cout

问题 Whenever I am writing this following code, I am getting garbage(unexpected) output in some online compiler, but if I use code block then getting satisfied output. So my question is why I am getting this type of output? for example, if I input 5 7 + 5 - 10 - 20 + 40 - 20 then I am getting 22 1 in the code block. But in the online compiler, it's something else. #include<iostream> #include<cstdlib> using namespace std; int main() { int have, n, i; int kid=0; cin>>n>>have; int line[n]; for(i=0;i

问题 I would like to make cout will always print additional string(above and under) whenever I call it. It is actually weird to me cause I use in Java and C# primaly . EDIT: Maybe we can define other value for "y" string or something like that? Example code: #include <iostream> int main(){std::cout<<"y\n";} result(printed): x y z I would not want to change int main() method just maybe overriding the << for the string type? OR MAYBE make cout invoke additional method? 回答1: I don't think you can do

问题 I am getting an error when I try to output the return value from one of my methods: Error: No operator "<<" matches these operands. Operand types are: std::ostream << std::string Main.cpp #include <iostream> using namespace std; #include "Book.h" int main() { book.setTitle("Advanced C++ Programming"); book.setAuthorName("Linda", "Smith"); book.setPublisher("Microsoft Press", "One Microsoft Way", "Redmond"); book.setPrice(49.99); cout << book.getBookInfo(); // <-= this won't compile because of

问题 I am getting an error when I try to output the return value from one of my methods: Error: No operator "<<" matches these operands. Operand types are: std::ostream << std::string Main.cpp #include <iostream> using namespace std; #include "Book.h" int main() { book.setTitle("Advanced C++ Programming"); book.setAuthorName("Linda", "Smith"); book.setPublisher("Microsoft Press", "One Microsoft Way", "Redmond"); book.setPrice(49.99); cout << book.getBookInfo(); // <-= this won't compile because of

问题 sorry for asking you a stupid question, but I just can't figure out why I keep on getting this output. So here is my code: #include <cstdio> #include <iostream> using namespace std; unsigned n = 4242; int getRemainderOf(int m, int n, int& quotient); static int l = 0; int main() { int quotient; // the value of which should be changed after calling the func. for(int i=-1; i<=1; ++i) { for(int j=-1; j<=1; ++j) { if( i && j ) { cout << "("<< i*3 << "," << j*7 << ") " <<( getRemainderOf(i*3, 7*j,

问题 I want a ostream with a prefix at the beginning of every line redirected on cout; I try this: #include <iostream> #include <thread> class parallel_cout : public std::ostream { public: parallel_cout(std::ostream& o):out(o){} template <typename T> std::ostream& operator<< (const T& val) { out << "prefix " << val; return *this; } std::ostream& out; }; int main() { parallel_cout pc(std::cout); pc<<"a\nb"<<"c\n"; } but i have in output prefix a b without c. why this? 回答1: The way you modify the

问题 This question already has answers here : Why is “using namespace std;” considered bad practice? (36 answers) Closed 6 years ago . Why do I need to type in using namespace std; in order to be able to use cout and endl ? Also what are these called; is cout a function? Is there cout in C? I heard it was implemented in C++ because it is better in many ways. 回答1: cout is a global object defined in the std namespace, and endl is a (stream manipulator) function also defined in the std namespace. If

问题 When using cout , what is the default formatter defined in the <iomanip> header? In other words, once I've set my formatter to fixed using cout << fixed << setPrecision(2) , how do I change it back? Or, what am I changing it back to ? 回答1: The opposite of std::fixed is std::scientific . (You find a nice list of manipulators in this great answer.) 回答2: The answer is std::defaultfloat in C++11. To achieve this in C++03 you can do cout.unsetf(std::ios_base::floatfield); See Really, what's the

问题 When using cout , what is the default formatter defined in the <iomanip> header? In other words, once I've set my formatter to fixed using cout << fixed << setPrecision(2) , how do I change it back? Or, what am I changing it back to ? 回答1: The opposite of std::fixed is std::scientific . (You find a nice list of manipulators in this great answer.) 回答2: The answer is std::defaultfloat in C++11. To achieve this in C++03 you can do cout.unsetf(std::ios_base::floatfield); See Really, what's the

问题 I'd like to wrap std::cout for formatting, like so: mycout([what type?] x, [optional args]) { ... // do some formatting on x first std::cout << x; } and still be able to use expressive syntax like mycout("test" << i << endl << somevar, indent) instead of being forced to be more verbose like mycout(std::stringstream("test") << i ...) How can I implement this? What type to make x ? Edit: added consideration for optional arguments 回答1: How about this: struct MyCout {}; extern MyCout myCout;