[algogeeks] C++ Doubts !!
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 (Eg: char , int , pointer
of any kind , float )
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Re: [algogeeks] C++ Doubts !!
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 ankurseth...@gmail.com wrote:
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 (Eg: char , int , pointer
of any kind , float )
--
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[algogeeks] C Doubts
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.
#includestdio.h
struct s1 {
char a;
};
char s2 {
char b;
int c;
};
printf(%d,sizeof(struct s1)); // output : 1
printf(%d,sizeof(struct s2)); // output : 8
please explain..
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Re: [algogeeks] C Doubts
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 aligned and appear at any byte
boundary.
'int' (4 byte) variables must be 4 byte aligned ( they can only appear at
byte boundarys that are a multiple of 4 bytes). So, here
char b --- At any available byte
int c -- 3 bytes from b.
So, a total of 1+3+4 = 8 bytes.
On Sun, Jul 17, 2011 at 11:38 PM, Abhi abhi123khat...@gmail.com wrote:
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.
#includestdio.h
struct s1 {
char a;
};
char s2 {
char b;
int c;
};
printf(%d,sizeof(struct s1)); // output : 1
printf(%d,sizeof(struct s2)); // output : 8
please explain..
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Re: [algogeeks] C Doubts
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 integer .This is more efficient
On Sun, Jul 17, 2011 at 11:38 PM, Abhi abhi123khat...@gmail.com wrote:
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.
#includestdio.h
struct s1 {
char a;
};
char s2 {
char b;
int c;
};
printf(%d,sizeof(struct s1)); // output : 1
printf(%d,sizeof(struct s2)); // output : 8
please explain..
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Re: [algogeeks] C Doubts
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 unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
