std::getline does not work inside a for-loop

问题

I'm trying to collect user's input in a string variable that accepts whitespaces for a specified amount of time.

Since the usual cin >> str doesn't accept whitespaces, so I'd go with std::getline from <string>

Here is my code:

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        string local;
        getline(cin, local); // This simply does not work. Just skipped without a reason.
        //............................
    }

    //............................
    return 0;
}

Any idea?

回答1:

You can see why this is failing if you output what you stored in local (which is a poor variable name, by the way :P):

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        string local;
        getline(cin, local);
        std::cout << "> " << local << std::endl;
    }

    //............................
    return 0;
}

You will see it prints a newline after > immediately after inputting your number. It then moves on to inputting the rest.

This is because getline is giving you the empty line left over from inputting your number. (It reads the number, but apparently doesn't remove the \n, so you're left with a blank line.) You need to get rid of any remaining whitespace first:

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
    int n;
    cin >> n;
    cin >> ws; // stream out any whitespace
    for(int i = 0; i < n; i++)
    {
        string local;
        getline(cin, local);
        std::cout << "> " << local << std::endl;
    }

    //............................
    return 0;
}

This the works as expected.

Off topic, perhaps it was only for the snippet at hand, but code tends to be more readable if you don't have using namespace std;. It defeats the purpose of namespaces. I suspect it was only for posting here, though.

回答2:

Declare a character to get in the carriage return after you have typed in the number.char ws;int n;cin>>n;ws=cin.get(); This will solve the problem.

Using cin>>ws instead of ws=cin.get(),will make first character of your string to be in variable ws,instead of just clearing '\n'.

回答3:

Are you hitting enter? If not get line will return nothing, as it is waiting for end of line...

回答4:

My guess is that you're not reading n correctly, so it's converting as zero. Since 0 is not less that 0, the loop never executes.

I'd add a bit of instrumentation:

int n;
cin >> n;
std::cerr << "n was read as: " << n << "\n"; // <- added instrumentation
for // ...

回答5:

Is n properly initialized from input?
You don't appear to be doing anything with getline. Is this what you want?
getline returns an istream reference. Does the fact that you're dropping it on the ground matter?

回答6:

On which compiler did you try this? I tried on VC2008 and worked fine. If I compiled the same code on g++ (GCC) 3.4.2. It did not work properly. Below is the versions worked in both compilers. I dont't have the latest g++ compiler in my environment.

int n;
cin >> n;
string local;
getline(cin, local); // don't need this on VC2008. But need it on g++ 3.4.2. 
for (int i = 0; i < n; i++)
{
    getline(cin, local);
    cout << local;
}

回答7:

The important question is "what are you doing with the string that gives you the idea that the input was skipped?" Or, more accurately, "why do you think the input was skipped?"

If you're stepping through the debugger, did you compile with optimization (which is allowed to reorder instructions)? I don't think this is your problem, but it is a possibility.

I think it's more likely that the string is populated but it's not being handled correctly. For instance, if you want to pass the input to old C functions (eg., atoi()), you will need to extract the C style string (local.c_str()).

回答8:

You can directly use getline function in string using delimiter as follows:

#include <iostream>
using namespace std;
int main()
{
    string str;
    getline(cin,str,'#');
    getline(cin,str,'#');
}

you can input str as many times as you want but one condition applies here is you need to pass '#'(3rd argument) as delimiter i.e. string will accept input till '#' has been pressed regardless of newline character.

来源：https://stackoverflow.com/questions/2039918/stdgetline-does-not-work-inside-a-for-loop

标签

c++

getline