C++ user enters floating point instead of integer - c++

Question

Im trying to get the program to only accept x as an integer then ask for another integer, y. However when i enter a floating point into x it takes the decimal part of the input and makes that the y value. i am unsure of my mistake here.
#include <iostream>
#include <string>
#include <limits>
using namespace std;
int getInt()
{
int x = 0;
while (!(cin >> x))
{
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << "Please input a proper 'whole' number: " << endl;
}
return (x);
}
int toobig()
{
cout << "Your number is too large, please enter something smaller: " << endl;
int x = getInt();
return (x);
}
int toosmall()
{
cout << "your number is negative, please enter a positive number: " << endl;
int x = getInt();
return (x);
}
int main()
{
cout << "your number please:-" << endl;
int x = getInt();
if (x>100000)
{
toobig();
}
else if (x<0)
{
toosmall();
}
int y = 0;
cout << "enter y " << endl;
cin >> y;
cout << "x = " << x << endl;
cout << "y = " << y << endl;
system("PAUSE");
return 0;
}

Answer 1

Most conversions to int stop as soon as they find something that can't be part of an int and only a few conversion functions tell you if they stop before parsing the whole string.
Let's use one of those few, shall we?
int getInt()
{
for ( ; ; ) // loop until user provides something we can use.
// This is dangerous. You probably want to give up after a while.
{
std::string input; // read in as string
if (std::cin >> input)
{
char * endp; // will be updated with pointer to where conversion stopped
errno = 0;
// convert string to int
long rval = std::strtol (input.c_str(), &endp, 10);
if (*endp == '\0') // check whole string was read
{
if (errno != ERANGE) // check converted number did not overflow long
{
if (rval >= std::numeric_limits<int>::min() &&
rval <= std::numeric_limits<int>::max())
// check converted number did not overflow int
// you could replace this min and max with your own passed-in
// min and max values if you want
{
return rval; // return the known-to-be-good int
}
}
}
}
else
{ // note: usually when cin fails to read a string, it's over.
// This is actually a good time to throw an exception because this
// just shouldn't happen.
std::cin.clear(); // but for now we'll just clear the error and
// probably enter an infinite loop of failure
}
// failed for any reason. Blow off all user input and re-prompt
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
std::cout << "Please input a proper 'whole' number: " << std::endl;
}
return 0; // to satisfy compiler because a non-void function must always return.
// Never reached because of infinite for loop.
}

Answer 2

since every console input on C++ is treated as a string, then under your getInt() method I would do the following:
int GetInt(istream &stream)
{
char obtainChar; //read a character from input
int x;
stream >> x;
while(stream.fail() || (stream.peek() != '\r' && stream.peek() != '\n'))
{
stream.clear(); //clear the fail state of stream
obtainChar = stream.get(); //read a character from input
while(obtainChar != '\n' && obtainChar != EOF) //while gotten char is not a return key or EOF
obtainChar = stream.get(); //read a character from input iterate up to '\n' or EOF
//displays an error message if there was a bad input (e.g. decimal value)
cerr << endl << "Please input a proper 'whole' number: " << endl;
cout << endl << "Please re-enter x: "; //re-prompt to re-enter a value
x = GetInt(stream); //Try again by calling the function again (recursion)
}
return x; //will return after the user enters ONLY if an integer was inputted
}
the first while is basically saying, if the stream (console input) does fails or the next stream char (.peek()) is not a \r or a \n the clear the stream and get the first character.
while that char is not a \n and not End Of File (EOF) then obtain the next char, so on and so forth.
if a problem occurred then display a error message to the user and re-prompt the user for the value of x.
then call the same function to re-test the input (recursively), if all is well then return the value of x.
you can now call this function to evaluate the value of Y.
NOTE: istream is part of the iostream library is basically cin
NOTE: call the function like so:
int x;
cout << "your number please:-" << endl;
x = GetInt(cin);