How to cin to a vector

问题:

I'm trying to ask the user to enter numbers thats put into a vector, then using a function call to cout the numbers, why is this not working? I am only able to cout the first number.

template <typename T>
void write_vector(const vector<T>& V)
{
   cout << "The numbers in the vector are: " << endl;
  for(int i=0; i < V.size(); i++)
    cout << V[i] << " ";
}

int main()
{
  int input;
  vector<int> V;
  cout << "Enter your numbers to be evaluated: " << endl;
  cin >> input;
  V.push_back(input);
  write_vector(V);
  return 0;
}

回答1:

As is, you're only reading in a single integer and pushing it into your vector. Since you probably want to store several integers, you need a loop. E.g., replace

cin >> input;
V.push_back(input);

with

while (cin >> input)
    V.push_back(input);

What this does is continually pull in ints from cin for as long as there is input to grab; the loop continues until cin finds EOF or tries to input a non-integer value. The alternative is to use a sentinel value, though this prevents you from actually inputting that value. Ex:

while ((cin >> input) && input != 9999)
    V.push_back(input);

will read until you try to input 9999 (or any of the other states that render cin invalid), at which point the loop will terminate.

回答2:

Other answers would have you disallow a particular number, or tell the user to enter something non-numeric in order to terminate input. Perhaps a better solution is to use std::getline() to read a line of input, then use std::istringstream to read all of the numbers from that line into the vector.

#include <iostream>
#include <sstream>
#include <vector>

int main(int argc, char** argv) {

    std::string line;
    int number;
    std::vector<int> numbers;

    std::cout << "Enter numbers separated by spaces: ";
    std::getline(std::cin, line);
    std::istringstream stream(line);
    while (stream >> number)
        numbers.push_back(number);

    write_vector(numbers);

}

Also, your write_vector() implementation can be replaced with a more idiomatic call to the std::copy() algorithm to copy the elements to an std::ostream_iterator to std::cout:

#include <algorithm>
#include <iterator>

template<class T>
void write_vector(const std::vector<T>& vector) {
    std::cout << "Numbers you entered: ";
    std::copy(vector.begin(), vector.end(),
        std::ostream_iterator<T>(std::cout, " "));
    std::cout << '\n';
}

You can also use std::copy() and a couple of handy iterators to get the values into the vector without an explicit loop:

std::copy(std::istream_iterator<int>(stream),
    std::istream_iterator<int>(),
    std::back_inserter(numbers));

But that’s probably overkill.

回答3:

You need a loop for that. So do this:

while (cin >> input) //enter any non-integer to end the loop!
{
   V.push_back(input);
}

Or use this idiomatic version:

#include <iterator> //for std::istream_iterator 

std::istream_iterator<int> begin(std::cin), end;
std::vector<int> v(begin, end);
write_vector(v);

You could also improve your write_vector as:

 #include <algorithm> //for std::copy

template <typename T>
void write_vector(const vector<T>& v)
{
   cout << "The numbers in the vector are: " << endl;
   std::copy(v.begin(), v.end(), std::ostream_iterator<int>(std::cout, " "));
}

回答4:

you have 2 options:

If you know the size of vector will be (in your case/example it's seems you know it):

vector<int> V(size)
for(int i =0;i<size;i++){
    cin>>V[i];
 }

if you don't and you can't get it in you'r program flow then:

int helper;
while(cin>>helper){
    V.push_back(helper);
}

回答5:

One-liner to read a fixed amount of numbers into a vector (C++11):

#include <algorithm>
#include <iterator>
#include <iostream>
#include <vector>
#include <cstddef>

int main()
{
    const std::size_t LIMIT{5};
    std::vector<int> collection;

    std::generate_n(std::back_inserter(collection), LIMIT,
        []()
        {
            return *(std::istream_iterator<int>(std::cin));
        }
    );

    return 0;
}

回答6:

You need a second integer.

int i,n;
vector<int> V;
cout << "Enter the amount of numbers you want to evaluate: ";
cin >> i;
cout << "Enter your numbers to be evaluated: " << endl;
while (V.size() < i && cin >> n){
  V.push_back(n);
}
write_vector(V);
return 0;

回答7:

cin is delimited on space, so if you try to cin "1 2 3 4 5" into a single integer, your only going to be assigning 1 to the integer, a better option is to wrap your input and push_back in a loop, and have it test for a sentinel value, and on that sentinel value, call your write function. such as

int input;
cout << "Enter your numbers to be evaluated, and 10000 to quit: " << endl;
while(input != 10000) {
    cin >> input;
   V.push_back(input);
}
write_vector(V);

回答8:

You probably want to read in more numbers, not only one. For this, you need a loop

int main()
{
  int input = 0;
  while(input != -1){
    vector<int> V;
    cout << "Enter your numbers to be evaluated: " << endl;
    cin >> input;
    V.push_back(input);
    write_vector(V);
  }
  return 0;
}

Note, with this version, it is not possible to add the number -1 as it is the "end signal". Type numbers as long as you like, it will be aborted when you type -1.

回答9:

You can simply do this with the help of for loop
->Ask on runtime from a user (how many inputs he want to enter) and the treat same like arrays.

int main() {
        int sizz,input;
        std::vector<int> vc1;

        cout<< "How many Numbers you want to enter : ";
        cin >> sizz;
        cout << "Input Data : " << endl;
        for (int i = 0; i < sizz; i++) {//for taking input form the user
            cin >> input;
            vc1.push_back(input);
        }
        cout << "print data of vector : " << endl;
        for (int i = 0; i < sizz; i++) {
            cout << vc1[i] << endl;
        }
     }

回答10:

#include<iostream>
#include<vector>
#include<string>
using namespace std;
int main()
{
    vector<string>V;
    int num;
    cin>>num;
    string input;
    while (cin>>input && num != 0) //enter any non-integer to end the loop!
{
    //cin>>input;
   V.push_back(input);
   num--;
   if(num==0)
   {
   vector<string>::iterator it;
    for(it=V.begin();it!=V.end();it++)
        cout<<*it<<endl;
   };

}
return 0;

};

回答11:

In this case your while loop will look like

int i = 0;
int a = 0;
while (i < n){
  cin >> a;
  V.push_back(a);
  ++i;
}

回答12:

The initial size() of V will be 0, while int n contains any random value because you don't initialize it.

V.size() < n is probably false.

Silly me missed the "Enter the amount of numbers you want to evaluate: "

If you enter a n that's smaller than V.size() at that time, the loop will terminate.

回答13:

Just add another variable.

int temp;
while (cin >> temp && V.size() < n){
    V.push_back(temp);
}

回答14:

#include<iostream>
#include<vector>
#include<sstream>
using namespace std;

int main()
{
    vector<string> v;
    string line,t;
    getline(cin,line);
    istringstream iss(line);
    while(iss>>t)
        v.push_back(t);

    vector<string>::iterator it;
    for(it=v.begin();it!=v.end();it++)
        cout<<*it<<endl;
    return 0;
}

回答15:

would be easier if you specify the size of vector by taking an input :

int main()
{
  int input,n;
  vector<int> V;
  cout<<"Enter the number of inputs: ";
  cin>>n;
  cout << "Enter your numbers to be evaluated: " << endl;
  for(int i=0;i<n;i++){
  cin >> input;
  V.push_back(input);
  }
  write_vector(V);
  return 0;
}

回答16:

#include<bits/stdc++.h>
using namespace std;

int main()
{
int x,n;
cin>>x;
vector<int> v;

cout<<"Enter numbers:\n";

for(int i=0;i<x;i++)
 {
  cin>>n;
  v.push_back(n);
 }


//displaying vector contents

 for(int p : v)
 cout<<p<<" ";
}

A simple way to take input in vector.