sizeof

[awaise17]

hello every one....i am facing a little problem...my instructor used sizeof(x)/sizeof(int) to find the size of an array x[7]....but i coudlnt get it...can any one please elaborate it for me???

[SirSmilesaLot]

say u wanna find out size of int X[7];

assume u donno its an array of seven integers.

sizof(x)/sizeof(int)
ie
sizeof(seven integers)/sizeof(one integer)=7
ie the size of array

[vlad from moscow]

Maybe you use a pointer instead of the array. For example consider the following code

#include <iostream>

void func( int array[7] )
{
   std::cout << "In func: " << sizeof( array ) / sizeof( int ) << std::endl;
}

int main()
{
   int array[7];

   std::cout << "In main: " << sizeof( array ) / sizeof( int ) << std::endl;

   return 0;
}

Here in the main the result of outputting will be 7 - size of the array. However in func the result of outputting will be usually equal to 1. What is the matter?
When an array is pased to a function which has a parameter declared as int array[7] or int array[] the array is converted to the pointer to its first element. So the declaration

void func( int array[7] );

is equivalent to

void func( int *p );


Size of a pointer to int usually is equal to 4, And size of int usually is equal to 4. So in the function you will get 1.

[R0mai]

If you pass the array by reference :

void func( int (&a)[7] )
{
   std::cout << "In func: " << sizeof( a ) / sizeof( int ) << std::endl;
}

then it will output the correct result (7).

Someone here, in this forum had a very nice solution to get the size for any static array :

template<class T, std::size_t N>
std::size_t asize(const T (&)[N]) {
	return N;
}

Of course, you should always use boost::array or std::array if you're using static arrays.


Edit : 666th ;)

[awaise17]

thanks guys....for your help....especially anirudh sn, you rocked :-)