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Verifying valid input in a loop

P: n/a
JR
Hey all,

I have read part seven of the FAQ and searched for an answer but can
not seem to find one.

I am trying to do the all too common verify the data type with CIN.
The code from the FAQ looks like this:

#include <iostream>

int main()
{
std::cout << "Enter a number, or -1 to quit: ";
int i = 0;
while (std::cin >> i) { // GOOD FORM
if (i == -1) break;
std::cout << "You entered " << i << '\n';
}
}
This works well except that I do not want to exit the loop unless a
valid input is entered. So if I enter an 'a' I want to stay in the
loop to allow for another chance at input not exit the loop. Can
someone please provide a modified version of this code that will allow
this?
Thanks,

James
Jul 22 '05 #1
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7 Replies


P: n/a

"JR" <JR*****@email.com> wrote in message
news:8b**************************@posting.google.c om...
Hey all,

I have read part seven of the FAQ and searched for an answer but can
not seem to find one.

I am trying to do the all too common verify the data type with CIN.
The code from the FAQ looks like this:

#include <iostream>

int main()
{
std::cout << "Enter a number, or -1 to quit: ";
int i = 0;
while (std::cin >> i) { // GOOD FORM
if (i == -1) break;
std::cout << "You entered " << i << '\n';
}
}
This works well except that I do not want to exit the loop unless a
valid input is entered. So if I enter an 'a' I want to stay in the
loop to allow for another chance at input not exit the loop. Can
someone please provide a modified version of this code that will allow
this?


It's difficult to validate input, it gets very fiddly. C++ cannot provide a
standard solution because what is valid input for one program is not valid
for another. And since you don't say what is valid input for your program
its impossible to give a detailed answer. I would guess that only positive
integers and -1 are valid for you.

However the general approach is simple enough. Read the input as a string,
check to see if the string is a valid number (whatever that means for your
program) and only then convert the string to an integer.

John
Jul 22 '05 #2

P: n/a
JR writes:
I am trying to do the all too common verify the data type with CIN.
The code from the FAQ looks like this:

#include <iostream>

int main()
{
std::cout << "Enter a number, or -1 to quit: ";
int i = 0;
while (std::cin >> i) { // GOOD FORM
if (i == -1) break;
std::cout << "You entered " << i << '\n';
}
}
This works well except that I do not want to exit the loop unless a
valid input is entered. So if I enter an 'a' I want to stay in the
loop to allow for another chance at input not exit the loop. Can
someone please provide a modified version of this code that will allow
this?


The overloaded >> specifies conversion. There is no back door way to get
the faulty input (for example, a letter), the stream goes into a fail state
if it gets bad input. So base your solution on getline() and do the
conversion "manually". Also, see ios::fail() and ios::clear(). The
simplest conversion would probably be strtol() in cstdlib. Using the stuff
in the iostreams package would be more elegant, but IMO harder.

You would have to give up the ability to display the faulty input to the
user if you wanted to persist in using cin.
Jul 22 '05 #3

P: n/a
In article <8b**************************@posting.google.com >,
JR <JR*****@email.com> wrote:

#include <iostream>

int main()
{
std::cout << "Enter a number, or -1 to quit: ";
int i = 0;
while (std::cin >> i) { // GOOD FORM
if (i == -1) break;
std::cout << "You entered " << i << '\n';
}
}
This works well except that I do not want to exit the loop unless a
valid input is entered. So if I enter an 'a' I want to stay in the
loop to allow for another chance at input not exit the loop. Can
someone please provide a modified version of this code that will allow
this?


First, let's just do the error checking for a single input, and not worry
about the "-1 to quit" yet:

int i;
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> i))
{
cin.clear(); // reset stream status flags
cin.ignore(1000,'\n'); // skip past next newline (ENTER), or 1000
// chars, whichever comes first
cout << "Hey dummy, I said enter an integer! Try again: ";
}

If you want to input repeatedly until the user enters -1, wrap this in
another loop that tests for end of data:

int i = 0;
while (i != -1)
{
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> i))
{
cin.clear();
cin.ignore(1000,'\n');
cout << "Hey dummy, I said enter an integer! Try again: ";
}
if (i != -1)
{
// process i normally
}
}

Or you can make the real processing-code a bit cleaner by writing a
function to do the input and error-checking:

bool GetNumber (int& num)
{
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> num))
{
cin.clear();
cin.ignore(1000,'\n');
cout << "Hey dummy, I said enter an integer! Try again: ";
}
return (num != -1); // returns true if not at end of input yet
}

// the real processing-code

int i;
while (GetNumber(i))
{
// process i normally
}

--
Jon Bell <jt*******@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
Jul 22 '05 #4

P: n/a
JR
jt*******@presby.edu (Jon Bell) wrote in message news:<ch**********@jtbell.presby.edu>...
<snip>>

First, let's just do the error checking for a single input, and not
worry
about the "-1 to quit" yet:

int i;
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> i))
{
cin.clear(); // reset stream status flags
cin.ignore(1000,'\n'); // skip past next newline (ENTER), or 1000
// chars, whichever comes first
cout << "Hey dummy, I said enter an integer! Try again: ";
}

If you want to input repeatedly until the user enters -1, wrap this in
another loop that tests for end of data:

int i = 0;
while (i != -1)
{
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> i))
{
cin.clear();
cin.ignore(1000,'\n');
cout << "Hey dummy, I said enter an integer! Try again: ";
}
if (i != -1)
{
// process i normally
}
}

Or you can make the real processing-code a bit cleaner by writing a
function to do the input and error-checking:

bool GetNumber (int& num)
{
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> num))
{
cin.clear();
cin.ignore(1000,'\n');
cout << "Hey dummy, I said enter an integer! Try again: ";
}
return (num != -1); // returns true if not at end of input yet
}

// the real processing-code

int i;
while (GetNumber(i))
{
// process i normally
}


Thanks All,

Jon I really like your solution. I came across is a couple hours
after I posted in this post .

http://groups.google.com/groups?hl=e...ng.c.moderated

This is essentially what I did and it is similar to the solution you
provided. It seems very elegant and easy to use. The ironic thing is
he posted that code to help him solve another problem. So his problem
was my solution. :)

Thank you again.
Jul 22 '05 #5

P: n/a

"JR" <JR*****@email.com> wrote in message
news:8b**************************@posting.google.c om...
jt*******@presby.edu (Jon Bell) wrote in message
news:<ch**********@jtbell.presby.edu>...
<snip>>

First, let's just do the error checking for a single input, and not
worry
about the "-1 to quit" yet:

int i;
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> i))
{
cin.clear(); // reset stream status flags
cin.ignore(1000,'\n'); // skip past next newline (ENTER), or
1000
// chars, whichever comes first
cout << "Hey dummy, I said enter an integer! Try again: ";
}

If you want to input repeatedly until the user enters -1, wrap this in
another loop that tests for end of data:

int i = 0;
while (i != -1)
{
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> i))
{
cin.clear();
cin.ignore(1000,'\n');
cout << "Hey dummy, I said enter an integer! Try again: ";
}
if (i != -1)
{
// process i normally
}
}

Or you can make the real processing-code a bit cleaner by writing a
function to do the input and error-checking:

bool GetNumber (int& num)
{
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> num))
{
cin.clear();
cin.ignore(1000,'\n');
cout << "Hey dummy, I said enter an integer! Try again: ";
}
return (num != -1); // returns true if not at end of input yet
}

// the real processing-code

int i;
while (GetNumber(i))
{
// process i normally
}


Thanks All,

Jon I really like your solution. I came across is a couple hours
after I posted in this post .


It has a weakness. Most people would consider 123abc to be erroneous input.
But the code above will read this as the integer 123, then loop and read abc
(which obviously it would consider an error).

If you want to avoid this problem, then you must read as a string and then
convert to an integer.

john

Jul 22 '05 #6

P: n/a

"JR" <JR*****@email.com> schrieb im Newsbeitrag
news:8b**************************@posting.google.c om...
Hey all,

I have read part seven of the FAQ and searched for an answer but can
not seem to find one.

I am trying to do the all too common verify the data type with CIN.
The code from the FAQ looks like this:

#include <iostream>

int main()
{
std::cout << "Enter a number, or -1 to quit: ";
int i = 0;
while (std::cin >> i) { // GOOD FORM
if (i == -1) break;
std::cout << "You entered " << i << '\n';
}
}
This works well except that I do not want to exit the loop unless a
valid input is entered. So if I enter an 'a' I want to stay in the
loop to allow for another chance at input not exit the loop. Can
someone please provide a modified version of this code that will allow
this?
Thanks,

James


I would do it like this:

#include <iostream>

using namespace std;

//clears an error in an input stream object like cin
//and makes it usable again by clearing the associated stream buffer
int ClearError(istream& isIn) // Clears istream object
{
streambuf* sbpThis;
char szTempBuf[20];
int nCount, nRet = isIn.rdstate();

if (nRet) // Any errors?
{
isIn.clear(); // Clear error flags
sbpThis = isIn.rdbuf(); // Get streambuf pointer
nCount = sbpThis->in_avail(); // Number of characters in buffer

while (nCount) // Extract them to szTempBuf
{
if (nCount > 20)
{
sbpThis->sgetn(szTempBuf, 20);
nCount -= 20;
}
else
{
sbpThis->sgetn(szTempBuf, nCount);
nCount = 0;
}
}
}

return nRet;
}
int main()
{
int pri = 0;
cout << "Enter a number: " << endl;
do
{
ClearError(cin);
cin >> pri;
} while (!cin);
return 0;
}
Jul 22 '05 #7

P: n/a
JR
"John Harrison" <jo*************@hotmail.com> wrote in message news:<2q************@uni-berlin.de>...
"JR" <JR*****@email.com> wrote in message
news:8b**************************@posting.google.c om...
jt*******@presby.edu (Jon Bell) wrote in message
news:<ch**********@jtbell.presby.edu>...
<snip>>

First, let's just do the error checking for a single input, and not
worry
about the "-1 to quit" yet:

int i;
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> i))
{
cin.clear(); // reset stream status flags
cin.ignore(1000,'\n'); // skip past next newline (ENTER), or
1000
// chars, whichever comes first
cout << "Hey dummy, I said enter an integer! Try again: ";
}

If you want to input repeatedly until the user enters -1, wrap this in
another loop that tests for end of data:

int i = 0;
while (i != -1)
{
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> i))
{
cin.clear();
cin.ignore(1000,'\n');
cout << "Hey dummy, I said enter an integer! Try again: ";
}
if (i != -1)
{
// process i normally
}
}

Or you can make the real processing-code a bit cleaner by writing a
function to do the input and error-checking:

bool GetNumber (int& num)
{
cout << "Please enter an integer, and press ENTER: ";
while (! (cin >> num))
{
cin.clear();
cin.ignore(1000,'\n');
cout << "Hey dummy, I said enter an integer! Try again: ";
}
return (num != -1); // returns true if not at end of input yet
}

// the real processing-code

int i;
while (GetNumber(i))
{
// process i normally
}


Thanks All,

Jon I really like your solution. I came across is a couple hours
after I posted in this post .


It has a weakness. Most people would consider 123abc to be erroneous input.
But the code above will read this as the integer 123, then loop and read abc
(which obviously it would consider an error).

If you want to avoid this problem, then you must read as a string and then
convert to an integer.

john

That is a good point. It also has the weakness of if 100,000 is
entered then only 100 is taken. It does seem that overall the input
as a string makes the most sense.

James
Jul 22 '05 #8

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