Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 5 years ago.
Improve this question
I'm quite a newbie (started to learn about coding just like 2 weeks ago) and I'd really appreciate some help to explain why my code isn't working.
I wrote a simple code to calculate the probability. That's not really important part. The code requires some user input so in order to make it kinda foolproof, I wrote two if statements for various kind of wrong input (wrong data type, number too high, number too low) to throw an error message and kill the program. And if the input is right (means none of the conditions of the if statements is true), it should call a function diceTwo(des).
However, if I input number 8 - which should be allright - for some reason that I can't figure out - it triggers the number too high/low if statement, so no matter what number I enter, the function never gets called.
cin >> des;
if (cin.fail())
{
cin.clear();
cin.ignore(100, '\n');
cout << "Invalid input! Use only whole positive numbers!" << endl;
cout << "Try again." << endl;
return -1;
}
if (des> 12 || des< 1);
{
cout << "Invalid input!" << endl;
cout << "Remember the only possible values are from 2 to 12!" << endl;
cout << "Try again." << endl;
return -1;
}
diceTwo(des);
cout << "The probability of this roll is " << x << "%" << endl;
This is the part that makes troubles. I already tried to split the problematic if statement into two (one for <1, one for >12), or adding a new if statement for des>1 && des<12 to call the function, none of it worked. I worked with if statements to provide foolproofness for user input few times before, always worked well, so I really can't seem to find what's wrong this time. Anyone could tell me how to fix it please?
You have a semi colon after the if statement. Look CLOSELY:
if (des> 12 || des< 1); //<----- semi colon
{
cout << "Invalid input!" << endl;
cout << "Remember the only possible values are from 2 to 12!" << endl;
cout << "Try again." << endl;
return -1;
}
The semi colon terminates the if statement, so that your program just continues to the next line, which, happens to be your invalid input line. Remove the semi-colon and you're golden :)
Mistakenly added ; in if
if (des> 12 || des< 1);
Related
This question already has answers here:
Uninitialized variable behaviour in C++
(4 answers)
Closed 2 years ago.
I'm a newbie in the programming world, and i've decided to start with C++ code a few days ago as my first programming language.
I just started to read an online course which i'm guiding on (and aply while i'm reading it).
The course in question assigns a serie of small optional exercises, wich go hand in hand with the topic that is being dealt at that moment.
One of this optional exercises is: "Create a program that multiplies two whole numbers in the following way: it will ask the user for a first whole number. If the number that you type is 0, it will write on the screen "The product of 0 by any number is 0". If a number other than zero has been entered, the user will be prompted for a second number and the product of both will be displayed."
How it says, I did my best to coding that program.
The code of what I did is:
#include <iostream>
using namespace std;
int main ()
{
int a;
int b;
int solve;
cout << "Enter a number: ";
cin >> a;
if (a!=0)
{
cout << "Enter another number: ";
cin >> b;
}
if (b!=0)
{
solve = a * b;
cout << "The result of your operation is: " << solve << endl;
}
else cout << "The product of 0 by any number is 0." << endl;
return 0;
}
So, I press F9 to compile, then F10 to run it.
I proceed to testing it.
I put and different number from zero, I put another one. Throws me a multiplication of both. Nice.
I put and different number from zero, I put another one that actually its zero. Throws me the message of "else" order. Nice.
**BUT
I put and number equal to zero, and throws me the message of the "cout" order of "if (b!=0)".**
I didn't really know what I had done wrong, so, I ask for help from a friend who has some more experience than me, and tells me that actually it's nothing wrong. In fact, he proved me sending to me a screen cap of his Dev C++ with my code in it, and how it runs just how it had to be.
Then, I opened an online compiler (https://www.onlinegdb.com/online_c++_compiler#) to get down my doubts, and yes, there runs correctly too.
So, here's my question?
What's the problem? Why that happens?
I have the DevC++ 5.11 TDM-GCC 4.9.2, and I'm using the deafult compiler.
Please, I would like some of help, I feel more comfortable compiling in PC than online, it's more quick.
Thank you anyway for reading until here.enter image description here
The reason why this happens, as #TrebledJ mentioned, is that you've not initialised the variable b. So you can get over this problem just by initialising b to any value of your choice (preferrably 1 or 0 for simplicity). But there's a workaround if you don't want to initialise the values. I would transform your code to something like this:
if (a!=0)
{
cout << "Enter another number: ";
cin >> b;
if (b!=0)
{
solve = a * b;
cout << "The result of your operation is: " << solve << endl;
}
else cout << "The product of 0 by any number is 0." << endl;
}
else cout << "The product of 0 by any number is 0." << endl;
Basically, what I'm doing is, if the value of a after input is 0, I'm not running the part for taking the user input of b.
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 2 years ago.
Improve this question
I am not a professional programmer but after getting some experience in easy languages like python or matlab, I need to make a little program in C++. For this I try to read user input until the user inputs something sensible - however, this loop never terminates due to my control variable (test2) never being reassigned, even though the corresponding code block is entered. Let me explain in detail:
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include "Header.h"
using namespace std;
int test2; //variable to stay 1 until user has entered suitable input
string input2[2]; //storage for input strings
int MinimalTest() {
test2 = 1; //control variable is set one
cout << "Enter a string." << endl;
do {
//we will enter a string at least once, and we exit this loop only when the input is suitable
std::string line; //complicated input part - a simple getline() command led to weird behaviour
std::getline(std::cin, line);
std::stringstream linestream(line);
linestream >> input2[i];
cout << "input: " << input2[i] << " test: " << test2 << "input bool: " << input2[i].empty() << endl; //this is just for troubleshooting
if (input2[i].empty()) {
//if the user entered an empty string, he needs to do it again
cout << "Your string is empty. Please enter a valid string (non-empty)." << endl;
}
else {
//if he entered a valid string, we can continue with the next input
cout << "I am here." << endl; //This is for trouble shooting. This code block is entered and executed since this gets printed.
test2 = 0;// this gets ignored for some reason. Hence, the loop never terminates.
}
}(while (test2 = 1);
}
So the first loop never terminates. Test2 never gets reassigned to 0, even though the else command is executed. This boggles my mind tbh - it is just a simple assignment operator on an int. Possible output looks like this (Note how I still got a second problem: strings with a space inside get cut off. I would appreciate any feedback on that as well, even though I try to trouble shoot one thing at a time and this post is not aimed at this problem):
Output example
Thank you very much for your consideration!
All the best,
A very confused newbie.
Change your while condition to test2 == 1
test2 = 1 is an assignment, setting the value to 1.
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 5 years ago.
Improve this question
Prerequisites: Atom as code editor with gpp-compiler(plugin), that uses g++ to compile and run cpp files in the editor.
Some not real example code to understand the problem:
int main()
{
int number;
while(cin >> number)
{
cout << "Your number is " << number << endl;
}
}
So this program can easily compile by g++ compiler, the problems appears at runtime, when compiled program starts in terminal and... It's just don't work. There is no something else but "Press any key to continue..." There is no even mistakes.
So the compiler cannot support this loop argument? (while(cin>>number))
And yes, gpp-compiler in Atom works fine with other types of scripts.
Sorry, if this question is stupid but i just want to know why this happens. Thank you!
Some edits:
So yeah. I can't competently explain my problem. So my problem is not the while loop argument, i just don't understand why program starts in empty terminal(with message above), while on my phone it also compiles by g++ and the program works perfectly .-.
The (cin >> number) condition always evaluates to true until you send an EOF character to it. On Windows it is Ctrl + Z. The reason you are not seeing anything on standard output is that the program waits for your to enter a value and press Enter. Afterwards it enters the endless loop. Modify your program to include some simple logic:
#include <iostream>
int main() {
char choice = 'y';
int number;
while (std::cin && choice == 'y') {
std::cout << "Enter the number: ";
std::cin >> number;
std::cout << "Your number is " << number << std::endl;
std::cout << "Repeat? y / n: ";
std::cin >> choice;
}
}
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 6 years ago.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
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.
Improve this question
#include <iostream>
using namespace std;
int main()
{
int i,t,x[20], even, odd, prime;
cout << "Enter 20 integer numbers from 0 to 99: "<<endl;
for (i=1;i<=20;i++)
{
cout << "Input " << i <<":";
cin >> x[i];
}
cout << "\nPrime numbers are: " << endl ;
prime=1;
for (i=2; i<=20 ; i++)
{
for(t=2;t<x[i];t++)
{
if(x[i]%t==0)
{
prime=0;
}
}
if(prime==1)
{
cout << x[i] << endl;
}
prime=1;
}
for(i=1; i<=20; i++) // this is where i have problem.
{
if(x[i]% 2 == 0)
{
even++;
}
else
{
odd++;
}
}
cout << "Number of odd numbers: " << odd << "\n";
cout << "Number of even numbers: " << even << "\n";
return 0 ;
}
When i compile it shows even (40) and odd (10) for input of 0 till 19. Where it should show even 10(including the 0) and odd (10). Im not sure where am i doing it wrongly. I hope someone can help me improve the code.
Variables even and odd are never set to a known value, so you are not formally allowed to read from them. Doing so invokes that most infamous Standardese concept: undefined behaviour. So the values of these variables could be right or could be wrong; the variables and all code trying to read them could be optimised entirely out of your program; or anything can happen. You cannot rely on these variables doing anything right. All attempts to read them make your program ill-formed, so now it can do anything, including things you would never have imagined.
You should search for the abundant background info about these concepts, but I like to think I made a fairly decent summary here: https://stackoverflow.com/a/38150162/2757035
Also, as Thomas points out in the comments, you appear not to understand how array indexing works: Indexes are 0-based. So, int i[20] declares 20 elements numbered from 0 to 19. You try to access index 20, which is not part of the array and hence is more undefined behaviour.
Closed. This question needs debugging details. It is not currently accepting answers.
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 7 years ago.
Improve this question
New to coding. The first part in the first link which is the character letter 'a' is correct. But then after that, its downhill from there.
Its showing me whats wrong, but I do not understand it.
You have declared several variables with the same name 'number' in the same scope. That is not allowed and leads to the error messages.
Within a scope (such as a function or loop or something), you can only declare a variable once. Otherwise, it would be ambiguous which one you were talking about.
The error is saying you've already declared a variable called number (as an int), and you cannot declare it again within the same scope.
Make the second variable called something else:
double dNumber = 1.11;
cout << "Please enter a double: " << dNumber << endl;
bool bNumber = 0;
cout << "Please enter a bool: " << bNumber << endl;
Note, it's usually more typical to set bool values to either true or false.
Now, if you really, really wanted to use the variable name number multiple times, you could put each section in curly braces:
{
double number = 1.11;
cout << "Please enter a double: " << number << endl;
}
{
bool number = 0;
cout << "Please enter a bool: " << number << endl;
}
In that case, you would no longer have access to that variable outside the curly braces, thus the reference is no longer ambiguous.