TheTechy15

Programmer/Tester- Aditya Gupta (aka Techy3.5 :-) I used g++ to compile these on linux 1. Type casting of floating point numeral to integer (either automatic or by type casting)- Conversion of floating point to integer is by taking the GIF (Greatest Integer Funtion) and NOT ROUNDING OFF Greatest Integer Function(Maths,12th) it gives the integer part when the original number is written as integer+fractional part (where fractional part is less than 1 but positive) 2. NEsted MultiLine Comment [actually wont work as expected?!] if you do this in C++.... /*  blah /*blah blah blah*/ blah*/ the last blah will be visible to the compiler!!! (Inference- Multiline-Comment ends whenever */ is encountered, no matter where!!)NEsted MultiLine Comment 3. Size of various types (of C++) on 'Linux' Mint - (and same as in later versions of windows, and most PCs nowadays) char - 1 int - 4  (and: short - 2, unsigned long - 8) long long - 8 float - 4 double - 8 (and: l...

1. To define a function outside the class, there MUST be a prototype inside the class. For eg. class Jg{ int num; public: Jg(){ num=0; } }; void Jg::Display() //Error will be here { cout<<num; } //This Program will give the error "no 'void Jg::Display()' member function declared in class 'Jg'" 2. What if one member function defined twice, once inside class and once outside? Will give error "[error]redefintion of int Jg::Display()" And then pointing on the first defintion it will give "[NOTE] int Jg::Display() previously defined here" PROGRAM - #include<iostream> using namespace std; class Jg{ int num; public: Jg(){ num=0; } Display(){ cout<<"Hello"; } }; int Jg::Display(){ //ERROR cout<<num; } main(){ Jg J1; J1.Display(); } 3. What if a member function declared twice inside the class? Will show error "int Ag::Display() cannot be overloaded...

1. strlen() FUNCTION OF string.h It stops counting where it finds uninitialised character (ie if Arr[i] is uninitialised, it gives length as i) for eg. char S[10]; S[0]='c'; cout<<strlen(S); //This prints '1' for eg. char S[10]; S[4]='c'; cout<<strlen(S); //This prints '0', even though there is one character NOTE- Though such case may not be ever encountered in actual use but i just found this out experimenting with the code Reason- See the function definition of strlen() in string (or string.h on older ones) header file 2. THE NULL CHARACTER The null character (that terminates strings, ie '\0') has the integer code '0'. (i. cout<<int('\0'); gives 0) '\n' has code 10 ' ' has code 32 txt files terminate with a character with code '-1', but when i tried to print the respective character using cout<<char(-1); it just printed a space!? //I used Dev C++ Ver 5.11...

When negative number is divided by its positive counterpart will give remainder 0; for eg. cout<<(-5)%(5); //will print 0 BUT, a problem is that it may return negative remainders also, which is mathematically wrong for eg. cout<(-5)%2; //will print -1 //actually it should have given remainder 1 not -1 NOTE.... C++ with its math.h (or cmath) provides a function called "remainder(x,y)", it corrects this problem as well as it is more advance since it can also give floating point remainders... You may use that IF you think such case/need can be encountered remainder() requires two arguments () of any arithmetic type

//I used Dev C++ Ver 5.11 (compiler) ... i told since some of this might run differently on other IDEs Some interesting observations in C++ ---------------------------------------- 1. About pow() function *pow() cant find 'fractional' powers, but can find decimal powers for eg. pow(4,1/2) gives 1; but, pow(4,0.5) gives 2;      pow(4,0.55) gives 2.14; *roots of negative numbers: pow(-4,0.5) gives "nan" 2. Time Battle between cout and for loop (UnExpEcTEd) In this program, see the time taken by them (individually) "on my computer" (on faster computers there may be more/less gap) #include<iostream> using namespace std; int main() { //for(int i=1; i<100000;i++);  //0.215 sec cout<<"hello";  //0.303 sec } 3. Dont ask user to enter 0, if you are going to use an IF statement, because 0 is treated as false; instead ask for -1 instead of 0 (*you may even as for any other number) For eg. cout<<"Enter 0 if...