How to compute size of an array parameter in a function?
Consider below C++ program:
The above output is for a machine where size of integer is 4 bytes and size of a pointer is 8 bytes.
The cout statement inside main prints 40, and cout in findSize prints 8. The reason is, arrays are always passed pointers in functions, i.e., findSize(int arr) and findSize(int *arr) mean exactly same thing. Therefore the cout statement inside findSize() prints size of a pointer. See this and this for details.
How to find size of array in function?
We can pass a ‘reference to the array’.
The above program doesn’t look good as we have hardcoded size of array parameter. We can do it better using templates in C++.
We can make a generic function as well:
40 40 80 80
Now the next step is to print the size of a dynamically allocated array. It’s your task man ! I’m giving you a hint.
This article is contributed Swarupananda Dhua Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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