Question's all in the title.
I'm making a calculator and I obviously need input. I use the cin>> function but I was wondering if there is a way to test the input to find out if it's a number.
If I enter anything that isn't a number the program crashes. Is there a built in function/operator? Please help!
The input operator will only read to an integer if the input is a number. Otherwise it will leave the characters in the input buffer.
Try something like this
int i;
if (cin >> i)
{
// input was a number
}
else
{
// input failed
}
atoi and sscanf are your friends, or just compare if the entered charcode is within the "0"-"9" range
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
first of all I'm sorry for my
bad english.
I'm trying to read some numbers and write them into a vector in C++.
This should go as long as the input is a double number and the
loop should be stopped if the user writes an 'a'.
My Question is how can I check if the input is 'a'.
Breaking the loop is not the problem
while(true){
if(!(cin>>userInput)){
//here i want to know if the input is 'a' or some other stuff//
//also i want to do some other stuff like printing everything//
//what already is in the vector//
//when everything is done; break//
}
else
//the input is a valid number and i push it into my vector//
'userInput' is defined as double so the loop will stop.
My Problem is, if the user write 'q' the loop stops but it's instantly stoping the whole program. My try look like this:
while(true){ //read as long as you can
cout<<"Input a number. With 'q' you can stop: "<<endl;
if(!(cin>>userInput)){ //here the progam stops when the input is anything but a number
cout<<"How many numbers do you want to add up?"<<endl; //there are numbers in a vector that should be added up
cin>>numberOfAdditions;
break;
}
So I have a vector with some numbers the users writes down (20,50,90,...)
When the input is equal to 'q' (in this example everything but numbers )
the loop stops and I want to ask the user how many numbers should be added.
The cout-command is displayed but the input is beeing skipped.
So my program is not reading how many valued from the vector I want to add.
I hope you know what I mean and I don't want to use two questions and two variables to save the input but if it's not working without it I'll change my program.
Have a nice Day :)
Because your Input variable is of type double you have to flush the Input from cin before reading again. Otherwise there is still a newline in the buffer.
Consider the following example:
#include <iostream>
using namespace std;
int main(){
double userInput;
int numberOfAdditions;
while(true){ //read as long as you can
cout<<"Input a number. With 'q' you can stop: "<<endl;
if(!(cin>>userInput)){ //here the progam stops when the input is anything but a number
cout<<"How many numbers do you want to add up?"<<endl; //there are numbers in a vector that should be added up
cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');
cin.clear();
cin.ignore(INT_MAX,'\n');
cin >> numberOfAdditions;
break;
}
}
return 0;
}
The two Statements:
cin.clear();
cin.ignore(INT_MAX,'\n');
are flushing the Input stream until the newline is encountered.
The first answer already explains how to flush the cin stream after the user types in a char.
If you want to determine which character it was, you should define userInput as std::string. If the string is not "q" or "a" or whatever you are looking for, you have to cast the string to a double, just like this:
std::string str;
cin >> str;
if (str == "j")
// User typed in a special character
// ...some code...
else
double d = atof(str.c_str()); // Cast user input to double
Notice that the result of the cast is zero, if the user typed in any other string than the ones you especially look for.
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'm an absolute beginner to programming and i'm just doing some exercises exercises for the beginning.
First of all, i'm using Visual C++ 2010 to compile C-Code. I just create a new project and choose an empty console application. After that, I create a ressource file named test.c and change in the file properties the elementype to C/C++ Compiler and compile as C++ Code, so that i can use #include <iostream> for the std::cin.get() command. Now the code:
#include <stdio.h>
#include <iostream>
int main()
{
int number1, number2;
int sum;
puts("Enter number 1 please:");
scanf_s("%d",&number1);
puts("Enter number 2 please:");
scanf_s("%d",&number2);
std::cin.get();
std::cin.get(); //(1)
sum = number1 + number2;
printf("The average is %f\n", sum/2);
return 0;
}
Now my problem ist that the "std::cin.get()" command is just ignored. Afer typing in the two numbers the program just stops and the console window closes.
Any idea where the problem is?
I have another question please.
Since my problem with holding the console open is solved (1), now my printf() gives me just zeros as output. I want to have a float number as output but no matter what i type in as number1 and number2 i always get "0.000000".
Since i'm still working on my little program to verify the input before it is accepted, i have another question please.
I want to use the following code just to check the input.
#include <stdio.h>
#include <iostream>
#include <ctype.h>
int main()
{
int number1, number2;
int sum;
puts("Enter number 1 please:");
scanf_s("%d",&number1);
if (isdigit(number1))
{
puts("Enter number 2 please:");
scanf_s("%d",&number2);
}
else
{
puts("Your input is not correct. Enter a number please.");
}
std::cin.get();
std::cin.get();
/*
sum = number1 + number2;
printf("The average is %f\n", sum/2); */
return 0;
}
Well it doensn't work. I type in a digit and my response is "Your input is not...". I have used the search and found the following: Check if User Inputs a Letter or Number in C. Unfortunately the suggestions doesn't help me.
It's not ignored. When you type your second number, then hit enter, it puts your number plus a newline character in the input stream. scanf removes the number, but leaves the newline character alone. When you call cin.get(), since there's a character in the stream, it doesn't wait for your input.
PigBen has already given you a good explanation of where you err. However, I have some additonla points to make about your program which won't fit into a comment:
You are mixing C and C++ input. Why? What's wrong with std::cin >> number1?
When you change number1 to double, you need to remember to change the formatting string in scanf(), too, while with IO streams the compiler will figure out everything for you. With streams and C++' strings, containers and other data structures, it's much harder to do something that compiles, but invokes the dreaded Undefined Behavior at run-time.
Also note that you do not check whether your inputting operations succeed. What happens if I invoke your program, and instead of passing it numbers, I enter non-digits? Never use input from users, files, or other externals sources unverified.
With IO streams, the input operator >> returns (a reference to) the stream, and you can use streams as if they were booleans, so you can do
if(std::cin >> number1)
// input succeeded
or
if( !(std::cin >> number2) ) // note the negation operator !
// input error
to check.
Streams enter a bad state internally after input/output errors. Any further IO operations will fail on a stream that had encountered an error. Therefore, if you want, you can delay input verification until all input operations are done:
std::cout << "Enter number 1 please:";
std::cin >> number1;
std::cout << "Enter number 2 please:";
std::cin >> number2;
if(!std::cin)
// input error
However, remember to always verify input before you first use it.
Note that I didn't check the output for errors. That's because it's hard to imagine something going wrong with output to the console. (And what would you do about it? Print an error message?) However, if you write into a file, remember to check output, too. It's easy for a file operation to go wrong.
In answer to your modified question it is because you are using ints for division. Change int sum to float sum and everything should be fine.
To answer your modified question: you're using the %f printf() format with an int, and that doesn't work. If you want to print out floating-point, you need to pass a double. You could print out (double)sum / 2 or even sum / 2.0, both of which yield doubles. (No, a float doesn't work the same for a variadic function like printf().) As it is, you're passing what is probably a four-byte type and telling printf() to treat it as an eight-byte type of different format, so it's no wonder you're not getting the expected results.
Alternately, you could switch to C++ iostreams, which save you the problem of matching types and knowing the default promotions. You'd still get an integer from sum/2, and that would drop any one-half, but it would be the right result.
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){}