is it possible, say your trying to do calculations so the primary variable type may be int... but as a part of the program you decide to do a while loop and throw an if statement for existing purposes.
you have one cin >> and that is to take in a number to run calculations, but you also need an input incase they want to exit:
Here's some code to work with
#include <iostream>
using namespace std;
int func1(int x)
{
int sum = 0;
sum = x * x * x;
return sum;
}
int main()
{
bool repeat = true;
cout << "Enter a value to cube: " << endl;
cout << "Type leave to quit" << endl;
while (repeat)
{
int input = 0;
cin >> input;
cout << input << " cubed is: " << func1(input) << endl;
if (input = "leave" || input = "Leave")
{
repeat = false;
}
}
}
I'm aware they wont take leave cause input is set to int, but is it possible to use a conversion or something...
another thing is there a better way to break the loop or is that the most common way?
One way to do this is read a string from cin. Check its value. If it satisfies the exit condition, exit. If not, extract the integer from the string and proceed to procss the integer.
while (repeat)
{
string input;
cin >> input;
if (input == "leave" || input == "Leave")
{
repeat = false;
}
else
{
int intInput = atoi(input.c_str());
cout << input << " cubed is: " << func1(intInput) << endl;
}
}
You can read the input as a string from the input stream. Check if it is 'leave' and quit.. and If it is not try to convert it to a number and call func1.. look at atoi or boost::lexical_cast<>
also it is input == "leave" == is the equal operator. = is an assignment operator.
int main() {
cout << "Enter a value to cube: " << endl;
cout << "Type leave to quit" << endl;
while (true)
{
string input;
cin >> input;
if (input == "leave" || input == "Leave")
{
break;
}
cout << input << " cubed is: " << func1(atoi(input.c_str())) << endl;
}
}
you can use like
int input;
string s;
cint>>s; //read string from user
stringstream ss(s);
ss>>input; //try to convert to an int
if(ss==0) //not an integer
{
if(s == "leave") {//user don't want to enter further input
//exit
}
else
{
//invalid data some string other than leave and not an integer
}
}
else
{
cout<<"Input:"<<input<<endl;
//input holds an int data
}
Related
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'm trying to check two separate inputs if they are integers or not. I'm able to error check one input but I'm not quite sure how to check two separate ones if I'm using the 'get' function and both inputs are from the 'cin' stream. Using c++.
My code for checking one integer is displayed below.
#include <iostream>
using namespace std;
int main() {
int input;
cout << "Enter an integer: ";
cin >> input;
char next;
int x=0;
int done = 0;
while (!done){
next = cin.get();
if (next == ' ' || next == '\n'){
cout << "The Integer that you have entered is: " << input << "\n";
done = 1;
}
else if (next == '.'){
cerr << "Error: Invalid Input. Not an Integer." << "\n";
done = 1;
}
else{
cerr << "Error: Invalid Input. Not a number." << "\n";
done = 1;
}
}
return 0;
}
Well you could use >> into an int all the way through, drop all that get() stuff and character handling, and check cin.fail(). For example (I'll leave working this into your program and repeating it in a loop as an exercise for you):
int x;
cin >> x;
if (cin.fail())
cout << "Not a valid integer." << endl;
You can handle all subsequent input in exactly the same way. There's no reason to only limit operator >> to the first input.
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;
}
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.
}
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);
I am trying to write a simple program in C++ that reads in an unspecified number of marks, then once the user inputs the character 'q', the program must calculate and display the average mark. However I am having some trouble. The approach I am taking is to save each value as a double, the I want to compare the double to the character 'q' and if they are the same character, end the loop, calculate and display the average.
However I think that the comparison between the char value 'q' and double value for the mark seem to be incomparable. This worked for me when I did the same using integer values for the mark but not doubles it seems. Any help would be appreciated.
Here is the code:
int main()
{
cout << "Please enter any number of marks. Enter 'q' to stop." << endl;
double total = 0;
int counter = 0;
bool repeat = true;
do
{
double mark;
cin >> mark;
if (mark != 'q')
{
total += mark;
counter++;
}
else
{
repeat = false;
}
}
while (repeat == true);
double average = total/counter;
cout << "Average: " << average << endl;
return 0;
}
you'll need to change the mark variable to string, and then compare it to 'q', else try to parse is as a number.
otherwise this entire code, does not make a lot of sense, because 'q' in ASCII is 113, which I guess is a possible value
Typecast double to int and then compare, It must work because it compares ASCII value of character
Here is the code:
int main()
{
cout << "Please enter any number of marks. Enter 'q' to stop." << endl;
double total = 0;
int counter = 0;
bool repeat = true;
do
{
double mark;
cin >> mark;
if ((int)mark != 'q')
{
total += mark;
counter++;
}
else
{
repeat = false;
}
}
while (repeat == true);
double average = total/counter;
cout << "Average: " << average << endl;
return 0;
}
you can not cast a double to char. You may need to use additional c++ library functions which convert string (char*) to double. There are different ways to do this.
Try this :
#include <iostream>
#include <sstream>
using namespace std;
int main()
{
cout << "Please enter any number of marks. Enter 'q' to stop." << endl;
double total = 0;
int counter = 0;
bool repeat = true;
do
{
char userinput[8];
cin >> userinput;
std::stringstream ss;
double mark;
if (userinput[0] != 'q')
{
ss << userinput;
ss >> mark;
total += mark;
counter++;
}
else
{
repeat = false;
}
}
while (repeat == true);
double average = total/counter;
cout << "total : " << total << " count : " << counter << endl;
cout << "Average: " << average << endl;
return 0;
}
You are doing it wrong. If you try to cin a char into a double variable, the input char stays in the input buffer and the double variable remains the same. So this will end in an infinite loop.
If you really want the user to enter a char to end input, you need to take the whole input in a string variable. Check the string for q. If not present, use atof() to convert it to double.
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.