Logical Size of Array

[kosilva]

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

double getLogicalSize( char a[] )
{
	int x;
	for( x = 0; a[x] != '\0'; x++ );
	return x;
}

[pata]

try it... lol

[Fransje]

It won't work for all arrays, because arrays don't automatically end with '\0'.

This is the thing I use:

template < typename T, int size >
std::size_t arraySize( const T ( &array ) [ size ] )
{
    return ( size );
}

And you can use it like this:

int array[ 50 ];
double array2[ 560 ];

std::cout << arraySize( array ) << ' ' << arraySize( array2 ) << std::endl;

50 560

[vlad from moscow]

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>.:)

[amchinese]

You return x(int), but function must return double because you write double getLogicalSize( char a[] )

[Peter87]

@amchinese
If you return an integer from a function with return type double, the integer will automatically be converted to a double.

[kosilva]

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]

Fransje isnt your code return the physical size only??

Yeah, it is. I misunderstood your question :)

[giblit]

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.