before you can find out its memory location?

rsnd · Feb 4th, 2006, 10:57 PM

it doesnot need to contain entered data or anything...but it needs to be existant. for the address of array[5] i think you can go &array[5] or array+5 (as array itself is pointer to the addr.)

b1g4L · Feb 4th, 2006, 11:03 PM

I tried that and I dont get a hex address value but instead funny symbols.

rsnd · Feb 4th, 2006, 11:10 PM

I think its because you are outputting it wrong...you are expected to get funny symbols if the adress bytes gets converted into chars.

b1g4L · Feb 4th, 2006, 11:25 PM

This worked. Thanks.

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cout << &array + 5;

Another question....Can a char pointer be set to point to a char array? For instance......

This would be in the main:

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char *array = new char [10];
Class.add(array);

This would be the function receiving 'array'

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add (char *startOfArray);

I want 'array' and 'startOfArray' to have the same memory address (point to the same object).

grumpy · Feb 5th, 2006, 2:33 AM

yes, it is possible. However, you need to ensure that the "array" in main() continues to exist, otherwise almost anything that Class does with pointer it receives will cause undefined behaviour.

For example;

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class MyClass
{
    public:
        void add(char *startOfArray) {start = startOfArray;};
        void manipulate() {start[5] = 42;};
    private:
        char *start;
};

int main()
{
    MyClass Class;
    char *array = new char [10];
    Class.add(array);
    Class.manipulate();
    delete [] array;          //  hence forth, array no longer exists
    Class.manipulate();     // undefined behaviour as this modifies array[5]
    return 0;
}

Feb 4th, 2006, 6:23 PM	#1
b1g4L Programmer Join Date: Dec 2005 Location: SC Posts: 38 Rep Power: 0	address memory in an array I have an array of chars (1 bit). (Toggle Plain Text) char array = new char [15]; char array = new char [15]; If I want to set a pointer to the 6th index in that array, does this code do that? (Toggle Plain Text) char address = array + 5 char address = array + 5

Feb 4th, 2006, 6:34 PM	#2
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Yes, but a char is not one bit, it's one byte (8 bits). While pointers to arrays and arrays appear to be the same thing in some contexts, it isn't strictly true. If this seems confusing to you, you might want to refer to the "Pointer Basics" link in my signature, in particular the section "What is not a pointer." __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 4th, 2006, 7:33 PM	#4
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	It's called 'array arithmetic.' Again, see my tutorial. For a char array it moves one byte. It moves one element. If the elements are integers, it moves the size of one integer, and so forth. That's one reason the pointer has a type. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 4th, 2006, 10:57 PM	#6
rsnd Hobbyist Programmer Join Date: Jun 2005 Location: Helltown Posts: 162 Rep Power: 4	it doesnot need to contain entered data or anything...but it needs to be existant. for the address of array[5] i think you can go &array[5] or array+5 (as array itself is pointer to the addr.) __________________ Spread your wings and fly! Chicken!

Feb 4th, 2006, 11:10 PM	#8
rsnd Hobbyist Programmer Join Date: Jun 2005 Location: Helltown Posts: 162 Rep Power: 4	I think its because you are outputting it wrong...you are expected to get funny symbols if the adress bytes gets converted into chars. __________________ Spread your wings and fly! Chicken!

Feb 4th, 2006, 7:30 PM	#3
b1g4L Programmer Join Date: Dec 2005 Location: SC Posts: 38 Rep Power: 0	Yea, 1 byte (sorry). Does the +1 move to the next spot in memory (in the array) then?

Feb 4th, 2006, 10:48 PM	#5
b1g4L Programmer Join Date: Dec 2005 Location: SC Posts: 38 Rep Power: 0	How can I get the memory address of a specific index in an array? Say I wanted to know the memory address of array [5]? Does array [5] need to contain data before you can find out its memory location?

Feb 4th, 2006, 11:03 PM	#7
b1g4L Programmer Join Date: Dec 2005 Location: SC Posts: 38 Rep Power: 0	I tried that and I dont get a hex address value but instead funny symbols.

Feb 4th, 2006, 11:25 PM	#9
b1g4L Programmer Join Date: Dec 2005 Location: SC Posts: 38 Rep Power: 0	This worked. Thanks. (Toggle Plain Text) cout << &array + 5; cout << &array + 5; Another question....Can a char pointer be set to point to a char array? For instance...... This would be in the main: (Toggle Plain Text) char array = new char [10]; Class.add(array); char array = new char [10]; Class.add(array); This would be the function receiving 'array' (Toggle Plain Text) add (char startOfArray); add (char startOfArray); I want 'array' and 'startOfArray' to have the same memory address (point to the same object). Last edited by b1g4L; Feb 4th, 2006 at 11:49 PM.

Feb 5th, 2006, 2:33 AM	#10
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,207 Rep Power: 5	yes, it is possible. However, you need to ensure that the "array" in main() continues to exist, otherwise almost anything that Class does with pointer it receives will cause undefined behaviour. For example; (Toggle Plain Text) class MyClass { public: void add(char startOfArray) {start = startOfArray;}; void manipulate() {start[5] = 42;}; private: char start; }; int main() { MyClass Class; char array = new char [10]; Class.add(array); Class.manipulate(); delete [] array; // hence forth, array no longer exists Class.manipulate(); // undefined behaviour as this modifies array[5] return 0; } class MyClass { public: void add(char startOfArray) {start = startOfArray;}; void manipulate() {start[5] = 42;}; private: char start; }; int main() { MyClass Class; char array = new char [10]; Class.add(array); Class.manipulate(); delete [] array; // hence forth, array no longer exists Class.manipulate(); // undefined behaviour as this modifies array[5] return 0; }

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Programming Forums > Application Development > C++

address memory in an array