ostream

At 50:40 of http://channel9.msdn.com/Events/GoingNative/2013/Writing-Quick-Code-in-Cpp-Quickly Andrei Alexandrescu makes a joke about how not efficient/slow istream is. I had an issue in the past with ostream being slow and fwrite being significantly faster (reducing many seconds when running the main loop once) but I never understood why nor looked into it. What makes istream and ostream slow in C++? or at least slow compared to other things (like fread/fget, fwrite) which would equally satisfied the needs. Dietmar Kühl Actually, IOStreams don't have to be slow! It is a matter of implementing

double x = 1500; for(int k = 0; k<10 ; k++){ double t = 0; for(int i=0; i<12; i++){ t += x * 0.0675; x += x * 0.0675; } cout<<"Bas ana: "<<x<<"\tSon faiz: "<<t<<"\tSon ana: "<<x+t<<endl; } this the output Bas ana: 3284.78 Son faiz: 1784.78 Son ana: 5069.55 Bas ana: 7193.17 Son faiz: 3908.4 Son ana: 11101.6 Bas ana: 15752 Son faiz: 8558.8 Son ana: 24310.8 Bas ana: 34494.5 Son faiz: 18742.5 Son ana: 53237 Bas ana: 75537.8 Son faiz: 41043.3 Son ana: 116581 Bas ana: 165417 Son faiz: 89878.7 Son ana: 255295 Bas ana: 362238 Son faiz: 196821 Son ana: 559059 Bas ana: 793246 Son faiz: 431009 Son ana: 1

I've been googling around and I just can't find a simple answer to this. And it should be simple, as the STL generally is. I want to define MyOStream which inherits publicly from std::ostream. Let's say I want to call foo() each time something is written into my stream. class MyOStream : public ostream { public: ... private: void foo() { ... } } I understand that the public interface of ostream is non-virtual, so how can it be done? I want clients to be able to use both operator<< and write() and put() on MyOStream and have use the extended ability of my class. It's not a simple question,

I have difficulties in understanding the sequence of calls in the code below. I was expecting to see the output below A1B2 While I can see that the output I get is BA12 I thought that the call std::cout<< b->fooA() << b->fooB() << std::endl was equivalent to call std::cout.operator<<( b->fooA() ).operator<< ( b->fooB() ) but I can see that this is not the case. Can you help me understanding better how this does it work and the relationship with the global operator<< ? Is this last ever called in this sequence? Regards AFAG #include <iostream> struct cbase{ int fooA(){ std::cout<<"A"; return 1;

It is well known that the user can define stream manipulators like this: ostream& tab(ostream & output) { return output<< '\t'; } And this can be used in main() like this: cout<<'a'<<tab<<'b'<<'c'<<endl; Please explain me how does this all work? If operator<< assumes as a second parameter a pointer to the function that takes and returns ostream & , then please explain my why it is necessary? What would be wrong if the function does not take and return ostream & but it was void instead of ostream & ? Also it is interesting why “dec”, “hex” manipulators take effect until I don’t change between

For educational purposes I want to create a ostream and stream buffer to do: fix endians when doing << myVar; store in a deque container instead of using std:cout or writing to a file log extra data, such as how many times I did <<, how many times I did .write, the amount of bytes I written and how many times I flush(). But I do not need all the info. I tried overloading but failed horribly. I tried overloading write by doing ostream& write( const char* s, streamsize n ) in my basic_stringstream2 class (I copied paste basic_stringstream into my cpp file and modified it) but the code kept using