C++: Why Does It Output On Alternative Basis? - c++

Consider the following C++ code: (codeblocks = IDE)
#include <iostream>
#include <stdio.h>
using namespace std;
int main() {
int a;
while(cin>>a) {
cout<<"You sir, Just Entered: "<<a<<endl;
if(!(cin>>a)) {
cout<<"Exiting"<<endl;
break;
}
}
}
If I remove the if(!(cin>>a)) { condition and just write cout<<"You sir, Just Entered: "<<a<<endl;, it outputs. Otherwise, currently, it doesn't output on (n+1)th input. See this image:

if(!(cin>>a)) { is catching your next input (e.g. 2, 4, 6 ,8).
Then, it doesn't go inside the if statement because of the negation !
It doesn't display your second input because you do not print a after your second cin

You're catching input twice per iteration : once in the while's conditions, and a second time in the if's condition. Thus, only one in two inputs get printed. That's because even if the if's condition fails, it has already read by then.
Your while loop already does the check : it won't iterate if the user enters a non-number. Your fixed code is simply :
int main() {
int a;
while(cin>>a) {
cout<<"You sir, Just Entered: "<<a<<endl;
}
cout<<"Exiting"<<endl;
}

if you want to check for non-integer input this will do.
int a;
while(cin >> a)
{
cout << "Sir you entered: " << a <<endl;
}
cout << "Exiting"
<<endl;
the checking is done automatically in the while loop because the input yields true if it is valid (integer in your case) this allows further inputs. if the input isn't valid, the program leaves the loop and the exiting message is displayed.

Related

Validating Integer input using booleans

In my c++ code, I would like to validate my user input to be an int between 1,10 using a do while loop. I am able to validated for integers outside of the range. However if user inputs a float or a letter, it becomes an infinite loop. My idea is to add a condition in my while loop for if the input is not an integer to keep asking for input.
the CAPITAL letters is where I am having trouble.
#include <iostream>
using namespace std;
int main(){
cout << "Welcome, ";
int steps;
int count=0;
do{
cout << "How many? \n";
cin >> steps;
IF (STEPS IS NOT INTEGER==TRUE){
COUNT=1;
}
if (steps <1)
{
cout << "not enough...\n";
}
if (steps > 10){
cout << "too many steps.\n Please pick a lower number of steps.\n\n";
}
} while (steps < 1|| steps >10 || COUNT==1);
//doing stuff with valid input
return 0;
}
Essentially I am trying to add another condition that just returns a boolean. and if the boolean implies that the input is not valid, then it reassigns count to make sure the do while loops continues until the input is valid.
The problem i am working on asks for a max and min steps, since all of them were having a similar problem i tried to simplify it and forgot some of the edits.
You can check whether the input failed, i.e. the user entered something that could not be read as an int like this:
if (cin.fail()) { // in place of IF (STEPS IS NOT INTEGER==TRUE)
cin.clear();
cin.ignore();
cout << "not an integer, try again\n";
continue;
}
This avoids the need for the COUNT variable.
Also, your while condition doesn't appear to match the checks inside the loop. What happens when step is either 9 or 10? You should be consistent with the checks inside the loop.
You could use the ! operator.
For example:
if ( !(std::cin >> steps) )
{
std::cin.clear();
std::cin.ignore();
std::cout << "Incorrect entry. Try again: ";
}
Also consider not using using namespace std;.

Unable to detect enter key in C++

Now, before this question gets marked for duplicate. I have already gone through most of the questions and their relative answers of C++. These are the links that I have tried and none of them work for me. It maybe because they are using an older version of C++, and I have the latest version of C++. Here are the links that I have tried:
Detecting ENTER key in C++
https://www.sololearn.com/Discuss/1863352/how-can-i-check-that-user-press-enter-key-in-c
http://www.cplusplus.com/forum/beginner/2624/
https://www.dreamincode.net/forums/topic/398680-detect-enter-key/
Now, with the duplicates out of the way. I am making an expression calculator. So, for example if the user input is: 2+2*6*9/9, then the output should be 14.
The code where I suspect that the problem lies is in:
#include <iostream>
#include <vector>
using std::cout;
using std::cin;
using std::string;
using std::vector;
void clear();
void error(string message);
int main() {
cout << "Enter an expression: ";
double l_Value = 0, r_Value = 0, result = 0, count = 0, previous_number;
char op;
while (cin >> l_Value) { // 1+2*3+6-4/2+3
if (!cin) {
error("Invalid operand entered!");
}
else {
bool is_Error = 0; // false
vector<double> numbers;
numbers.push_back(l_Value);
previous_number = l_Value;
while (cin >> op) {
if (op == '\0') {
break;
}
cin >> r_Value;
switch (op)
{
case '+':
numbers.push_back(r_Value);
previous_number = r_Value;
break;
case '-':
numbers.push_back((-1 * r_Value));
previous_number = (-1 * r_Value);
break;
case '*':
numbers.pop_back(); // take out the number
r_Value *= previous_number;
numbers.push_back(r_Value);
previous_number = r_Value;
break;
case '/':
if (r_Value == 0) {
error("Sorry, division by zero has occured. Please re-evaluate your expression!\n");
is_Error = 1; // true
break;
}
else {
numbers.pop_back(); // take out the number
previous_number /= r_Value;
numbers.push_back(previous_number);
break;
}
}
}
if (!is_Error) {
for (int i = 0; i < numbers.size(); i++) {
result += numbers[i];
}
cout << result << '\n';
}
numbers.clear();
result = 0;
l_Value = 0;
r_Value = 0;
}
cout << "Enter an expression: ";
}
clear();
return 0;
}
None of the links above seemed to work for me.
When I press the Enter key, it expects me to give another input, and that is not supposed to happen. So when I used cin.get() == 'n' or cin.get() == (int)'\n', it expects for another input. But, when I have an 'x' at the end of the expression, it works perfectly fine. So, I need the "cin" operator to help me detect an Enter character at the end of the expression and then terminate the program.
Here, is a sample run of a program with 'x':
[![running as x-terminator][1]][1]
[1]: https://i.stack.imgur.com/ORPQa.png
When I try the above solution such as "cin.get() == '\n':
Then, I thought that maybe it is reading the null character and so, I tried if (op == '\0'):
For the enter key and null character I had to press Ctrl+Z to terminate the program. Please help!
As, mentioned by user #idclev, I already have a string program that works, but I am trying to avoid using string to calculate any expressions! So, if I could detect an enter key pressed using a character datatype that would be great!
I avoided strings to avoid parsing through the text
That argument is moot. What you can read from cin you can also read from a std::string, no difference whatsoever. You just need to add one step:
#include <iostream>
#include <string>
#include <sstream>
int main( ){
std::string x;
std::cin >> x;
if (x == "") {
std::cout << "user pressed enter (and nothing else)";
} else {
double y;
std::stringstream ss{x};
ss >> y;
std::cout << y;
}
}
This will read one std::string. If user only hit enter then the string will be empty. If the user entered something the else branch will be taken and you can extract the number from the string in the same way you did extract the number from cin (via using a std::stringstream).
If you have more than one number in the input you need to use getline to read the string, because cin will (by default) only read till the next whitespace.
Again...
If I used a string, I would have a tough time in extracting single-digit and two-digit or n-number of digits in a string. The double data type does that for me
You can read single-digit or any number of digits from a stringstream in exactly the same way as you read them from cin.
I already made a program with string in it. I was trying to avoid string to see how much faster would it be without string.
It won't be any faster. Constructing the string and the stringstream is maybe in the order of microseconds. A user entering input is in the order of seconds, maybe milliseconds when they are typing very fast.
Your approach cannot work because hitting enter is not considered as a character. Trying to read a character when there is none in the stream will fail. It will not set the character to \n or \r.
On the outer loop, you are trying to read a double, but you keep pressing enter. There is no input to evaluate, so it keeps trying to read a double. You can get out of it by ending the input stream with ^Z, or you can give it any actual content to read, and it will try to make it into a double (which is what your code explicitly told it to wait for).
Basically, when you press enter, it's ignoring it because
http://www.cplusplus.com/reference/istream/istream/operator%3E%3E/
Extracts as many characters as possible from the stream and inserts them into the output sequence controlled by the stream buffer object pointed by sb (if any), until either the input sequence is exhausted or the function fails to insert into the object pointed by sb.
Try experimenting with this to see what is happening.
#include <iostream>
#include <vector>
#include <string>
int main() {
double x;
std::cin >> x;
std::cout << "read this value: " << x << std::endl;
// this is what while or if will look at
bool success = !std::cin.fail();
if (success)
std::cout << "success" << std::endl;
else
std::cout << "failure, loop will exit" << std::endl;
return 0;
}
What you should want (in my opinion) is a function that takes the expression as a string, and returns the result, so you can write unit tests, and make sure the function works. You can use this function with any expression that you can put in a string. It doesn't HAVE to be typed in by a user.
If you want the user to type in the experession, it's a lot easier to just use getline() then pass the string to your function. The big problem with using cin on each variable and character is that the user has no idea which datetype is expected right then. Granted, it's not hard to guess with an expression, but you wrote it and debugged it and still didn't know which cin you were failing to get the right datatype to. (this is normal, btw -- been there, which is why I getline and parse separately)

C++: Why does this code run an endless loop if the input was not integral?

I have this simple code that duplicate a string by using either a reference or a pointer (I'm a beginner), and I would like to make it determine whether the input is integral or not because if it wasn't, it keeps looping for some reason. I've tried using cin.clear() but it doesn't seem to work..
Here is the code:
#include <iostream>
#include <string>
#include "test.h"
using namespace std;
int main() {
while (true) {
cout << "Type something: ";
string hi;
cin >> hi;
cout << "Choose a method to change it:\n1. By reference.\n2. By using pointers.\n";
start:
int i;
cin >> i;
if (cin.fail())
cin.clear();
switch (i) {
case 1:
changestringref(hi);
cout << hi << '\n';
break;
case 2:
changestringptr(&hi);
cout << hi << '\n';
break;
default:
cout << "Choose by typing either 1 or 2: ";
goto start;
break;
}
}
}
Whenever I type a string (at the second input), it keeps looping this sentence: "Choose by typing either 1 or 2: Choose by typing either 1 or 2: Choose by typing either 1 or 2: "etc.
How can I solve this problem? And why isn't cin.clear working? isn't it supposed to clear the input from the last time cin was used or am I wrong? O.o
Thanks in advance :D
And why isn't cin.clear working? isn't it supposed to clear the input from the last time cin was used or am I wrong?
Not quite. It's supposed to Set[s] a new value for the stream's internal error state flags.. clear() without argument sets goodbit and clears the fail() state, but leaves any unprocessed input untouched.
Let's assume you entered 'x'. Since 'x' is not a number, cin >> i sets failbit, and does not consume input (actually, it pushes the already consumed x back into the input buffer, but that's another story).
This is behavior documented somewhat obscurely in num_get::get(). The returned iterator (i.e. the stream) points right before the first character that could not be parsed.
So next time, after the goto start; you read x again. Same input, same result, hence the endless loop.

how to allow multiple inputs in case of an incorrect data type entry in cpp?

I have a program which generates random number and asks user to keep guessing it until he/she gets it right. I want it to keep accepting new values even if i incorrectly enter any other data type by handling the error cases.
My problem is that when i am trying to run the below program, as soon i input a character and hit enter, it goes into an infinite loop. I tried using cin.ignore() and cin.clear() but that just makes the program stop after the first entry.
Can anyone please help me understand what is going on and how to achieve the desired output? Thanks in advance.
#include <iostream>
#include <cstdlib>
#include <time.h>
using namespace std;
int main()
{
int secret_num, guess;
srand(time(NULL));
secret_num=rand() % 101 + 0;
cout<<"Enter your guess between 0 and 100: ";
do
{
if(!(cin>>guess))
{
cout<<" The entered value is not an integer"<<endl;
}
else if( isnumber(guess))
{
if(guess>secret_num)
cout<<"Too high";
else if(guess<secret_num)
cout<<"too low";
cout<<endl;
}
}
while(secret_num!=guess);
if((guess==secret_num)| (isnumber(guess)))
{
cout<<"yes the correct number is "<<secret_num<<endl;
}
return 0;
}
Edit: Here is a screenshot of what the output looks like with cin.clear() and cin.ignore(1000,'\n') in my code, when i enter a number after entering character twice.
if (!(cin >> guess))
{
cout << " The entered value is not an integer" << endl;
cin.clear(); // clear must go before ignore
// Otherwise ignore will fail (because the stream is still in a bad state)
cin.ignore(std::numeric_limits<int>::max(), '\n');
}
By default cin.ignore will ignore a single character. If they type more than 1 char, it won't be enough, that's why I've modified it a bit.
if ((guess == secret_num) | (isnumber(guess)))
| is a bitwise operator [OR]
|| is the logical operator [OR]
But I think what you actually want is && (AND)
if ((guess == secret_num) && (isnumber(guess)))
There're several problems.
You should use cin.clear() and cin.ignore() as #José suggested.
What's isnumber()? I guess it's returning false so no hint message (i.e. "Too high" and "too low") is printed out, looks like it stops although it's just waiting the next input. And isnumber() doesn't make sense to me. guess has been declared as an int, it has to be a number, doesn't it?
if((guess==secret_num)| (isnumber(guess))) is unnecessary here. The loop won't end until the user input the correct number, this condition should have been statisfied.
You can use clear and flush
if(!(cin>>guess))
{
cout<<" The entered value is not an integer"<<endl;
cin.clear();
fflush(stdin);
}
This works if you are reading from console. Otherwise you can go with #José answer.
I would change the logic inside your loop as there are some useless tests. This works for me:
#include <iostream>
#include <limits>
#include <cstdlib> // You may take a look at <random> and <chrono>
#include <time.h>
using std::cout;
using std::cin;
int main() {
srand(time(NULL));
int secret_num = rand() % 101;
cout << secret_num << '\n';
cout << "Enter your guess between 0 and 100:\n";
int guess = -1;
do {
cin >> guess;
if ( cin.eof() )
break;
if ( cin.fail() ) {
cout << "The entered value is not an integer, please retry.\n";
// clear the error flag
cin.clear();
// ignore the rest of the line
cin.ignore(std::numeric_limits<int>::max(),'\n');
// clear the value of the variable
guess = -1;
continue;
}
// now we know that guess is a number
if ( guess > secret_num )
cout << "Too high\n";
else if ( guess < secret_num )
cout << "Too low\n";
else {
cout << "Yes the correct number is " << secret_num << std::endl;
break;
}
} while ( true );
return 0;
}

Making an if condition return if its true or while return if its true

When the condition is true or false, how can I make it return back and ask the question again, making the user re-enter the value?
Here is what I want to implement:
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
int n;
cout<<"Enter numbers. Press 5 to stop: ";
cin>>n;
bool tr=true;
while(tr)
{
if(n!=5)
cout<<"You entered "<<n; //How to make it return again, since its false? I keep getting infinite loops :( ;
else
tr=false;
}
return 0;
}
You need to prompt the user in the while loop, so that it occurs in each iteration:
int n;
bool tr = true;
while(tr)
{
cout << "Enter numbers. Press 5 to stop: ";
cin >> n;
if(n!=5) {
cout << "You entered " << n;
} else {
tr = false;
}
}
Just put all your code (except 'n' and 'tr' definition) in while loop as follow:
int main()
{
int n;
bool tr=true;
while(tr)
{
cout<<"Enter numbers. Press 5 to stop: ";
cin>>n;
if(n!=5)
cout<<"You entered "<<n;
else
tr=false;
}
return 0;
}
The other answers all work, and there is something to be learned about improving program flow from them, but I believe the trick you're asking for is the continue keyword, which skips the remainder of this iteration of the loop.
bool tr = true;
int n;
while (tr)
{
cout << "Enter numbers...";
cin >> n;
if (n != 5)
continue;
else
tr = false;
}
EDIT Part 1: On the continue keyword.
You want to make your code as readable as possible. In this example, its use is unnecessary (as the other posters have shown); but it is the answer to the question "How do I skip the rest of processing in this iteration of my loop and continue to the next iteration?". Usually, such flow-breaking directives actually make code harder to read; but sometimes the opposite is true. Anything (or, at least, almost anything) that can be accomplished with continue or break, can be accomplished without them, so if you're going to use them, you want to have a definite reason for doing so. Usually, when I use continue, it's because I'm looping through a collection of inputs and I want to skip processing the loop whenever the input isn't in the format I'm expecting. Something like this (pseudo-code)...
foreach (Input i in ReceivedInputs)
{
if (i.isBad())
{
cout << "Bad input";
continue;
}
// Do massive block of calculating here.
}
is easier to read than this...
foreach (Input i in ReceivedInputs)
{
if (i.isBad())
cout << "Bad input";
else
{
// Do massive block of calculating here.
}
}
because the second version makes it harder to track what scope you're in, if you're looking toward the end of the massive block of calculating. In this case, I gain code readability by continue, so I use it. But simple code probably shouldn't use it. The break keyword is similar, though it's a lot easier to come up with examples where break is beneficial.
EDIT Part 2: On multiple iterations
This is just an issue of setting up the loop; there are no magic keywords here. The shortest way I can come up with, is probably something like this:
int n = 0;
int numberToTake = 10;
for ( int numbersTaken = 0; numbersTaken < numberToTake; ++numbersTaken)
{
cout << "Enter numbers...";
int n = 0;
for (cin >> n; n != 5; cin >> n)
cout << "Try again.";
// Do whatever processing on n you want to do here.
}
Though I should point out that, doing it this way, the only value you will ever get from the user will be 5, and if he inputs anything that doesn't fit in an integer, you will get unexpected behavior.
EDIT 3: After reading the comment more thoroughly, I think you're just looking for is the more traditional use of the for loop.
No need for the exra bool variable.
The idiom can be: Infinitely loop until the user enters 5:
for(;;) { // Loops infinitely
cout << "Enter numbers. Press 5 to stop: ";
cin >> n;
if(n == 5)
break; // Exits the loop
cout << "You entered " << n; // Before the if if you want to print 5 as well
}