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how to initialize 3d array

14
hi,
how do i initialize all the elements of 3d array.

bool isValid[5][5][5];

this is a member variable of a class. how to set all the elements of this member variable to FALSE in the constructor of that class.

thanks
rsennat
Jan 20 '08 #1
2 12273
Ganon11
3,652 Expert 2GB
How would you do this if it were a 1 dimensional array?

If it were a 2 dimensional array?
Jan 20 '08 #2
weaknessforcats
9,207 Expert Mod 8TB
Read this and post again if you still have problems:
First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:
Expand|Select|Wrap|Line Numbers
  1. int array[5];
  2.  
That is an array of 5 elements each of which is an int.

Expand|Select|Wrap|Line Numbers
  1. int array[];
  2.  
won't compile. You need to declare the number of elements.

Second, this array:
Expand|Select|Wrap|Line Numbers
  1. int array[5][10];
  2.  
is still an array of 5 elements. Each element is an array of 10 int.

Expand|Select|Wrap|Line Numbers
  1. int array[5][10][15];
  2.  
is still an array of 5 elements. Each element is an array of 10 elements where each element is an array of 15 int.


Expand|Select|Wrap|Line Numbers
  1. int array[][10];
  2.  
won't compile. You need to declare the number of elements.

Third, the name of an array is the address of element 0
Expand|Select|Wrap|Line Numbers
  1. int array[5];
  2.  
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*.

Expand|Select|Wrap|Line Numbers
  1. int array[5][10];
  2.  
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:
Expand|Select|Wrap|Line Numbers
  1. int array[5][10];
  2.  
  3. int (*ptr)[10] = array;
  4.  
Fourth, when the number of elements is not known at compile time, you create the array dynamically:

Expand|Select|Wrap|Line Numbers
  1. int* array = new int[value];
  2. int (*ptr)[10] = new int[value][10];
  3. int (*ptr)[10][15] = new int[value][10][15];
  4.  
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:

Expand|Select|Wrap|Line Numbers
  1. int** ptr = new int[value][10];    //ERROR
  2.  
new returns the address of an array of 10 int and that isn't the same as an int**.

Likewise:
Expand|Select|Wrap|Line Numbers
  1. int*** ptr = new int[value][10][15];    //ERROR
  2.  
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***.

With the above in mind this array:
Expand|Select|Wrap|Line Numbers
  1. int array[10] = {0,1,2,3,4,5,6,7,8,9};
  2.  
has a memory layout of

0 1 2 3 4 5 6 7 8 9

Wheras this array:
Expand|Select|Wrap|Line Numbers
  1. int array[5][2] = {0,1,2,3,4,5,6,7,8,9};
  2.  
has a memory layout of

0 1 2 3 4 5 6 7 8 9

Kinda the same, right?

So if your disc file contains

0 1 2 3 4 5 6 7 8 9

Does it make a difference wheher you read into a one-dimensional array or a two-dimensional array? No.

Therefore, when you do your read use the address of array[0][0] and read as though you have a
one-dimensional array and the values will be in the correct locations.
Jan 20 '08 #3

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