How to ask for Age and set everything else as invalid? - c++

Question

I'm trying to set a basic program that asks for Age, And if the Number entered by the user is less than 99, It will say "Perfect". If the number is MORE than 99, it will say "You Can't be that old, Try again". Additionally, if the user enters something that is not a number (like a letter "m, r" or anything else like "icehfjc") then it will say "That is not a number."
This is my code so far:
#include <iostream>
#include <string>
using namespace std;
int main()
{
int age;
backtoage:
cout << "How old are you?\n";
cin >> age;
if (age < 99)
{
cout << "Perfect!\n";
system("pause");
}
if (age > 99)
{
cout << "You can't be that old, Try again.\n";
system("pause");
system("cls");
goto backtoage;
}
Else
{
cout << "That is not a number, Please Enter a Valid Number\n";
system("pause");
system("cls");
goto backtoage;
}
}
I know "Else" doesn't work because C++ treats letters as integers as well, so
if I write "m" it will take it as a >99 number (because of the integer value of "m") therefore displaying the "you can't be that old" message. but how can I fix this so the program displays "Please enter a number" when a letter is entered? (If anyone could fix the code and write it in a way that works, I'd
be forever grateful).
Any suggestions, tips or hints are very welcome.

Answer 1

so if I write "m" it will take it as a >99 number (because of the integer value of "m")
No, "m" can't be inputted into an int, cin will fail here. So what you should do is to check the status of cin, such as
if (cin >> age) {
// ok
if (age < 99)
{
...
} else
{
...
}
}
else
{
// failed
cout << "That is not a number, Please Enter a Valid Number\n";
system("pause");
system("cls");
cin.clear(); // unset failbit
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n'); // skip bad input
goto backtoage;
}
Check the behavior of std::basic_istream::operator>>
If extraction fails (e.g. if a letter was entered where a digit is expected), value is left unmodified and failbit is set.
BTW: goto is almost obsolete in modern c++ programming. It should be easy to implement the same logic with a loop.
LIVE

Answer 2

You can Try it.It will validate numeric input in C++.Here cin.good() function is return true if the input is valid, if it is not valid it will return fase. cin.ignore() is used to ignore the rest of the
input buffer, which contains the erroneous input and cin.clear() is
used to clear the flag.
#include <iostream>
#include<string>
#include <limits>
using namespace std;
int main() {
backtoage:
int age = 0;
cout << "How old are you?\n";
cin >> age;
if(cin.good()){
if (age < 99){
cout << "Perfect!\n";
system("pause");
}
else if (age > 99){
cout << "You can't be that old, Try again.\n";
system("pause");
system("cls");
goto backtoage;
}
}
else{
cout << "That is not a number, Please Enter a Valid Number\n";
system("pause");
system("cls");
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
goto backtoage;
}
return 0;
}
Input/Output:
How old are you?
k
That is not a number, Please Enter a Valid Number
How old are you?
120
You can't be that old, Try again.
How old are you?
10
Perfect!

Answer 3

first of all , if the users can enter characters like 'm' 'k' or anything else , I suggest using string , then if you want to change to number just substract '0' , so you can work with numbers, secondly programmers , usually do not use goto statement because it can be dangerous and make underfiend behaivor appear into your program.
Sorry for my bad english.
#include<iostream>
#include <string>
using namespace std;
int main()
{
string input;
getline(cin, input);
unsigned int lenght = input.size(),age=0;
for (int i = 0; i < lenght; ++i) {
if (input[i] >= '0' && input[i] <= '9') {
age = age*10+(int)input[i]-'0';
}
}
if (age > 99)
cout << "Nice try , you can`t be that old\n";
else
cout << "perfect!\n";
return 0;
}