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It is because array name implies its address and if you want to access first element of it pointer address logic is as below:
Arrays first element address = array base address + 0
Arrays second element address = array base address + 1
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Full defecation of array?
Full representation of an array begins from the index 0 and ends at n-1 where n is the number of variables of the array.
Why the array is starting from zero?
By design; it makes the compiler's work easier. 1-based array's addressing-function: Address (array, index) = Address (array) + (index-1)*Elemsize(array) 0-based array's addressing-function: Address (array, index) = Address (array) + index*Elemsize (array)
What is the Program to find the largest element in an array and position occurrence?
#include<stdio.h> #include<conio.h> main() { int a[100]; int n,largest,index,position; printf("enter the number of elements in the array"); scanf("%d",&n); printf("enter %d elements",n); for(index=0;index<n;index++) scanf("%d",&a[index]); largest=a[0]; position=0; for(index=1;index<n;index++) if(a[index]>largest) { largest=a[index]; position=index; } printf("largest element in the array is %d\n",largest); printf("largets element's position in the array is %d\n",position+1); getch(); }
What is the definition of a square array?
A square array has the same number of columns and rows the array [1] is a square array (a trivial example) the array [1 0] [0 1] is a square array the array [1 0 0 0] [0 1 0 0] [0 0 1 0] [0 0 0 1] is a square array the array [1 0 0 0] [0 1 0 0] [0 0 1 0] is not a square array
What is the index number of the last element of an array with 9 elements?
(array.length - 1) will find the index of the last element in an array (or -1 if the array is empty).
If a CBT is stored using array , then what is the parent node of element stored at index 11?
In a complete binary tree (CBT) stored using an array, the parent of an element at index i can be found at index (i-1)/2, assuming the array is 0-indexed. So for an element stored at index 11, the parent node would be stored at index (11-1)/2 = 5.
How do you write a C program to find the largest of 10 integers in an array?
#include<stdio.h> // returns the index of the largest value in the array int max (int a[], unsigned size) { if (!size) return -1; // invalid array int index = 0; // assume index 0 holds the largest value (so far) for (int i=1; i<size; ++i) { // traverse the remainder of the array if (a[i]>a[index]) { // compare with the current largest index = i; // the current value is larger } } return index; // a[index] holds the largest value } int main (void) { int x[10] = {7, 4, 3, 9, 5, 2, 1, 8, 6}; printf ("The largest value in the array is %d\n", max (x, 10)); return 0; }
How do you start array list from 1?
Array indices are zero-based because the first element is at offset zero from the start of the array. There is no such thing as a one-based array in C. Some other languages do allow you to specify the base of an array, but converting a non-zero-based index to a zero-based index adds a runtime overhead to the implementation.
Why in c plus plus index always start with 0?
C++ array indices are zero-based because the first element in any array is offset 0 elements from the start address. The second element is offset by 1 element and the third by 2 elements, and so on. To put it another way, the index refers to the number of elements that come before the desired element. The first element has zero elements before it, so it is index 0. For an array of n elements, the last element is at index n-1.
How can you fix an ArrayIndexOutOfBoundsException in java?
This is very common exception and name of the exception class tells exactly what kind of problem you have. If you would look into stack trace, which should have been generated too you would be able to find exact place where it happened. The problem is that "Array Index Out Of Bounds". This means that you used index on array which is invalid. That could negative number, because all arrays starts from 0. If you array has N items and you will try to get item with index N or higher you will get this exception too. Only available indexes are from 0 to N - 1, where 0 points to the first item and N - 1 to the last one.
Why you will declare top as -1 for stacks to be empty in c program?
You would do this if you implement a stack using an array. Using a zero-based index to keep track of the top of the stack (the end of the array) means we must use the value -1 to indicate an empty stack. This is because an array of n elements will have indices 0 through n-1, where index n-1 represents the element at top of the stack. An empty stack has 0 elements, therefore the top of the stack is represented by index -1.
How To find a maximum number in a 2D array?
// Pseudocode int findMax( int[][] data ) { // Return if data is empty if( data.length 0 ) { return 0; } int max = data[0][0]; // Iterate through each element in the array for( int r = 0; r < data.length; ++r ) { for( int c = 0; c < data[0].length; ++c ) { // If we find a value greater than the current max, update max if( data[r][c] > max ) { max = data[r][c]; } } } return max; }
