problem with sizeof()

[kikirikou]

Hello ,in the following code i use a double variable and a double pointer and i use the sizeof() function to examine their length.


#include<iostream>
using namespace std;

int main()
{
double some;
double* thing;

cout<<"size of some is:"<<sizeof(some)<<endl;
cout<<"size of thing is:"<<sizeof(thing)<<endl;


return 0;
}

But i can't understant the output:
size of some is:8
size of thing is:4

Why the size function returns less leng for the pointer although both the variable "some" and the pointer "thing" are declared as doubles?Thank you

[gcampton]

One is a double one is a pointer.

The fact that you declared as "double* thing;" simply means it's a pointer, that points to a double. It's still just a pointer though.

This is why we use pointers, it's cheap, less memory.

[stereoMatching]

because thing only store the address of "double" type
And I guest your os is 32bit's,right?
that is why the size of pointer would be 4byte = 32bits

[kikirikou]

yeap,my os is 32.That's why however i declared i pointer(int,double.etc) the sizeof returns me the same size ,thus 4. Ok got it,thanks you all.

[sohguanh]

Actually unlike Java, the sizeof values can be different depending you compile your C/C++ programs on which platform. Of cuz from what I "play" around with, 4 bytes seem a common value for pointer datatype using Linux gcc/g++.

Anyone report see-ing 8 bytes? Say on a 64bit OS maybe?

[jsmith]

Try storing a pointer to a virtual method declared in a class that multiply inherits.
Size of such a pointer will be at least 6.