is there a difference between nested and multi-level array according to their running time?
actually what is the differences between them?
1 5961
Nope. Just the pointer arithmetic.
Read this:
First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:
That is an array of 5 elements each of which is an int.
won't compile. You need to declare the number of elements.
Second, this array:
is still an array of 5 elements. Each element is an array of 10 int.
is still an array of 5 elements. Each element is an array of 10 elements where each element is an array of 15 int.
won't compile. You need to declare the number of elements.
Third, the name of an array is the address of element 0
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*.
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.
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