How do I manually throw an std::iostream::failure?
I have a try-catch loop that catches an exception when the user tries to input a non-integer string, however it does not throw an exception if the user tries to input a float since it will try to read everything before the decimal point in a float value. My solution is to manually throw the exception if there is still data remaining in the stream, how do I do that?
/*
Sample Implementation Code in C++
Handling Inputs from User in C++
This code only stops running when the user
inputs the appropriate values. Otherwise, the program
will continue asking the user for input.
*/
#include <iostream>
#include <limits> //numeric_limits
#include <stdexcept>
int main() {
std::cin.exceptions(std::ios::failbit); // set exceptions to be thrown when a failbit is set
int num = 0;
int den = 0;
while (true) {
try {
std::cout << "Enter numerator: ";
std::cin >> num;
if(std::cin.peek() != '\n') {
//HOW TO DO THIS PART?
std::iostream::failure e;
throw e;
}
std::cout << "Enter denominator: ";
std::cin >> den;
std::cout << "The quotient is " << num/den << std::endl;
} catch (std::iostream::failure& e){
std::cout << "Input should be an integer." << std::endl;
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
}
return 0;
}
Apparently it's as simple as:
throw std::iostream::failure("");
the important thing I forgot is the empty string ("") since it has a constructor that takes a string argument but not a void argument.
Related
As I am very new to exceptions, I am having some trouble with the below memory exception code. Basically, I would need to calculate and show sum of 10 positive inputs. During the input, my program should use exception mechanism to display a message that it doesn't allow negative numbers / not-a-numbers (NaN) if so is the input and then exit the program. I am told to use std::cin.fail() to detect whether the value entered fits the variable type(but I am not sure how to implement it). Would appreciate your help thanks!
#include <iostream>
int main() {
int number;
int a=-2;
try {
for(int i=0; i<10;i++) {
std::cin>>number;
}
}
catch (...) {
number==-number?
std::cout << "Its negative"<<std::endl;
number==a?
std::cout << "Its NaN"<<std::endl;
}
}
You say "10 positive inputs", but you have int number;. If your number is going to be an integer, then it can't take fractions like 33.44 for instance, and it also can't be NaN (so you wouldn't need to check for that). If you want to allow fractions, do double number instead. That aside, the check could look like this:
for (int i = 0; i < 10; i++) {
std::cin >> number;
if (std::cin.fail()) {
std::cin.clear();
std::string input;
std::cin >> input;
std::cout << "input failed! Invalid input: " << input << std::endl;
return -1;
}
After every inputted number, you do the std::cin.fail() check. If it failed, you can just return. In this example, I also made it print the invalid input. For that you need to call std::cin.clear(); to reset that error flag, and then you can put that invalid input into an std::string (which should work, unlike with the number). You need to include <string> for that. If you don't need to print anything, you can just return and forget about the clearing, the string and the output. Also, in this example, since the mess made by the invalid input is neatly cleaned up, you could take more (valid) inputs again if you don't return instead.
Next, checking for negative numbers:
if (number < 0) {
std::cout << "Error: Negative number!" << std::endl;
return -1;
}
Finally, the NaN check (for double):
if (isnan(number)) {
std::cout << "Error: number is NaN!" << std::endl;
return -1;
}
You need to include <math.h> for the isnan.
Putting it all together, it could look like this:
#include <iostream>
#include <string> // std::istream >> std::string
#include <math.h> // isnan
int main() {
double number;
for (int i = 0; i < 10; i++) {
std::cin >> number;
if (std::cin.fail()) {
std::cin.clear();
std::string input;
std::cin >> input;
std::cout << "input failed! Invalid input: " << input << std::endl;
return -1;
}
if (number < 0) {
std::cout << "Error: Negative number!" << std::endl;
return -1;
}
if (isnan(number)) {
std::cout << "Error: number is NaN!" << std::endl;
return -1;
}
}
return 0;
}
I was typing this and it asks the user to input two integers which will then become variables. From there it will carry out simple operations.
How do I get the computer to check if what is entered is an integer or not? And if not, ask the user to type an integer in. For example: if someone inputs "a" instead of 2, then it will tell them to reenter a number.
Thanks
#include <iostream>
using namespace std;
int main ()
{
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}
You can check like this:
int x;
cin >> x;
if (cin.fail()) {
//Not an int.
}
Furthermore, you can continue to get input until you get an int via:
#include <iostream>
int main() {
int x;
std::cin >> x;
while(std::cin.fail()) {
std::cout << "Error" << std::endl;
std::cin.clear();
std::cin.ignore(256,'\n');
std::cin >> x;
}
std::cout << x << std::endl;
return 0;
}
EDIT: To address the comment below regarding input like 10abc, one could modify the loop to accept a string as an input. Then check the string for any character not a number and handle that situation accordingly. One needs not clear/ignore the input stream in that situation. Verifying the string is just numbers, convert the string back to an integer. I mean, this was just off the cuff. There might be a better way. This won't work if you're accepting floats/doubles (would have to add '.' in the search string).
#include <iostream>
#include <string>
int main() {
std::string theInput;
int inputAsInt;
std::getline(std::cin, theInput);
while(std::cin.fail() || std::cin.eof() || theInput.find_first_not_of("0123456789") != std::string::npos) {
std::cout << "Error" << std::endl;
if( theInput.find_first_not_of("0123456789") == std::string::npos) {
std::cin.clear();
std::cin.ignore(256,'\n');
}
std::getline(std::cin, theInput);
}
std::string::size_type st;
inputAsInt = std::stoi(theInput,&st);
std::cout << inputAsInt << std::endl;
return 0;
}
Heh, this is an old question that could use a better answer.
User input should be obtained as a string and then attempt-converted to the data type you desire. Conveniently, this also allows you to answer questions like “what type of data is my input?”
Here is a function I use a lot. Other options exist, such as in Boost, but the basic premise is the same: attempt to perform the string→type conversion and observe the success or failure:
template <typename T>
auto string_to( const std::string & s )
{
T value;
std::istringstream ss( s );
return ((ss >> value) and (ss >> std::ws).eof()) // attempt the conversion
? value // success
: std::optional<T> { }; // failure
}
Using the optional type is just one way. You could also throw an exception or return a default value on failure. Whatever works for your situation.
Here is an example of using it:
int n;
std::cout << "n? ";
{
std::string s;
getline( std::cin, s );
auto x = string_to <int> ( s );
if (!x) return complain();
n = *x;
}
std::cout << "Multiply that by seven to get " << (7 * n) << ".\n";
limitations and type identification
In order for this to work, of course, there must exist a method to unambiguously extract your data type from a stream. This is the natural order of things in C++ — that is, business as usual. So no surprises here.
The next caveat is that some types subsume others. For example, if you are trying to distinguish between int and double, check for int first, since anything that converts to an int is also a double.
There is a function in c called isdigit(). That will suit you just fine. Example:
int var1 = 'h';
int var2 = '2';
if( isdigit(var1) )
{
printf("var1 = |%c| is a digit\n", var1 );
}
else
{
printf("var1 = |%c| is not a digit\n", var1 );
}
if( isdigit(var2) )
{
printf("var2 = |%c| is a digit\n", var2 );
}
else
{
printf("var2 = |%c| is not a digit\n", var2 );
}
From here
If istream fails to insert, it will set the fail bit.
int i = 0;
std::cin >> i; // type a and press enter
if (std::cin.fail())
{
std::cout << "I failed, try again ..." << std::endl
std::cin.clear(); // reset the failed state
}
You can set this up in a do-while loop to get the correct type (int in this case) propertly inserted.
For more information: http://augustcouncil.com/~tgibson/tutorial/iotips.html#directly
You can use the variables name itself to check if a value is an integer.
for example:
#include <iostream>
using namespace std;
int main (){
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
if(firstvariable && secondvariable){
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}else{
cout << "\n[ERROR\tINVALID INPUT]\n";
return 1;
}
return 0;
}
I prefer to use <limits> to check for an int until it is passed.
#include <iostream>
#include <limits> //std::numeric_limits
using std::cout, std::endl, std::cin;
int main() {
int num;
while(!(cin >> num)){ //check the Input format for integer the right way
cin.clear();
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
cout << "Invalid input. Reenter the number: ";
};
cout << "output= " << num << endl;
return 0;
}
Under C++11 and later, I have the found the std::stoi function very useful for this task. stoi throws an invalid_argument exception if conversion cannot be performed. This can be caught and handled as shown in the demo function 'getIntegerValue' below.
The stoi function has a second parameter 'idx' that indicates the position of the first character in the string after the number. We can use the value in idx to check against the string length and ascertain if there are any characters in the input other than the number. This helps eliminate input like 10abc or a decimal value.
The only case where this approach fails is when there is trailing white space after the number in the input, that is, the user enters a lot of spaces after inputting the number. To handle such a case, you could rtrim the input string as described in this post.
#include <iostream>
#include <string>
bool getIntegerValue(int &value);
int main(){
int value{};
bool valid{};
while(!valid){
std::cout << "Enter integer value: ";
valid = getIntegerValue(value);
if (!valid)
std::cout << "Invalid integer value! Please try again.\n" << std::endl;
}
std::cout << "You entered: " << value << std::endl;
return 0;
}
// Returns true if integer is read from standard input
bool getIntegerValue(int &value){
bool isInputValid{};
int valueFromString{};
size_t index{};
std::string userInput;
std::getline(std::cin, userInput);
try {
//stoi throws an invalid_argument exception if userInput cannot be
//converted to an integer.
valueFromString = std::stoi(userInput, &index);
//index being different than length of string implies additional
//characters in input that could not be converted to an integer.
//This is to handle inputs like 10str or decimal values like 10.12
if(index == userInput.length()) isInputValid = true;
}
catch (const std::invalid_argument &arg) {
; //you could show an invalid argument message here.
}
if (isInputValid) value = valueFromString;
return isInputValid;
}
You could use :
int a = 12;
if (a>0 || a<0){
cout << "Your text"<<endl;
}
I'm pretty sure it works.
I am beginning C++ programming, and have to do a lot of input validation. I have found this function that seems universally applicable, but am having trouble with one aspect; If I were to type -90, the program doesn't give an error. my question(s) are:
1. How can I add the circumstance that input cannot be <= 0?
2. Is there a better way to limit users input? Maybe a library within C++?
Thank you for any help, or advice.
#include <ios> // Provides ios_base::failure
#include <iostream> // Provides cin
template <typename T>
T getValidatedInput()
{
// Get input of type T
T result;
cin >> result;
// Check if the failbit has been set, meaning the beginning of the input
// was not type T. Also make sure the result is the only thing in the input
// stream, otherwise things like 2b would be a valid int.
if (cin.fail() || cin.get() != '\n')
{
// Set the error state flag back to goodbit. If you need to get the input
// again (e.g. this is in a while loop), this is essential. Otherwise, the
// failbit will stay set.
cin.clear();
// Clear the input stream using and empty while loop.
while (cin.get() != '\n')
;
// Throw an exception. Allows the caller to handle it any way you see fit
// (exit, ask for input again, etc.)
throw ios_base::failure("Invalid input.");
}
return result;
}
Usage
inputtest.cpp
#include <cstdlib> // Provides EXIT_SUCCESS
#include <iostream> // Provides cout, cerr, endl
#include "input.h" // Provides getValidatedInput<T>()
int main()
{
using namespace std;
int input;
while (true)
{
cout << "Enter an integer: ";
try
{
input = getValidatedInput<int>();
}
catch (exception e)
{
cerr << e.what() << endl;
continue;
}
break;
}
cout << "You entered: " << input << endl;
return EXIT_SUCCESS;
}
You can use functions to validate
template <typename T>
T getValidatedInput(function <bool(T)> validator) {
T tmp;
cin >> tmp;
if (!validator(tmp)) {
throw ios_base::failure("Invalid input.");
}
return tmp;
}
Usage
int input = getValidatedInput<int>([] (int arg) -> bool {
return arg >= 0;
});
std::istream::operator >> is defined in terms of strtol, strtoul, and cousins*, which unfortunately all invariably accept a minus sign even for unsigned types.
Essentially all you can do is accept signed int input and compare the result to zero. std::cin.setf( std::ios::failbit ) artificially raises a conversion exception, so you can sort-of emulate how the conversion function should behave on error, but that might not really be much help.
* operator >> is defined in terms of std::num_get, which is defined in terms of scanf, which is defined in terms of strto*. Everyone just passed the buck, but strtoul is pretty surely defective.
Use unsigned int as a template parameter.
Only you can setup a rules about what input is valid and what is not.
I hope this is what you're after, it exit's upon entering zero, but will display negative numbers. It throws an exception error due to the input catch method.
#include "stdafx.h"
#include <iostream>
using namespace std;
void inputcatch()
{
cin.clear();
cin.ignore(cin.rdbuf()->in_avail());
}
int main()
{
int input;
bool quit = false;
while (!quit)
{
cout << "Enter number" << endl;
cin >> input;
if (cin.fail())
{
inputcatch();
cout << "incorrect input" << endl;
}
else if (input == 0)
{
quit = true;
}
else
{
cout << "your number: " << input << endl;
}
}
return 0;
}
I have the following simple code:
#include <iostream>
int main()
{
int a;
std::cout << "enter integer a" << std::endl;
std::cin >> a ;
if (std::cin.fail())
{
std::cin.clear();
std::cout << "input is not integer, re-enter please" <<std::endl;
std::cin >>a;
std::cout << "a inside if is: " << a <<std::endl;
}
std::cout << "a is " << a <<std::endl;
std::cin.get();
return 0;
}
When I run the above code and input: 1.5, it outputs: a is 1. FYI: I compile and run the code with gcc 4.5.3.
This means that if cin expects an integer but sees a float, it will do the conversion implicitly. So does this mean that when cin sees a float number, it is not in fail() state? Why this is the case? Is it because C++ does implicit conversion on >> operator?
I also tried the following code to decide whether a given input number is integer following idea from this post: testing if given number is integer:
#include <iostream>
bool integer(float k)
{
if( k == (int) k) return true;
return false;
}
int main()
{
int a;
std::cout << "enter integer a"<< std::endl;
std::cin >> a ;
if (!integer(a))
{
std::cout << "input is not integer, re-enter please" ;
std::cin.clear();
std::cin >> a;
std::cout << "a inside if is: " << a <<std::endl;
}
std::cout << "a is " << a <<std::endl;
std::cin.get();
return 0;
}
This block of code was also not able to test whether a is integer since it simply skip the if block when I run it with float input.
So why this is the case when getting user input with cin? What if sometimes I want the input to be 189, but typed 18.9 by accident, it will result in 18 in this case, which is bad. So does this mean using cin to get user input integers is not a good idea?
thank you.
When you read an integer and you give it an input of 1.5, what it sees is the integer 1, and it stops at the period since that isn't part of the integer. The ".5" is still in the input. This is the reason that you only get the integer part and it is also the reason why it doesn't seem to wait for input the second time.
To get around this, you could read a float instead of an integer so it reads the whole value, or you could check to see if there is anything else remaining on the line after reading the integer.
When reading user input I prefer not to use operator>> as user input is usally line based and prone to errors. I find it best to read a line at a time and validate:
std::string line;
std::getline(std::cin, line);
This also makes it easy to check for different types of numbers.
std::stirngstream linestream(line);
int val;
char c;
if ((linestream >> val) && !(linestream >> c))
{
// Get in here if an integer was read.
// And there is no following (non white space) characters.
// i.e. If the user only types in an integer.
//
// If the user typed any other character after the integer (like .5)
// then this will fail.
}
Of course boost already supports this:
val = boost::lexical_cast<int>(linestream); // Will throw if linestream does
// not contain an integer or
// contains anything in addition
// to the integer.
Boost of course will convert floats as well.
I have some snippet which is kind a poor coding, but it works.
This method is pretty simple, but doesn't handle case when input value is invalid.
See more: https://en.cppreference.com/w/cpp/string/byte/atof
static float InputFloat(std::string label)
{
std::string input;
std::cout << label;
std::cin >> input;
return atof(input.c_str());
}
int main()
{
float value = InputFloat("Enter some float value: ");
std::cout << "value = " << value;
return 0;
}
Say I have a program that takes in integers. How do I stop the program from falling apart if the user enters an out of range number, or a letter or something?
The cin's base class is std::basic_istream. The input stream indicates a recoverable error in case it cannot extract the requested data from the stream. In order to check for that error bit, std::basic_istream::fail() method must be used — it returns true if there was a failure or false if everything is alright. It is important to remember that if there is an error, the data is left in the stream and, of course, the error bit(s) must also be cleared using std::basic_istream::clear(). Also, a programmer must ignore incorrect data, or otherwise an attempt to read something else will fail again. For that purpose, std::basic_istream::ignore() method can be used. As for the valid range of values, it must be checked manually. Okay, enough theory, here is a simple example:
#include <limits>
#include <iostream>
int main()
{
int n = 0;
for (;;) {
std::cout << "Please enter a number from 1 to 10: " << std::flush;
std::cin >> n;
if (std::cin.fail()) {
std::cerr << "Sorry, I cannot read that. Please try again." << std::endl;
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
continue;
}
if (n < 1 || n > 10) {
std::cerr << "Sorry, the number is out of range." << std::endl;
continue;
}
std::cout << "You have entered " << n << ". Thank you!" << std::endl;
break;
}
}
Hope it helps. Good Luck!
I prefer reading the input as strings, and then sanitizing them with boost::lexical_cast<>:
#include <boost/lexical_cast.hpp>
#include <iostream>
#include <string>
int main () {
std::string s;
while( std::cin >> s) {
try {
int i = boost::lexical_cast<int>(s);
std::cout << "You entered: " << i << "\n";
} catch(const std::bad_cast&) {
std::cout << "Ignoring non-number: " << s << "\n";
}
}
}
Postscript: If you are allergic to Boost, you can use this implementation of lexical_cast:
template <class T, class U>
T lexical_cast(const U& u) {
T t;
std::stringstream s;
s << u;
s >> t;
if( !s )
throw std::bad_cast();
if( s.get() != std::stringstream::traits_type::eof() )
throw std::bad_cast();
return t;
}
something like this should do you need to clear the buffer after checking aswell if i remember right
if (cin.fail())
{
cout<<"need to put a number"<<endl;
cin.clear();
cin.ignore();
}
If You dont want to add libraries to your code you could also use do..while() statements.
in your do while you will ask for user input and then receive it to your variable then in the while part you will be able to check that this is the data you are expecting if not continue to ask for the data.
just another option ....even though the answer already mentioned should work more than adequately
You can use the following code for simplest and fast checking of valid input in int :
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int intb;
while( !( cin>>intb ) ){
cin.clear ();
cin.ignore (1000, '\n');
cout<<"Invalid input enter again: "<<endl;
}
cout<<"The value of integer entered is "<<b<<endl;
return 0;
}
The while loop keeps on iterating until it gets the right input.
cin.clear() changes the error control state.
cin.ignore() removes clear the input stream so that new input can be taken again. If not done thw while loop will be in infinite state.