Logical Size of Array

Hi, I have a doubt, and wanted to clarify something. Can I get the logical size of an array by doing this:

 ``123456`` ``````double getLogicalSize( char a[] ) { int x; for( x = 0; a[x] != '\0'; x++ ); return x; }``````
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try it... lol
It won't work for all arrays, because arrays don't automatically end with '\0'.

This is the thing I use:
 ``12345`` ``````template < typename T, int size > std::size_t arraySize( const T ( &array ) [ size ] ) { return ( size ); }``````

And you can use it like this:
 ``1234`` ``````int array[ 50 ]; double array2[ 560 ]; std::cout << arraySize( array ) << ' ' << arraySize( array2 ) << std::endl;``````

 ``` 50 560 ```
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It is not clear why you are returning double because "logical size of an array" usually is unsigned int.:)

What you are trying to do is already done and named as std::strlen that is declared in header <cstring>.:)
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You return x(int), but function must return double because you write double getLogicalSize( char a[] )
@amchinese
If you return an integer from a function with return type double, the integer will automatically be converted to a double.
I will change it to int function then.

and vlad from moscow I forgot to meantion I cannot use strlen

Fransje isnt your code return the physical size only?? because sometimes an array may not be loaded completly, so thats the size I want to get
 Fransje isnt your code return the physical size only??

Yeah, it is. I misunderstood your question :)
You could always use a vector then and get the size of that?
and why not just use the function .size() or .length() you can use that on strings, arrays, vectors ect.
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