using sizeof operator the size of whole array can be calculated, then divide it with the zise of the datatype(E.g. for int =4, char =2....etc.)
Example:
#define Narray (sizeof(array)/sizeof(array[0]))
It really depends on the language. In Java, you can use the .length property.
in c simply add three lines in the begining of your program: int x; printf("enter the size of the array to be entered :"); scanf("%d",&x); after that use x as your maximum limit of array in your program. in c++ just replace above printf & scanf statements by cout<<"enter the size of the array to be entered :"; & cin>>x; respectively and do not use brackets.
The obvious answer is that one has a constant size while the other does not. More specifically, a fixed-size array is one where the size is known at compile time and does not change at runtime. By contrast, the size of a variable-sized array may or may not be known at compile time but may change at runtime. We often refer to a variable-size array as being a dynamic array, however some people (myself included) incorrectly refer to a fixed-size array as being a static array. The misunderstanding largely comes from the fact that we often refer to the heap (or free store) as being dynamic memory because all dynamic variables are allocated there (including variable-size arrays). But the term dynamic array does not refer to the memory, it refers to the dynamic -- as in changeable -- nature of the array itself. By contrast, a fixed-size array is only deemed static if it is statically allocated, in which case it will be allocated in the program's data segment along with all other static variables, global variables and constants. But a local fixed-size array is allocated on the program's stack and is therefore, by definition, non-static. Moreover, you can allocate a fixed-size array on the heap!
/* PROGRAM TO SORT ARRAY ELEMENTS USING BUBBLE SORT*/ #include #include void main() { int i,j,n,t,a[50]; clrscr(); printf("ENTER THE ARRAY SIZE:\n"); scanf("%d",&n); printf("ENTER THE ARRAY ELEMENTS:\n"); for(i=0;i
Yes. The array name is a reference to the array, so you can use sizeof (name) / sizeof (name[0]) to determine the number of elements. Note that sizeof (name) alone gives the length of the array in bytes.
you can write by two ways 1 by giving the size of array at declaration 2 by checking condition
#include "stdio.h" #define SIZE 100; void main() { int array[SIZE], i, size; printf("\nEnter the Size off Array :- "); scanf("%d", &size); printf("\nEnter the Elements of Array :- ")' for(i = 0; i < size; i++) scanf("%d", &array[i]; printf("\nThe Elements of entered Array :- "); for(i = 0; i < size; i++) printf("%7d", array[i]); }
It really depends on the language. In Java, you can use the .length property.
! variable to declase the size of an array in True Basic ! set up a dummy value for array - any initial value > 0 is fine. DIM array$(999) ! ask the user for the length of the array INPUT PROMPT "Enter array size " :size ! resize the array with user defined length MAT REDIM array$(size) ! program end END
To determine the size of an array in C using the keyword sizeof, you would use the syntax: sizeof(array) / sizeof(array0).
#include <stdio.h> #include <conio.h> #define size 10 void main() { int a[size],del,j,i; clrscr(); for(i=0;i<size;i++) {j=i+1; printf(" Enter the %d element of array",j); scanf("%d",&a[i]);} printf(" Enter the element to be deleted"); scanf("%d",&del); del--; for(i=del;i<size-1;i++) a[i]=a[i+1]; a[size-1]=0; printf(" Element has been deleted"); printf(" Array now is..... "); for(i=0;i<size;i++) printf("%d ",a[i]); getch(); }
in c simply add three lines in the begining of your program: int x; printf("enter the size of the array to be entered :"); scanf("%d",&x); after that use x as your maximum limit of array in your program. in c++ just replace above printf & scanf statements by cout<<"enter the size of the array to be entered :"; & cin>>x; respectively and do not use brackets.
The obvious answer is that one has a constant size while the other does not. More specifically, a fixed-size array is one where the size is known at compile time and does not change at runtime. By contrast, the size of a variable-sized array may or may not be known at compile time but may change at runtime. We often refer to a variable-size array as being a dynamic array, however some people (myself included) incorrectly refer to a fixed-size array as being a static array. The misunderstanding largely comes from the fact that we often refer to the heap (or free store) as being dynamic memory because all dynamic variables are allocated there (including variable-size arrays). But the term dynamic array does not refer to the memory, it refers to the dynamic -- as in changeable -- nature of the array itself. By contrast, a fixed-size array is only deemed static if it is statically allocated, in which case it will be allocated in the program's data segment along with all other static variables, global variables and constants. But a local fixed-size array is allocated on the program's stack and is therefore, by definition, non-static. Moreover, you can allocate a fixed-size array on the heap!
The results of that programming error is undefined. You must NEVER EVER write, or EVEN READ an array element beyond the allocated size of the array. Period.I would flunk a student that consistently did this, and I would fire a programmer that did the same.
/* PROGRAM TO SORT ARRAY ELEMENTS USING BUBBLE SORT*/ #include #include void main() { int i,j,n,t,a[50]; clrscr(); printf("ENTER THE ARRAY SIZE:\n"); scanf("%d",&n); printf("ENTER THE ARRAY ELEMENTS:\n"); for(i=0;i
No. An array is a collection of objects of any type, such as doubles, not just characters. You can even have arrays of arrays, or arrays of structs. In C, the size of an array is fixed, but it is possible to write code that will allow you to manually make it variable in size.
Yes. The array name is a reference to the array, so you can use sizeof (name) / sizeof (name[0]) to determine the number of elements. Note that sizeof (name) alone gives the length of the array in bytes.