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I'm only in 10th grade and there is an exercise where I have to get an input like (input = 1/2) and this input needs to be stored in 2 different integers and I don't know how to do that.
Any help is appreciated.
Sorry if this question sounds dumb, but this is my first post here and I don't know the etiquette.
Thanks in advance!
Read two integers and ignore/match the character between them. The nicest way to match the character would be a manipulator:
std::istream& slash(std::istream& in) {
if ((in >> std::ws).peek() == '/') {
in.ignore();
}
else {
in.setstate(std::ios_base::failbit);
}
return *this;
}
int main() {
int numerator, denominator;
if (std::cin >> numerator >> slash >> denominator) {
std::cout << " read " << numerator << '/' << denominator << '\n';
}
}
Simply use the existing iostream functionality. There is nothing complicated about that. String parsing is not necessary. The inserter operator will skip whit spaces. Additionally, it will convert your given values into an integer variable. The slash will be read, checked and then discarded.
#include <iostream>
int main() {
int number1{}, number2{};
char slash{};
std::cout << "\nEnter a string: ";
if (std::cin >> number1 >> slash >> number2 && slash == '/') {
std::cout << "\n\nNumber 1: " << number1 << "\nNumber 2: " << number2 << "\n\n";
}
else {
std::cerr << "\nError: invalid input\n\n";
}
return 0;
}
Solution:
Take the input as a std::string and then parse it.
Full code:
#include <iostream>
std::pair<int,int> inputNumbers()
{
std::string input;
std::cout << "Input: ";
std::cin >> input;
size_t slashPos = input.find('/');
if(slashPos == input.npos || slashPos == 0 || slashPos == input.size() - 1)
{
std::cout << "Error! Please try again.\n";
exit(0);
}
std::pair<int,int> ret;
try
{
ret.first = std::stoi(input.substr(0,slashPos));
ret.second = std::stoi(input.substr(slashPos+1,input.size()-slashPos-1));
}
catch(std::invalid_argument&)
{
std::cout << "Error! Please try again.\n";
exit(0);
}
return ret;
}
int main() {
std::pair<int,int> inputs = inputNumbers();
std::cout << "First number: " << inputs.first << ". Second number: " << inputs.second << ".\n";
}
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 have the following code:
qstn:
cout << "Input customer's lastname: ";
getline(cin, lname);
if (lname.find_first_not_of("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ ") != string::npos) {
cout << "You can only input alpha here!\n";
cin.clear();
goto qstn;
} else if (lname.empty()) {
cout << "Please enter your firstname!\n";
cin.clear();
goto qstn;
}
int lnamel = lname.length();
int strl = str.length();
int is = 0;
for (int i = 1; i < strl;) {
i++;
is++;
if (lname[i] == lname[is] && lname[i] == ' ' || lname[0] == ' ') {
cin.clear();
cout << "Please input your lastname properly!\n";
goto qstn;
}
}
// next question here
I'm having a hard time on thinking what will be the proper logic to avoid the
goto statement, since I was college I was using it but someone here said that
it's not good to use it at all cause it might ruin my code.
I tried using the do while loop but it's not smooth as goto.
Please help!
Here's an idiom I like to use:
int i;
if (std::cin >> prompt("enter an integer: ", i))
{
std::cout << "Read user input: " << i << "\n";
} else {
std::cout << "Input failed (too many attempts). Eof? " << std::boolalpha << std::cin.eof() << "\n";
}
Here, prompt is a smart input manipulator, that takes care of handling parse errors or stream failures and retrying.
It's quite generic so actually do many things, but you don't need to indicate all the options. When the manipulator is inserted into the stream it relays to the do_manip member:
template <typename Char, typename CharT>
friend std::basic_istream<Char, CharT>& operator>>(std::basic_istream<Char, CharT>& is, checked_input<T, Prompter>& manip) {
return manip.do_manip(is);
}
The do_manip handles all the logic without any gotos :) :
std::istream& do_manip(std::istream& is) {
auto attempt = [this] { return infinite() || retries_ > 0; };
while (attempt()) {
if (!infinite())
retries_ -= 1;
prompter_(out_);
if (is >> value_) {
if (!run_validators(out_))
is.setstate(is.rdstate() | std::ios::failbit);
else
break;
} else {
out_.get() << format_error_ << "\n";
}
if (attempt()) {
is.clear();
if (flush_on_error_)
is.ignore(1024, '\n');
}
}
return is;
}
You can see that there is a possibility to have validations run before accepting the input.
Here's a somewhat full-blown demo:
Live On Coliru
int main() {
using namespace inputmagic;
int i;
if (std::cin >> prompt("enter an integer: ", i)
.retries(3)
.flush_on_error(false)
.format_error("I couldn't read that (Numbers look like 123)")
.output(std::cerr)
.validate([](int v) { return v > 3 && v < 88; }, "value not in range (3,88)")
.validate([](int v) { return 0 == v % 2; })
.validate([](int v) { return v != 42; }, "The Answer Is Forbidden")
.multiple_diagnostics())
{
std::cout << "Read user input: " << i << "\n";
} else {
std::cout << "Input failed (too many attempts). Eof? " << std::boolalpha << std::cin.eof() << "\n";
}
}
You can see it will only accept valid integers
that are >3 and <88,
that are even
except 42 (forbidden number)
When entering the numbers 21, 42 and 10 on subsequent retries, you get: live
enter an integer: 21
Value not valid
enter an integer: 42
The Answer Is Forbidden
enter an integer: 10
Read user input: 10
However, if you enter 1 all the time you get this: live
enter an integer: 1
value not in range (3,88)
Value not valid
enter an integer: 1
value not in range (3,88)
Value not valid
enter an integer: 1
value not in range (3,88)
Value not valid
Input failed (too many attempts). Eof? false
Or if you read from a single line file: live
enter an integer: value not in range (3,88)
Value not valid
enter an integer: I couldn't read that (Numbers look like 123)
enter an integer: I couldn't read that (Numbers look like 123)
Input failed (too many attempts). Eof? true
Use a function:
bool getLastName(string & lname,
string & str)
{
cout << "Input customer's lastname: ";
getline(cin, lname);
if (lname.find_first_not_of("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ ")
!= string::npos)
{
cout << "You can only input alpha here!\n";
cin.clear();
return false;
}
else if (lname.empty())
{
cout << "Please enter your firstname!\n";
cin.clear();
return false;
}
int lnamel = lname.length();
int strl = str.length();
int is = 0;
for (int i = 1; i < strl;)
{
i++;
is++;
if (lname[i] == lname[is] && lname[i] == ' ' || lname[0] == ' ')
{
cin.clear();
cout << "Please input your lastname properly!\n";
return false;
}
}
return true;
}
All I've done here is replace the gotos with return false. If the program makes it to the end of the function, return true. Make the function call in a while loop:
while (!getLastName(lname, str))
{
// do nothing
}
Not only does this de-spaghettify the code, but it breaks it up into nice, small, easy to manage pieces. This is called procedural programming.
A While loop looks like your best bet. You can redo the loop with the continue keyword.
int incorrect = 0;
while(!incorrect) {
cout<<"Input customer's lastname: ";
getline(cin,lname);
if(lname.find_first_not_of("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ ")!=string::npos)
{
cout<<"You can only input alpha here!\n";
cin.clear();
continue;
}
else if(lname.empty())
{
cout<<"Please enter your firstname!\n";
cin.clear();
continue;
}
int lnamel=lname.length();
int strl=str.length();
int is=0;
for(int i=1; i<strl;)
{
i++;
is++;
if(lname[i]==lname[is]&&lname[i]==' '||lname[0]==' ')
{
cin.clear();
cout<<"Please input your lastname properly!\n";
continue;
}
incorrect = 1;
}
You need to use some do while loops
for instance this:
qstn:
cout<<"Input customer's lastname: ";
getline(cin,lname);
if(lname.find_first_not_of("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ ")!=string::npos)
{
cout<<"You can only input alpha here!\n";
cin.clear();
goto qstn;
}
else if(lname.empty())
{
cout<<"Please enter your firstname!\n";
cin.clear();
goto qstn;
}
could be re-written as this:
int flag;
do{
flag = 1;
cout<<"Input customer's lastname: ";
getline(cin,lname);
if(lname.find_first_not_of( "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ ")!=string::npos)
{
flag = 0;
cout<<"You can only input alpha here!\n";
}
else if(lname.empty())
{
flag = 0;
cout<<"Please enter your firstname!\n";
}
cin.clear();
} while( flag !=1 );
feel free to use a boolean type flag, it doesn't really matter
It seems to me that your code suffers from lack of clarity of purpose.
You apparently don't want the string that's entered to include a leading space, nor multiple consecutive spaces. Other than that, only alphabetic characters should be accepted.
If the user does enter multiple consecutive spaces, I'd probably just ignore all but the first. I'd probably write the code something like this:
#include <string>
#include <iostream>
#include <algorithm>
#include <cctype>
#include <sstream>
bool non_alpha(std::string const &s) {
return !std::all_of(s.begin(), s.end(), [](unsigned char c) { return std::isalpha(c) || std::isspace(c); });
}
std::string get_name(std::string const &prompt) {
std::string result;
std::string line;
do {
std::cout << prompt;
std::getline(std::cin, line);
} while (non_alpha(line));
std::istringstream words(line);
std::string word;
while (words >> word)
result += word + ' ';
return result;
}
int main() {
auto res = get_name("Please enter last name, alpha-only\n");
if (res.empty())
std::cout << "Oh well, maybe some other time";
else
std::cout << "Thanks Mr(s). " << res << "\n";
}
I'd be tempted to consider doing roughly the same for non-alphabetic characters--rather than asking the user to re-enter from the beginning, assume it's a mistake and just ignore it.
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Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 8 years ago.
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Evening All,
My following code:
#include <iostream>
#include <vector>
#include <fstream>
using namespace std;
const char NEWLINE = 0x0D + 0x0A;
int main() {
int headerCount;
char response;
int x = 0;
int i = 0;
string header[headerCount];
cout << "How many columns appear in you file: ";
cin >> headerCount;
cout << "There are " << headerCount << " columns in your file? (Y/N): ";
cin >> response;
if(response == 'Y'){
cout << endl << "Okay" << endl;
ofstream myfile;
myfile.open("example.csv", ios::binary | ios::out);
string vals;
vector< vector<string> > buff;
// To set values
while(i != 3){
vector<string> temp;
while(x != headerCount){
if (i == 0){
cout << "What is header number " << (x + 1) << endl;
} else {
cout << "What is person " << i << "'s " << header[x] << ":";
}
cin >> vals;
if (i == 0) header[x] = vals;
temp.push_back("\"" + vals + "\"");
if (x != (headerCount - 1)) { temp.push_back(","); }
x++;
}
buff.push_back(temp); // Store the array in the buffer
x = 0;
i++;
}
//To access values
for(vector<vector<string> >::iterator it = buff.begin(); it != buff.end(); ++it){
//it is now a pointer to a vector<int>
for(vector<string>::iterator jt = it->begin(); jt != it->end(); ++jt){
// jt is now a pointer to an integer.
myfile << *jt;
}
myfile << "\r\n";
}
myfile.close();
} else {
cout << endl << "oh";
}
return 0;
}
Crashes on its fourth iteration.
E.g. value[0][4]
Any ideas? I am reasonably new to vectors. This code causes Code::Blocks to display my application as "not responding". Really Lost at the moment!
I know there are probably plenty of similar posts but I have been trawling the web for hours.
First, you must thank #40two who quickly spotted the main issues.
So add the missing include. Then, define your variable header as a vector to allow for dynamic sizing. However, do this AFTER your cin reading, in order to make sure the variable is initialised and not random:
cin >> headerCount; // existing line
vector<string> header(headerCount); // definition in a new place
Then it should run without crashing (tested with VC2013)
As #40two mentioned, the real issue is that you cannot variable sized arrays in C++. There is huge scope to improve the code significantly. To get you going quickly, see the fixed up code below.
#include <iostream>
#include <vector>
#include <fstream>
using namespace std;
const char NEWLINE = 0x0D + 0x0A;
int main() {
int headerCount;
char response;
int x = 0;
int i = 0;
cout << "How many columns appear in you file: ";
cin >> headerCount;
cout << "There are " << headerCount << " columns in your file? (Y/N): ";
cin >> response;
if(response == 'Y'){
cout << endl << "Okay" << endl;
ofstream myfile;
myfile.open("example.csv", ios::binary | ios::out);
string vals;
vector< vector<string> > buff;
// To set values
while(i != 3){
vector<string> temp;
while(x != headerCount)
{
string header;
if (i == 0){
cout << "What is header number " << (x + 1) << endl;
} else {
cout << "What is person " << i << "'s " << header << ":";
}
cin >> vals;
if (i == 0)
header = vals;
temp.push_back("\"" + vals + "\"");
if (x != (headerCount - 1)) { temp.push_back(","); }
x++;
}
buff.push_back(temp); // Store the array in the buffer
x = 0;
i++;
}
//To access values
for(vector<vector<string> >::iterator it = buff.begin(); it != buff.end(); ++it){
//it is now a pointer to a vector<int>
for(vector<string>::iterator jt = it->begin(); jt != it->end(); ++jt){
// jt is now a pointer to an integer.
myfile << *jt;
}
myfile << "\r\n";
}
myfile.close();
} else {
cout << endl << "oh";
}
return 0;
}
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
}
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.