cout

问题 I know there are many buffer questions on here but I can't seem find a clear answer on this. std::cout << "write to screen" << std::endl; I know this code will write to the screen and flush the buffer because of the "endl", but if I wrote this: std::cout << "write to screen"; Wouldn't the buffer be flushed regardless since the text has been outputted to the screen? 回答1: Wouldn't the buffer be flushed regardless since the text has been outputted to the screen? Assuming that you have seen the

问题 Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 7 years ago . Improve this question I am an intermediate programmer, writing a program that's probably much to complicated for me. The programs aim is to construct certain 2-d arrays, and has a few different class objects that are communicating with each other in a not-so-simple way. In order to

问题 This may be a duplicate of this question, but I don't feel it was actually answered correctly. Observe: #include <iostream> #include <iomanip> using namespace std; int main () { float p = 1.00; cout << showpoint << setprecision(3) << p << endl; } Output: 1.00 Now if we change that line to: cout << fixed << showpoint << setprecision(3) << p << endl; we get: 1.000 And if we use the "opposite" of fixed we get something totally different: cout << scientific << showpoint << setprecision(3) << p <<

问题 This may be a duplicate of this question, but I don't feel it was actually answered correctly. Observe: #include <iostream> #include <iomanip> using namespace std; int main () { float p = 1.00; cout << showpoint << setprecision(3) << p << endl; } Output: 1.00 Now if we change that line to: cout << fixed << showpoint << setprecision(3) << p << endl; we get: 1.000 And if we use the "opposite" of fixed we get something totally different: cout << scientific << showpoint << setprecision(3) << p <<

问题 Let's say, we have: char x = 'a'; int y = 1; So, if you run: std::cout << x + y; It prints 98 instead of 'b'. As i see from here <<operator has only int parameter implementation. From now on i have 2 questions: After char + int operation what type is returned? Why there is no char parameter implementation, but std::cout << x still works as expected and prints char value? 回答1: Thanks to Fefux, Bo Persson and Matti Virkkunen answers are: From CPP Reference: Implicit conversions: arithmetic

问题 Let's say, we have: char x = 'a'; int y = 1; So, if you run: std::cout << x + y; It prints 98 instead of 'b'. As i see from here <<operator has only int parameter implementation. From now on i have 2 questions: After char + int operation what type is returned? Why there is no char parameter implementation, but std::cout << x still works as expected and prints char value? 回答1: Thanks to Fefux, Bo Persson and Matti Virkkunen answers are: From CPP Reference: Implicit conversions: arithmetic

问题 Let's say, we have: char x = 'a'; int y = 1; So, if you run: std::cout << x + y; It prints 98 instead of 'b'. As i see from here <<operator has only int parameter implementation. From now on i have 2 questions: After char + int operation what type is returned? Why there is no char parameter implementation, but std::cout << x still works as expected and prints char value? 回答1: Thanks to Fefux, Bo Persson and Matti Virkkunen answers are: From CPP Reference: Implicit conversions: arithmetic

问题 This question already has answers here : uint8_t can't be printed with cout (8 answers) Closed 5 years ago . In the following code, it appears that std::cout does not print the variable of type uint8_t properly. int main() { uint8_t var = 16; std::cout << "value: " << var << std::endl; } output: value: I don't have problem with similar uintX_t types. I know I'm missing something here, but I can't figure what it is. 回答1: uint8_t maps to unsigned char on most systems. As the result, the

问题 This question already has answers here : uint8_t can't be printed with cout (8 answers) Closed 5 years ago . In the following code, it appears that std::cout does not print the variable of type uint8_t properly. int main() { uint8_t var = 16; std::cout << "value: " << var << std::endl; } output: value: I don't have problem with similar uintX_t types. I know I'm missing something here, but I can't figure what it is. 回答1: uint8_t maps to unsigned char on most systems. As the result, the

问题 I calculated a total of floats and I got a number like 509990e-405 . I'm assuming this is the short version; how can I cout this as a full number? cout << NASATotal << endl; is what I have now. 回答1: You can force the output to be not in scientific notation, and to have the sufficient precision to show your small number. #include <iomanip> // ... long double d = 509990e-405L; std::cout << std::fixed << std::setprecision(410) << d << std::endl; Output: 0