Dude google C++ Faqs. You will find all your answers. You can also buy some
books
1. C++ Common Knowledge: Essential Intermediate Programming
2. Effective and More effective C++
3. C++ gotchas
On Sat, Sep 24, 2011 at 11:52 AM, Decipher wrote:
> Q1) What does the compiler does if I declare a bas
Q1) What does the compiler does if I declare a base class VIRTUAL ??
Q2) In the below test code ,
class A : public B, public C
{
};
The order of constructor invocation is :
B
C
A
but if C is virtual base class the it changes to :
C
B
A
Why ??
Q3) Write a macro that swaps any data given to it
That was really helpful.
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To view this discussion on the web visit
https://groups.google.com/d/msg/algogeeks/-/KvxrTcPf104J.
To post to this group, send email to algogeeks@googlegroups.com.
To u
1) you cannot change the const variable .
2) generally sizeof behaviour in case of structure is just undefined. Either
it follows the summation of size of members or it uses padding concept ie in
your example 4 byte each for character(1 byte for char rest of the 3 bytes
is padded up) and 4 byte for
1. Yes, you cannot modify a const variable. This itself means that it is
read-only.
2.Google structure padding. It is done to make sure that variables start in
memory at addresses that are a multiple of their size. This is more
efficient at hardware level.
'char' (1 byte) variables can be byte al
1.When I declare a variable as const then any subsequent assignment of it
gives an error "assignment of read only variable". Is a const variable
treated as a read only variable?
2.
#include
struct s1 {
char a;
};
char s2 {
char b;
int c;
};
printf("%d",sizeof(
