How to cin Space in c++?

问题:

Say we have a code:

int main()
{
   char a[10];
   for(int i = 0; i < 10; i++)
   {
       cin>>a[i];
       if(a[i] == ' ')
          cout<<"It is a space!!!"<<endl;
   }
   return 0;
}

How to cin a Space symbol from standard input? If you write space, program ignores! :( Is there any combination of symbols (e.g. '\s' or something like this) that means "Space" that I can use from standard input for my code?

回答1:

It skips all whitespace (spaces, tabs, new lines, etc.) by default. You can either change its behavior, or use a slightly different mechanism. To change its behavior, use the manipulator noskipws, as follows:

 cin >> noskipws >> a[i];

But, since you seem like you want to look at the individual characters, I'd suggest using get, like this prior to your loop

 cin.get( a, n );

Note: get will stop retrieving chars from the stream if it either finds a newline char (\n) or after n-1 chars. It stops early so that it can append the null character (\0) to the array. You can read more about the istream interface here.

回答2:

#include <iostream>
#include <string>

int main()
{
   std::string a;
   std::getline(std::cin,a);
   for(std::string::size_type i = 0; i < a.size(); ++i)
   {
       if(a[i] == ' ')
          std::cout<<"It is a space!!!"<<std::endl;
   }
   return 0;
}

回答3:

To input AN ENTIRE LINE containing lot of spaces you can use getline(cin,string_variable);

eg:

string input;
getline(cin, input);

This format captures all the spaces in the sentence untill return is pressed

回答4:

Use cin.get() to read the next character.

However, for this problem, it is very inefficient to read a character at a time. Use the istream::read() instead.

int main()
{
   char a[10];
   cin.read(a, sizeof(a));
   for(int i = 0; i < 10; i++)
   {
       if(a[i] == ' ')
          cout<<"It is a space!!!"<<<endl;
   }
   return 0;
}

And use == to check equality, not =.

回答5:

Using cin's >> operator will drop leading whitespace and stop input at the first trailing whitespace. To grab an entire line of input, including spaces, try cin.getline(). To grab one character at a time, you can use cin.get().

回答6:

I thought I'd share the answer that worked for me. The previous line ended in a newline, so most of these answers by themselves didn't work. This did:

string title;
do {
  getline(cin, title);
} while (title.length() < 2);

That was assuming the input is always at least 2 characters long, which worked for my situation. You could also try simply comparing it to the string "\n".

回答7:

Try this all four way to take input with space :)

#include<iostream>
#include<stdio.h>

using namespace std;

void dinput(char *a)
{
    for(int i=0;; i++)
    {
        cin >> noskipws >> a[i];
        if(a[i]=='\n')
        {
            a[i]='\0';
            break;
        }
    }
}


void input(char *a)
{
    //cout<<"\nInput string: ";

    for(int i=0;; i++)
    {
        *(a+i*sizeof(char))=getchar();

        if(*(a+i*sizeof(char))=='\n')
        {
            *(a+i*sizeof(char))='\0';
            break;
        }

    }
}



int main()
{
    char a[20];

    cout<<"\n1st method\n";
    input(a);
    cout<<a;

    cout<<"\n2nd method\n";
    cin.get(a,10);
    cout<<a;

    cout<<"\n3rd method\n";
    cin.sync();
    cin.getline(a,sizeof(a));
    cout<<a;

    cout<<"\n4th method\n";
    dinput(a);
    cout<<a;

    return 0;
}

回答8:

I have the same problem and I just used cin.getline(input,300);.

noskipws and cin.get() sometimes are not easy to use. Since you have the right size of your array try using cin.getline() which does not care about any character and read the whole line in specified character count.