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How can you quickly find the number of elements stored in a
a) static array b) dynamic array ?

Answers were Sorted based on User's Feedback



How can you quickly find the number of elements stored in a a) static array b) dynamic array ?..

Answer / santhosh kumar sahukari

1.for static array we can find the no.of elements of its by using sizeof() operator and data type of that array.
2.we can find the no.of elements by its subscript in static array.
eg: int arr[10]
we can use sizeof(arr) to find the no.of elements it v l print 20,then we can divide it by 2 as size of int is 2.
2.subscript tells us it has 10 elements.

for dynamic array we cant use sizeof() operator.
we cant say dat how many elements it will have..it will be decided at runtime only.

Is This Answer Correct ?    4 Yes 0 No

How can you quickly find the number of elements stored in a a) static array b) dynamic array ?..

Answer / jaroosh

a) sizeof(array)/sizeof(element)
b) you cant (well, you can, eg. by using realloc, counting
from 0 up, and checking the result etc. but this is absurd),
its best to keep track somehow how many elements are there
in the array, but the BEST solution is...do NOT use dynamic
arrays at all, use std containers instead like vector.

Is This Answer Correct ?    3 Yes 0 No

How can you quickly find the number of elements stored in a a) static array b) dynamic array ?..

Answer / murali

for static array start from highest index to lowest (
normal conditions apply*)
if dynamic array is a Container type, size() / length()
will do. if not process the nodes.

Is This Answer Correct ?    0 Yes 1 No

How can you quickly find the number of elements stored in a a) static array b) dynamic array ?..

Answer / brainless

b) use pointer of last element - pointer of first element + 1

However, we should know the last element's pointer...

Is This Answer Correct ?    0 Yes 2 No

How can you quickly find the number of elements stored in a a) static array b) dynamic array ?..

Answer / guest

static array

Is This Answer Correct ?    1 Yes 5 No

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