Fully Initializing an array

[alfons]

Hey,

In TC/C++

I know that if we intialize just the 1st digit of an Integer Array ( int a[3][3]={1} )
its output with will be:
100
000
000

But what if we want the output of a 2-D array a[3][3] like that.
***
***
***
Or with any other character or number .

[oler1s]

Well, if your array holds characters rather than integers, perhaps you should make it a char array?

[Savage]

You can do it like:

int a[3][3]={{0,0,0},{0,0,0},{0,0,0}};
 
//or 
 
int a[3][3]={0,0,0,0,0,0,0,0,0};

same with chars you can do:

char a[3][3]={{"ab"},{"cd"},{"ef"}}
 
 
//or 
 
char a[3][3]={{'a','b'},{'c','d'},{'e','f'}};

and two more combinations without inner '{' and '}'.

[weaknessforcats]

And you migh as well read this:

First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:

int array[5];
 

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

int array[];
 

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

Second, this array:

int array[5][10];
 

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

int array[5][10][15];
 

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

int array[][10];
 

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

Third, the name of an array is the address of element 0

int array[5];
 

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*.

int array[5][10];
 

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:

int array[5][10];
 
int (*ptr)[10] = array;
 

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

int* array = new int[value];
int (*ptr)[10] = new int[value][10];
int (*ptr)[10][15] = new int[value][10][15];
 

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:

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

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

Likewise:

int*** ptr = new int[value][10][15];    //ERROR
 

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:

int array[10] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Wheras this array:

int array[5][2] = {0,1,2,3,4,5,6,7,8,9};
 

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.

[alfons]

Yea..
But the catch is that i want to initialize a single entry , so that it can fill it up with the same .
like i did with the 0 one

[weaknessforcats]

Then you will need to write code for that.