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Given that I'm trying to acquire the information of the size of this struct:
Logically thinking, it should be the size of 2 ints. However, when I checked in the debugger, it gives me a value of 0 when checking its size using sizeof(). Is this normal, or is there something wrong with how I used the struct to get a sizeof = 0?Code:struct Point2D {int x,y;};
Thanks.
So you are saying that this prints 0 for you?
Code:#include <iostream>
struct Point2D {int x,y;};
int main()
{
std::cout << sizeof(Point2D) << std::endl;
}
I checked with a debugger again by assigning its value to a variable. That variable isn't 0. But when I checked "sizeof(Point2D)" in the debugger, it gives me a value of 0. My confusion is with what the debugger is doing.
Maybe the problem is that the debugger is (or you are, when using the debugger) a little confused by the fact that sizeof is a compile time operator rather than a function.
But how does that explain the case when "sizeof(int)" does not return 0 rather than a 4 when I evaluated that value in the debugger?
Post the smallest and simplest compilable program that demonstrates this problem. It might also be a good idea to post about which debugger you used and how you went about using it.
I was using Visual Studio debugger. I just typed in sizeof(whatever) in the watch window. Not sure how posting a simple compilable program would help.
I have a feeling that this might be the problem: it just does not make sense to me to try and run the debugger when you do not have a compilable program to run. Presumably sizeof(int) works because that is hard coded into the debugger.Quote:
Originally Posted by stanlvw