setw

来源： https://stackoverflow.com/questions/9673412/i-need-something-in-c-sharp-which-work-like-the-setw-in-c

来源： https://stackoverflow.com/questions/9673412/i-need-something-in-c-sharp-which-work-like-the-setw-in-c

问题 So, i'm attempting to set the precision for input values in my code. I want the values to be printed with two decimal points afterwards of precision though i'm not exactly sure how. Here's my code. #include <iostream> #include <iomanip> float uphill, wellD, waterLvl, buckVol; float buckAscRate, downHill, volume; float last; float timeReq; int scene = 1; void timeRequired() { std::setw(2); std::setprecision(2); std::cout << "Scenario " << scene << ":" << std::endl; std::cout << "up hill" << "

问题 I want to justify the output text like this 0x29823d80 0x10019b8 / 0 00000000000000000000000000000001 0x37449e60 0x10dfc / 12 00000000000000000001000000000000 However with this statement fout << std::setw(5) << std::hex << "0x" << (*it).addr << " " << std::setw(5) << std::hex << "0x" << (*it).pc << std::setw(10) << "/" << std::setw(5) << std::dec << (*it).off << "\t" << std::setw(5) << (*it).layout << "\n"; I get this: 0x29823d80 0x10019b8 / 0 00000000000000000000000000000001 0x37449e60

问题 I want it so that when the user inputs more than 5 characters, something will happen, instead of just skipping the rest. In this code, if you type in more than 5 characters, it will only show the first 5 characters. I want to put an "if" statement here that if the user inputs more than 5 characters, it will show an error or something, and not just show the first 5 characters. #include <iostream> #include <iomanip> int main() { using namespace std; string nationname; int maxchar = 5; cout <<

问题 My problem is shown in the following minimal example: #include <iostream> #include <string> #include <iomanip> int main() { int width = 15; std::cout << std::left; std::cout << std::setw(width) << "Prints well" << std::setw(width) << "This too" << '\n'; std::cout << std::setw(width) << "\u221E" << std::setw(width) << "This not?" << '\n'; std::cout << std::setw(width+2) << "\u221E" << std::setw(width) << "This is good" << '\n'; } Compiled using g++, it prints: Prints well This too ∞ This not?

问题 I recently was bitten by the fact that ios_base::width and/or the setw manipulator have to be reset with every item written to the stream. That is, you must do this: while(whatever) { mystream << std::setw(2) << myval; } Rather than this: mystream.width(2); while(whatever) { mystream << myval; } Ok, fine. But does anyone know why this design decision was made? Is there some rationale that I'm missing, or is this just a dark corner of the standard? Other stream formatting modifiers (as

问题 I am trying to use set width setw for string output to an output file, however, I am not able to make it work. I have with me the following example. // setw example #include <iostream> #include <iomanip> #include <fstream> int main () { std::ofstream output_file; output_file.open("file.txt"); output_file << "first" <<std::setw(5)<< "second"<< std::endl; output_file.close(); return 0; } Edit: With the above lines I expected to have many spaces between first and second , something like first

问题 I just want to format a string and an integer value with right justify. There is no problem to do this without leading space before the integer value. bytes.....................123981 total bytes..............1030131 But it should look like this: bytes ................... 123981 total bytes ............ 1030131 Unfortunately the example below wont work, because setw (right justify) relates only to the next stream element. int iBytes = 123981; int iTotalBytes = 1030131; cout << setfill('.');

问题 Is there a way that I can force setw to truncate? Say that I want to get the output: blah blah blee le bu blah blah blee Is there a way to make this work: string foo{"bu blah blah blee le"}; cout << setw(foo.size() - 3) << foo.data() + 3 << setw(foo.size() - 3) << foo << endl; 回答1: No, not really. You can switch to unformatted output for this example, though: assert(foo.size() > 3); cout.write(&foo[3], foo.size() - 3); cout.write(&foo[0], foo.size() - 3); 回答2: Not directly. In the printf