cin快读
ios::sync_with_stdio(false); \\取消同步，cin,cout的速度就不慢了！！ 来源： https://www.cnblogs.com/WestJackson/p/11384978.html
cin
Press Enter to Continue
This doesn't work: string temp; cout << "Press Enter to Continue"; cin >> temp; rlbond cout << "Press Enter to Continue"; cin.ignore(); or, better: #include <limits> cout << "Press Enter to Continue"; cin.ignore(std::numeric_limits<streamsize>::max(),'\n'); Try: char temp; cin.get(temp); or, better yet: char temp = 'x'; while (temp != '\n') cin.get(temp); I think the string input will wait until you enter real characters, not just a newline. Replace your cin >> temp with: temp = cin.get(); http://www.cplusplus.com/reference/iostream/istream/get/ cin >> will wait for the EndOfFile. By default,
Reading getline from cin into a stringstream (C++)
问题 So I'm trying to read input like this from the standard input (using cin ): Adam English 85 Charlie Math 76 Erica History 82 Richard Science 90 My goal is to eventually store each data piece in its own cell in a data structure I have created, so basically I want to parse the input so each piece of data is individual. Since each row of input is inputted by the user one at a time, each time I get an entire row of input that I need to parse. Currently I am trying something like this:
Getting input from user using cin [duplicate]
This question already has an answer here: C++ “cin” only reads the first word [duplicate] 5 answers I am using Turbo C++ 3.0 Compiler While using the following code .. char *Name; cin >> Name; cout << Name; When I gave input with space ... its only saving characters typed before space .. like if I gave input "QWERT YUIOP" ... Name will contain "QWERT"; Any explaination why ?? You need to allocate space for the char array into which you want to read the Name. char *Name; will not work as it only declares a char pointer not a char array. Something like char Name[30]; Also the cin << only allows
cin >> fails with bigger numbers but works with smaller ones?
Why does cin fail, when I enter numbers like: 1 3999999999 but it works for smaller numbers like: 1 5 ? int main() { int N, X; cout << sizeof(int); cout << "Please enter two numbers: "; cin >> N >> X; vector <int> numbers = vector<int>(); int currentNumber; cout << "Please enter list of numbers: "; for ( int i = 0; i < N; i++ ) { cin >> currentNumber; if (cin.fail()) cout << "Something sucks!"; numbers.push_back(currentNumber); } sort(numbers.begin(), numbers.end(), Compare(X)); cout << "The list is now: " << endl; for (int i = 0; i < N; i++) { cout << numbers[i] << " "; } cout << endl; return
What's the difference between getline and std::istream::operator>>()?
#include <iostream> #include <string> using namespace std; int main() { string username; cout<< "username" ; cin >> username; } So I was curious on what's the difference between these two codes, I heard it's the same thing but if it is then why two ways of doing it then? #include <iostream> #include <string> using namespace std; int main() { string username; cout << "username" ; getline (cin,username) ; } The difference is that std::getline — as the name suggests — reads a line from the given input stream (which could be, well, std::cin ) and operator>> reads a word 1 . That is, std::getline
Difference between cin and cin.get() for char array
问题 I have these 2 codes: char a[256]; cin>>a; cout<<a; and char a[256]; cin.get(a,256);cin.get(); cout<<a; and maybe, relative to the second one without cin.get(); char a[256]; cin.get(a,256); cout<<a; My question is (first one) : for a char array, what should i use? cin or cin.get()? And why should i use cin.get(); with no parameter after my char initialisation? And my second question is: my c++ teacher taught me to use every time cin.get() for initialisation chars and AFTER every
cin>> not work with getline()
问题 #include <iostream> #include <string> using namespace std; int main () { string str; int age; cout << "Please enter age: "; cin>>age; cout << "Please enter full name: "; getline (cin,str); cout << "Thank you, " << str << ".\n"; } Why function getline() not work when I using uperator >> to input integer ? What is better use for int input ? 回答1: You still have a newline in the stream after cin>>age; , which is giving you an empty string for the name. You could solve it by just adding another
C++ cout cin string manipulation
问题 I'm trying to get a line as input from the command line. My problem is that I'm not getting the whole line, but it's being tokenized by space. So if I entered something such as "I like Math a lot" instead of getting "you enterend: I like Math a lot" I get the follwoing: EDITING MODE: Enter a command i like Math a lot you entered i EDITING MODE: Enter a command you entered like EDITING MODE: Enter a command you entered Math EDITING MODE: Enter a command you entered a EDITING MODE: Enter a
Directly capturing cin without a variable - C++ [closed]
问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 5 months ago . More of a curiosity question than anything else, but is it actually possible to pass whatever goes through cin to a function without having to waste a variable? 回答1: You can easily define a wrapper function to do this. template<class T> T get(std::istream& is){ T result; is >> result; return