I am new to C++ and am in a class. I am trying to finish the first project and so far I have everything working correctly, however, I need the user to input a number to select their level, and would like to validate that it is a number, and that the number isn't too large.
while(levelChoose > 10 || isalpha(levelChoose))
{
cout << "That is not a valid level" << endl;
cout << "Choose another level:";
cin >> levelChoose;
}
That is the loop I made, and it sometimes works. If I type in 11 it prints the error, and lets me choose another level. However if the number is large, or is any alpha character it floods the screen with the couts, and the loop won't end, and I have to force exit. Why does it sometimes stop at the cin and wait for user input, and sometimes not? Thanks for the help!
This is an annoying problem with cin (and istreams in general). cin is type safe so if you give it the wrong type it will fail. As you said a really large number or non-number input it gets stuck in an infinite loop. This is because those are incompatible with whatever type levelChoose may be. cin fails but the buffer is still filled with what you typed so cin keeps trying to read it. You end up in an infinite loop.
To fix this, you need to clear the fail bit and ignore all the characters in the buffer. The code below should do this (although I haven't tested it):
while(levelChoose > 10 || isalpha(levelChoose))
{
cout << "That is not a valid level" << endl;
cout << "Choose another level:";
if(!(cin >> levelChoose))
{
cin.clear();
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
}
Edit: numeric_limits<> is located in the limits include:
#include<limits>
From your description, it seems likely (nearly certain) that levelChose is some sort of numeric type, probably an integer.
When you use operator>> to read a number, anything that couldn't be part of a number (e.g., most letters) will be left in the input buffer. What's happening is that you're trying to read the number, it's failing and leaving the non-digit in the buffer, printing out an error message, then trying to read exactly the same non-digit from the buffer again.
Generally, when an input like this fails, you want to do something like ignoring everything in the input buffer up to the next new-line.
levelChoose appears to be an integer type of some form (int, long, whatever).
It's not valid to input a character into an integer directly like that. The input fails, but leaves the character in the incoming buffer, so it's still there when the loop comes around again.
Here's a related question: Good input validation loop using cin - C++
I suspect the part while(levelChoose > 10..... This does not restrict level to less than 10 (assuming greater than 10 is a large number in your context). Instead it probably should be while(levelChoose < 10...
To check that an expression is not too large, the following could be a possibility to validate (brain compiled code!!)
const unsigned int MAX = 1000;
unsigned int x;
cin >> x;
while(x < MAX){}
Related
I've been working on a calculator and I am very close to getting it working, but I need to find a way to have the amount of numbers and operators that the user puts in to be up to the user. This is a simpler test version of what I have so far that I need to apply this to. I left out a few things but this is the exact part of the code I need to apply it to. The user should be able to input any amount of numbers and operands as long as it is under 20 of each. I don't want to stick to a rigid form of input, like cin >> numIn[n]; cin >> operator1[n], because then you would HAVE to end with an operator, for example.
#include <iostream>
float numIn[20];
char operator1[20];
int main(){
int n = 0;
while (n < 20){
std::cin >> numIn[n] >> operator1[n];
switch (operator1[n]){
case '+':
std::cout << numIn[n] + numIn[n];
break;
default:
std::cout << "does not work";
}
}
}
I edited this question to be more focused and clear. I added the part of the text above about how I need to have a 'fluid' way of input. If this doesn't make it more clear I don't know what does.
Use getline to get a whole line of input from the user.
Then use that string to construct a stringstream, which you can use your >> on to extract the numbers and single-char operators. But now you get an error when you try to read past the end of the original string. It knows you are done, because it has the complete input committed before it started to read, unlike an always-open terminal where the user could always type more.
use std::cin >> numIn[n] and std::cin >> operator1 [n]
as #Ivan suggested in the replies
cout << "Would you like to make another transaction? (y/n)" << endl;
cin >> repeat_transaction;
static_cast<char>(repeat_transaction);
while (repeat_transaction != 'y' && repeat_transaction != 'n')
{
cout << "Invalid selection: Please enter y or n";
cin >> repeat_transaction;
static_cast<char>(repeat_transaction);
}
During the Invalid selection loop, I once accidentally pressed "mn". I noticed the console read out Invalid selection..., So, it did in fact finish and re-enter the while loop. However, after this the console terminated the program. If you enter a single character 'a' or 'y' or 'n' it acts just as it should. Ending or not ending. This was before I attempted to use static_cast to force the truncation of the user input.
Since you managed to get this program to compile I can only assume that repeat_transaction was specified as a char and not a std::string.
When you use cin to get a character it only gets one character but it doesn't flush the buffer. I believe you understand this issue since you wrote This was before I attempted to use static_cast to force the truncation of the user input. . You can attempt to use cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n'); instead of static_cast<char>(repeat_transaction); after each call to cin >> repeat_transaction; . There are downsides to this. If you enter 'mn' it will work as expected. It reads the m which is not y or n and then flushes the extra characters until it finds end of line \n. If you do nm, n will match and the m will be thrown away. So in that case it will accept nm as valid and exit the loop.
There are other ways that may be easier and give you the effect closer to what you are looking for. Instead of reading a character at a time you can read an entire line into a string using getline (See the C++ documentation for more information). You can then check if the length of the string is not equal to 1 character. If it's not length 1 then it is invalid input. If it is 1 then you want to check for y and n. Although basic (and not overly complex) this code would do a reasonable job:
#include <iostream>
#include <string>
using namespace std;
int main()
{
string repeat_transaction;
cout << "Would you like to make another transaction? (y/n)" << endl;
getline(cin, repeat_transaction);
while (repeat_transaction.length() != 1 || (repeat_transaction != "y" && repeat_transaction != "n"))
{
cout << "Invalid selection: Please enter y or n";
getline(cin, repeat_transaction);
}
return 0;
}
I said reasonable job since one deficiency you might see is that you want to trim white spaces from the beginning and end. If someone enters n or y with a space or tab in front it will be seen as invalid (whitespace at the end would be similar). This may not be an issue for you, but I thought I would mention it.
On a final note, you may have noticed I used using namespace std;. I did so to match what was in the original question. However, this is normally considered bad practice and should be avoided. These StackOverflow answers try to explain the issues. It is better to not do it and prepend all standard library references with std::. For example string would be std::string, cin would be std::cin etc.
I have a simple little script I am coding, and I am trying to not allow people to enter a string, or if they do make it revert to the beginning of the function again. Here's the input code I have:
int main()
{
cout << "Input your first number" << endl;
cin >> a;
cout << "Input your second number" << endl;
cin >> b;
}
The rest of the code beyond this part works just fine for what's going on, although if a string is entered here it obviously doesn't work.
Any help would be appreciated.
You may find this post useful,
How to check if input is numeric in C++
Basically you can check the input, whether it is numeric value or not. After checking whether the given input is numbers, then you can add a while loop in main to ask user to repeat if input is not a valid number.
Every input is a string. If you want to know if a entered string can convert to a number, you have to read in a string and try to convert it yourself (eg with strol).
An alternative would be to check if the reading from cin failed, but personally i don't like it because cin.fail() covers more error situations than just a failed type conversion.
There's a library function may help,you can check it after input:
int isdigit(char c);
Tips:
1.You should include such files :
# include <ctype.h>
2.If c in 0 ~ 9 ,return 1 ; else return 0.
I have this code snippet that is supposed to test whether the user enters an integer or not. This works if the user enters letters, but not decimals and I'm left wondering why that is. Here's my code snippet:
Student student;
int id;
while(!(cin >> id))
{
cout << "\nERROR: Please enter a Positive Whole Number" << endl;
cin.clear();
cin.ignore ();
cout << "Enter Student ID: ";
}
Entering A will make it iterate through the while loop, but if I enter 12.5 it drops out of the while loop and keeps going. Isn't it testing whether it will parse to integer or not? Why is it accepting 12.5 but not characters?
cin>>id will succeed as long as it finds something it can convert to an int ("12", in this case). When it reaches something it can't convert, it stops, but if it's read an int already, that counts as success.
To check that everything it read was digits, you might want to do something like using std::getline to read a line of input into a string, then use std::isdigit to test whether those are all digits. Testing a conversion to int (by itself) will only tell you that it found something that could be read as an integer, but won't tell you if that was followed by other things that couldn't be converted to an int.
I rewrote this loop in several ways, with nested Ifs and do whiles, yet behavior is the same. It behaves as expected as long as the user does not enter a character or a string. Once the user does it just goes on spinning the loop until I CTRL+C it.
From what I have researched, when a variable is a number and the user inputs a char or a string, they just get converted into their ASCII numbers, in which case the while check should work. The number should be larger than allowed and the user should be prompted for a new value right? Why does it keep looping infinitely?
Width is declared as a float.
void setWidth ()
{
std::cout << "\nPlease enter the width (use numbers greater than 0 and no greater than 20.0).\n";
std::cin >> width;
while (width <= 0 || width > 20)
{
std::cin.clear();
std::cin.ignore();
std::cout << "You have entered a number outside of the allowed range.\nPlease enter a number greater than 0 and no greater than 20.\n";
std::cin >> width;
}
}
Like I said, for numbers it works great, doubles, negatives, whatever. But something like "asdf" or "a" will put it in infinitely spinning loop.
It seems like I've tried everything. Why does this happen? I mean I know why it loops, it's because the number is not between 0 and 20, but why does it not ask user for input? I do clear the buffer.
The line std::cin >> width; fails because the input isn't a number. It also doesn't consume any of the input, so you are stuck in an infinite loop.
To avoid this, you should read the input using std::getline(), then try to convert it (std::ostringstream is one option), handling and reporting failures accordingly.
The default for cin.ignore() is to ignore just a single character.
If you want to ignore longer strings, you have to add extra parameters for that, perhaps cin.ignore(1000, '\n') which skips up to 1000 characters or the next newline (whichever comes first).
cin::clear() "Sets a new value for the error control state" 1, but the remaining input is still here and still read.
Then I guess the actual behavior depends on the compiler since when I compile it with g++ 4.6.3 and type the input "abc", it only loops three times and then wait for another input.
To empty the cin buffer you may rather see How do I flush the cin buffer?
Try checking the failbit on cin
Ok, thanks for all the help guys... I finally managed to get it to work with cin (not getline), by doing exactly what ive been doing, except I made a clearBuffer() function. So instead of clearing the buffer from within the getWidth function, the getWidth function calls another function.. thereby leaves the getWidth function to execute some code... then comes back to run the rest of it...
For some reason when it goes outside of the function it works fine and strings and chars trigger the error.. but if the cin.clear and cin.ignore are kept within the function then I have that problem.
So the final code looks like this.
void clearBuffer()
{
std::cin.clear();
std::cin.ignore(80, '\n'); //Ignore the first 80 characters up to an Enter character.
}
void setWidth ()
{
std::cout << "\n\t\tPlease enter the width.\n(use numbers greater than 0 and no greater than 20.0).\n";
float temp = NULL; //Using temp here so that we dont write invalid characters to an actual variable.
std::cin >> temp;
clearBuffer();
while (temp <= 0 || temp > 20)
{
std::cout << "\nERROR: You have entered width outside of the allowed range.\nPlease enter a number greater than 0 and no greater than 20.\n";
std::cin >> temp;
clearBuffer();
}
if(temp > 0 && temp <= 20)
width=temp;
}