Inputting a character value into an int array?

Question

This is the program that I wrote:

    #include<iostream>
    using namespace std;

    int main()
    {
        int hex[16];
        cout << "enter the 16 hexadecimal numbers\n";
        for(int i = 0;i < 16;i++)
        {
            cin >> hex[i];
            cout << "input worked\n";
        }
    }

Can you please explain to me why the "input worked" output is again being displayed after I inputted 'a'. Shouldn't the program stop if I did that?

What I mean is why does "input worked" line output after cin stopped taking input?I mean if there is a problem in taking input from cin why is cout still working?

The character `a` is not an `int` and you are trying to extract an `int`. This creates an *error* in the input `cin` and it stops working. — Galik, Sep 19 '15 at 09:59

score 2 · Answer 1 · answered Sep 19 '15 at 09:59

You must pipe it through std::hex, otherwise it accepts just an integer and your input stream switches internally to an error state and the inputs of the next iteration are skipped. That's why you see the outputs of the remaining iterations.

cin >> std::hex >> hex[i];

πάντα ῥεῖ · Answer 2 · 2015-09-19T11:10:31.627

1

You should write

if(cin >> hex >> hex[i]) {
    cout << "input worked\n";
}
else {
    string dummy;
    cin.clear();
    cin >> dummy;
}

to determine if input worked. For the hex IO manipulator you must #include <iomanip>.

edited Sep 19 '15 at 11:10

answered Sep 19 '15 at 09:57

πάντα ῥεῖ

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yeah but why do I get the "input worked" line after cin stopped taking input?That is my main question – user2546419 Sep 19 '15 at 10:00
@Galik oh okay so the cin just stops from taking bad input but the for loop still works .Thats why cout is working right? – user2546419 Sep 19 '15 at 10:07
1

@user2546419 yes, after an error `cin` just returns immediately without doing anything. – Galik Sep 19 '15 at 10:10
@Galik: It's not an infinite loop because it terminates anyway after 16 iterations. – Christian Hackl Sep 19 '15 at 10:12
@Galik THX for pointing out. Fixed. – πάντα ῥεῖ Sep 19 '15 at 10:13

Galik · Answer 3 · 2015-09-19T10:28:25.453

By default input streams like std::cin expect integer input to be in decimal (characters '0'-'9'). If you want to input in hexadecimal you need to use std::hex:

#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
    int hex[16];
    cout << "enter the 16 hexadecimal numbers\n";
    for(int i = 0;i < 16;i++)
    {
        // hexedicimal
        cin >> std::hex >> hex[i];
        cout << "input worked\n";
    }
}

Also you should always check the stream for errors after reading from it.

#include <iostream>
#include <iomanip>
#include <string>
using namespace std;

int main()
{
    int hex[16];
    cout << "enter the 16 hexadecimal numbers\n";
    for(int i = 0;i < 16;i++)
    {
        if(cin >> std::hex >> hex[i])
            cout << "input worked\n";
        else
        {
            cout << "bad input try again\n";
            cin.clear(); // remove the error
            std::string skip;
            cin >> skip; // read past bad input
        }
    }
}

score 1 · Answer 4 · answered Sep 19 '15 at 10:09

The fact that you named your array hex does not matter to the compiler. It expects decimal input.

When you enter a, the input cannot be parsed as a decimal integer, so std::cin is set to an error mode. The subsequent call to >> then no longer prompts the user for text input but immediately returns, leaving the stream in its error state, until the loop finishes.

You can solve this problem in several ways:

Check the stream's state after the user has entered something. if (!cin) { /* error handling */ }.
Additionally, allow hexadecimal input with std::hex.

Or use a different strategy altogether. Instead of reading ints directly, each loop iteration could read a whole line of input into an std::string. That string can then be converted to an int with C++11's std::stoi, specifying base 16:

#include <iostream>
#include <string>

int main()
{
    int hex[16];
    std::cout << "enter the 16 hexadecimal numbers\n";
    for (int i = 0; i < 16; ++i)
    {
        std::string line;
        std::getline(std::cin, line);
        try
        {
            hex[i] = std::stoi(line, 0, 16);
            std::cout << "input worked: " << hex[i] << "\n";
        }
        catch (std::exception const& exc)
        {
            std::cout << "input failed\n";
        }
    }
}

Inputting a character value into an int array?

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