Method 1, you can input in multiple lines, with no limit on the number in each line, and whitespace characters (blank lines, spaces, tabs) in the input will be automatically ignored.

std::vector<double> v;
for (double d; std::cin >> d; v.push_back(d)) {}

Method 2: Enter only one line and automatically ignore whitespace characters (blank lines, spaces, tabs) in the input.

#include <sstream>
std::vector<double> v;
std::string line;
std::getline(std::cin, line);
std::istringstream iss(line);
for (double d; iss >> d; v.push_back(d)) {}

Method three has the same effect as method one.

#include <iterator>
std::istream_iterator<double> in(std::cin), eof;
std::vector<double> v(in, eof);

Note: Replacing std::cin in the above method with any std::istream type variable or a variable of its subclass has the same effect. For example, change to a variable of type std::ifstream to read the file.

If other circumstances are not considered:

1. First use [space] to separate each decimal and put it in a vector<char*>

2. Convert to double one by one using atof()

3. Convert one and put it into vector<double>

Optimization:

Parse and convert at the same time, push_back into vector at the same time, all in one loop.

C++ streaming input style

double x;
while(!cin.eof())
{
    cin>>x;
    v.push_back(x);
}

C style input

double x;
while(scanf("%lf",&x)!=EOF) v.push_back(x);
/*
    EOF是个库中的宏定义
    #define EOF (-1)
*/

The above method is suitable for reading from a file, not suitable for console input (because the console cannot know whether you have finished typing)
If it is console input, you can read a whole line first, and then read it one by one Read in:

#include <cstdio>
#include <cstdlib>
#include <vector>
#include <iostream>
using namespace std;

int main()
{
    #define SIZE_BUFF (100)
    char s[SIZE_BUFF],*p=s;
    cin.getline(s+1,SIZE_BUFF);
    vector<double> v;
    double x;
    for(s[0]=' ';*p;++p)
    {
        if(*p!=' ') continue;
        while(*++p==' ');
        sscanf(p,"%lf",&x);
        v.push_back(x);
    }
    for(vector<double>::iterator i=v.begin();i!=v.end();++i)
        printf("%lf\n",*i);
    return 0;
}

The implementation is not very elegant, so let’s make do with it.