Dev c++😅😅

don’t use this, anything else is better than this

Isnt is good??

No its actually fine. Especially for beginners. Its light weight, good ide, portable and also has cmake project support I think. So all you really need. Honestly IDEs are 99% preference, but i gave my gf this IDE also to learn C++ and it looks very good. Personally i dont use IDEs

I assume he can use notepad and g++ for trivial learning tasks

And it even could be better if he works with the compiler directly

And it even could be better if he works with the compiler directly

Can u tell me the difference between reference and pointer😅😅in simply

Hey guys ,I have a concern in my program .So if I passed two objects for a function like func(obj1,obj2) which one is supposed to get destructed first??!

are INT_MAX and INT_MIN some constants in c++?

Learn the stack’s mechanism. Push first, pop last. Push last, pop first

Sure. <numeric_limits>. std::numeric_limits<T> E.g.: std::numeric_limits<int>::max() // instead of INT_MAX

is it really defined like that? like the order of initialization of the parameters

what is that <numric_limits...

#include <iostream> #include <cmath> using namespace std; class clockType { public: ~clockType(); clockType(int =0,int =0 ,int =0); void setTime(int h, int m, int s); void getTime(int& h, int& m, int& s) const; void printTime() const; void tick(); private: int hour; int minut; int second; }; clockType::~clockType() { cout << "destruct : " ; printTime(); } clockType::clockType(int h, int m, int s) :hour(h), minut(m), second(s) { cout << "constract : " << endl;; } void clockType::setTime(int h, int m, int s) { hour = h; minut = m; second = s; } void clockType::getTime(int& h, int& m, int& s)const { h = hour; m = minut; s = second; } void clockType::printTime()const { cout << hour << " : " << minut << " : " << second << endl; } void clockType::tick() { second++; minut += second / 60; second = second % 60; hour += minut / 60; minut = minut % 60; hour = hour % 24; } int fun (clockType OPJ) { int hr, min, sec; OPJ.getTime(hr, min, sec); return (sec + 60 * min + 60 * 60 * hr); } bool isGreater(clockType OPJ1, clockType OPJ2) { int second1 = fun(OPJ1); int second2 = fun(OPJ2); if (second1 > second2) return true; else return false; } int main() { clockType clk1(2, 10, 10); clockType clk2(2, 20, 10); cout << isGreater(clk1, clk2); }

https://en.cppreference.com/w/cpp/types/numeric_limits

For this code in IsGreater function ,OBJ1 got destructed first .why that happened?!

Can anyone give some great c++ idea inspiration

but if you don't know the order of initialization you don't know the order of destructing also no?

Why it didnt?I dont understand how clk2 has anything to do with this

The order of initialization can be anything, but we can guarantee that the order of destruction will be reversed.

Write own vector

Can u tell me the difference between reference and pointer😅😅in simply

don’t use this, anything else is better than this

I assume you don't include TurboC++ in anything else :)

if n is int your math wil be truncted. For exmple, 99/10 returns 9. To return 9.9 use 99/10.0

if n is int your math wil be truncted. For exmple, 99/10 returns 9. To return 9.9 use 99/10.0

Who can.please

Hey everyone Anybody can help me in the understanding of operator overloading

Imagine you wrote a class Student. Imagine you have a vector of Students and wish to sort it. Alphabetically, let us say. But how do you compare them? You can embed the comparison inside your sorting algorithm. But that does not look nice at all. So much finer would it be, if only you could write something like this: if(students[i]> students[i+1]){ //student at position i should alphabetically be after student at position i+1 //Swap 'em } That's when our friend operator overloading comes in.

We write it like this: class Student { char* name; //code bool operator > (Student s1, Student s2){ if (strcmp(s1.name, s2.name)>0){ //string comparison from C, for char* return true; } //same for < operator }

you can convert it into: return strcmp(s1.name, s2.name)>0

Fine correction, that was my fault.

I am implementing a stack which contains characters. I have defined a function pop which returns void and doesn't take in anything as parameter. But while executing the code, i accidentally gave a character while calling pop(), and the code worked just fine without throwing an error. How is this possible?

Assuming we're talking about C, void pop(); declares a function that takes an unspecified (but not variable) number of parameters (and returns nothing). So all of the below calls are valid pop(3); pop("Hello"); pop(); If you want to make sure no parameter is passed you should use void pop(void);

HI, I'm confused about the following syntax while reading source code of an open source project. typedef Pair *(*PairCreator)(LAMMPS *); typedef std::map<std::string, PairCreator> PairCreatorMap; PairCreatorMap *pair_map; template <typename S, typename T> static S *style_creator(LAMMPS *lmp) { return new T(lmp); } pair_map = new PairCreatorMap(); (*pair_map)["lj/cut"] = &style_creator<Pair, Class> I can't understand what is "(*pair_map)["lj/cut"]". I believe it is a function pointer of some sort as it is assigned memory of a function "style_creator". But I don't understand how a class instance "pair_map" is indexed with string. Is there any material or reference where I can look for this syntax? I searched a whole day on google but can't find anything similar.

Hi I have a problem with the code written below: #include <stdio.h> int main(){ int a,b; scanf("%d %d",&a,&b); int c = b>>(a-1); c=c%10; printf("%d",c); For example if a=3 and b=1056 It should return c as 0 but it returns 4 Any idea how to fix it ?

why should it return 0? What do you want to do? 1056 >> 2 = 264; 264 mod 10 = 4

I want int c to return the 2 digits that are left after right shift

right shifts are performed in binary, every right shift is basically dividing by two and rounding down. Hence 1056 >> 2 = 264

So how should I get that two digits ?

(1056 / 100) % 10 ? or more general (b / (pow(10, a - 1))) % 10