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• LSP vs OCP / Liskov Substitution VS Open Close

  - by Kolyunya

  I am trying to understand the SOLID principles of OOP and I've come to the conclusion that LSP and OCP have some similarities (if not to say more). the open/closed principle states "software entities (classes, modules, functions, etc.) should be open for extension, but closed for modification". LSP in simple words states that any instance of Foo can be replaced with any instance of Bar which is derived from Foo and the program will work the same very way. I'm not a pro OOP programmer, but it seems to me that LSP is only possible if Bar, derived from Foo does not change anything in it but only extends it. That means that in particular program LSP is true only when OCP is true and OCP is true only if LSP is true. That means that they are equal. Correct me if I'm wrong. I really want to understand these ideas. Great thanks for an answer.

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• Liskov substitution and abstract classes / strategy pattern

  - by Kolyunya

  I'm trying to follow LSP in practical programming. And I wonder if different constructors of subclasses violate it. It would be great to hear an explanation instead of just yes/no. Thanks much! P.S. If the answer is no, how do I make different strategies with different input without violating LSP? class IStrategy { public: virtual void use() = 0; }; class FooStrategy : public IStrategy { public: FooStrategy(A a, B b) { c = /* some operations with a, b */ } virtual void use() { std::cout << c; } private: C c; }; class BarStrategy : public IStrategy { public: BarStrategy(D d, E e) { f = /* some operations with d, e */ } virtual void use() { std::cout << f; } private: F f; };

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• Stream buffering issue

  - by Kolyunya

  The mod_rewrite documentation states that it is a strict requirement to disable in(out)put buffering in a rewrite program. Keeping that in mind I've written a simple program (I do know that it lacks the EOF check but this is not an issue and it saves one condition check per loop): #include <stdio.h> #include <stdlib.h> int main ( void ) { setvbuf(stdin,NULL,_IOLBF,4200); setvbuf(stdout,NULL,_IOLBF,4200); int character; while ( 42 ) { character = getchar(); if ( character == '-' ) { character = '_'; } putchar(character); } return 42 - 42; } After making some measurements I was shocked - it was over 9,000 times slower than the demo Perl script provided by the documentation: #!/usr/bin/perl $| = 1; # Turn off I/O buffering while (<STDIN>) { s/-/_/g; # Replace dashes with underscores print $_; } Now I have two related questions: Question 1. I believe that the streams may be line buffered since Apache sends a new line after each path. Am I correct? Switching my program to setvbuf(stdin,NULL,_IOLBF,4200); setvbuf(stdout,NULL,_IOLBF,4200); makes it twice as fast as Perl one. This should not hit Apache's performance, should it? Question 2. How can one write a program in C which will use unbuffered streams (like Perl one) and will perform as fast as Perl one?

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