[source#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<int> n;
vector<int>::iterator pd;
unsigned long temp;
unsigned long ch = 1000001;
//enter numbers into the vector
while (cin >> temp)
{
n.push_back(temp);
}
return 0;
}
How do I terminate input?
Author
For example, let's say I can enter 1,000,000 integers into a vector. But, in reality, that value could be anywhere from 0 - 1,000,000. What do I do?
I can't do anything like "Press q to quit".
If you want to intentionally limit the size of a vector, you can check vector::size() before you do a push_back.
Not sure what you are asking? Do you wan't a terminate the loop when something
unique is inputted?
unique is inputted?
Author
The user is supposed to be able to enter 0 to 10^6 integers. Then I have to sort the integers from least to greatest.
My question was how do I know if the user stops before he reaches 10^6...for example, if he enters 500 integers and wants to stop.
But I think I have to use a different method than this. ;o
My question was how do I know if the user stops before he reaches 10^6...for example, if he enters 500 integers and wants to stop.
But I think I have to use a different method than this. ;o
Quote:I can't do anything like "Press q to quit".
Why not? If you want to avoid a lot of complex checking of input and you trust the user to follow instructions, you can prompt to enter something invalid and use cin.fail().
cout << "Enter 'q,enter' to terminate input."; // I think ctrl-Z works as wellcin >> temp;while( !cin.fail() ) { n.push_back(temp); cin >> temp;}// do your sorting - n.size() gives you the number of inputs
There are also a couple of commands that are OS specific:
Ctrl+z on Windows
Ctrl+d on Linux/Mac
If I remember correctly. Then press enter.
The size() function on the vector will give you the number of elements it contains.
Ctrl+z on Windows
Ctrl+d on Linux/Mac
If I remember correctly. Then press enter.
The size() function on the vector will give you the number of elements it contains.
This topic is closed to new replies.
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