Question about arrays and pointers, please help

[QiongZ]

I came across this when I was reading a book on Java.

In the book, there is a note trying to explain what's equivalent in C++ to

double[][] balance = new double[10][6];  // Java

So my question is what's the difference between

double balance[10][6];

and

double (*balance)[6] = new double[10][6];

and

double** balance = new double* [10];

I am learning c++ without a teacher and have no where to turn to. Arrays and pointers are getting very confusing. Your help is highly appreciated.

[puneetsardana88]

double balance[10][6];

Its an array oy type double with dimension 10X6


double (*balance)[6] = new double[10][6];

Its an array of pointers which points to array of type double with dimension 10X6.


double** balance = new double* [10];

Balance is a pointer to pointer which points to an array of pointers of size 10.

[Banfa]

double (*balance)[6] = new double[10][6];

Its an array of pointers which points to array of type double with dimension 10X6.

double (*balance)[6];

is not an array of pointers, it is a pointer to an array of size 6. The pointer to the array of size 6 is pointed to an array of size 10 of arrays of size 6.

An array of pointers would be

double *balance[6];

</pedantic>

[puneetsardana88]

@Banfa

You are right

[weaknessforcats]

Read this:

http://bytes.com/topic/c/insights/77...rrays-revealed

[QiongZ]

Thanks everyone and what a great post, weaknessforcats!

Now I understood

double (*balance)[6] = new double[10][6];

but still don't know what

double** balance = new double* [10];

means in C++?

In your article, you said

double** balance = new double [10][6];

would be error.

Is

new double [10][6];

equivalent to

new double* [10];

?

Thanks very much!

[Banfa]

double** balance = new double* [10];

balance is a pointer to a pointer to a double. This is actually a very basic array assignment. If you want a dynamically allocated array of some abstract type T then you declare a pointer to T and new an array of T like so

T *array = new T[<dimension>];

In this case T is double *, a pointer to double (and <dimension> is 10) giving

double **array = new double*[10];

Just the same as allocating a normal int array, setting T to int gives

int *array = new int[<dimension>];

new double [10][6];

is in no way equivalent to

new double* [10];

The first allocates a 2 dimensioned array of doubles, or rather an array of arrays of doubles. The second allocates an array of pointers to double. Very very different.

[QiongZ]

Thanks a lot Banfa for taking time to explain in detail. Also you explained it in a way very easy to understand.

Every time I post here, I get excellent answers. I can't thank you guys enough.

Qiong