This code I wrote is supposed to subtract one from the number inputed, or divide by 2 based on whether it is a multiple of 3 or not. However, every time I try to run the code, It outputs the numbers I want but then doesn't stop running. I am new to coding and not sure how to fix this.
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main() {
int n;
cout << "Enter a positive number: " << endl;
cin >> n;
if (n < 0) {
cout << "Invalid input." << endl;
}
while (n >= 1) {
if (n % 3 == 0) {
n = n-1;
cout << n << endl;
}
else if (n % 3 != 0) {
n = n / 2;
cout << n << endl;
}
}
return 0;
}
This is a screenshot of the output I get. Instead of giving me the opportunity to run the code again it just stays like this:
I may be misunderstanding what you're asking, however, traversing through the code you can identify that nothing is being done to make the code run again. You would need add what you have inside another while loop. This new while loop would be something like while (input != 0) then run everything you have. In your input statement you could say "Please enter a positive number or enter 0 to exit". This is just an example of an approach, but the premise is that you need something to keep this loop running.
Related
I'm trying to make this code work:
#include <iostream>
using namespace std;
int main()
{
int i;
do
{
cout << ("please enter a number between 1 and 10");
cin >> i;
} while(i > 10 && i < 1)
cout << "the square of the number you have entered is " << i*i;
}
Basically, the idea is that a user enters a number between 1 and 10. While the number is not between 1 and 10, it keeps asking the user to enter a number between the values. Then, when the number is between the values, it is squared and returned to the user.
I can't see why this isn't working
Any help is appreciated
You have:
while (i > 10 && i < 1)
You want:
while (i > 10 || i < 1)
while (i > 10 && i < 1)
Your condition is logically faulty; if reinterpreted, it says:
while i is greater than 10 AND i is less than 1
Judging from your code, the || operator should be used:
} while (i > 10 || i < 1);
As others mentioned, your condition is faulty.
a number can't obviously be under 1 AND above 10 at the same time, so the while loop exits immediately after the do statement.
#include <iostream>
using namespace std;
int main()
{
int i;
do
{
cout << ("please enter a number between 1 and 10");
cin >> i;
} while (i < 1 || i > 10)
cout << "the square of the number you have entered is " << i*i;
}
You should use an Or ||, that condition with && will never be true.
The loop condition is wrong and will never loop, as i cannot be less than 1 && greater than 10 at the same time. You should use the logical OR (||) operator instead. In addition, there must be a semicolon placed after the do-while statement. And you probably want and end of line placed after the prompt. Also, you don't want to start the bad habit of polluting the global namespace, even with the awesomeness of std. So:
#include <iostream>
int main()
{
int i;
do {
std::cout << "please enter a number between 1 and 10\n";
std::cin >> i;
} while (i > 10 || i < 1);
std::cout << "the square of the number you have entered is " << i*i << std::endl;
}
The following code is supposed to do as follows:
create list specified by the user
ask user to input number
3.a) if number is on the list , display number * 2, go back to step 2
3.b) if number isn't on the list, terminate program
HOWEVER step 3.a) will also terminate the program, which is defeating the purpose of the while loop.
here is the code :
#include <iostream>
#include <array>
using namespace std;
int main()
{
cout << "First we will make a list" << endl;
array <int, 5>list;
int x, number;
bool isinlist = true;
cout << "Enter list of 5 numbers." << endl;
for (x = 0; x <= 4; x++)
{
cin >> list[x];
}
while (isinlist == true)
{
cout << "now enter a number on the list to double" << endl;
cin >> number;
for (x = 0; x <= 4; x++)
{
if (number == list[x])
{
cout << "The number is in the list. Double " << number << " is " << number * 2 << endl;
}
else
isinlist = false;
}
}
return 0;
}
Please can someone help me to resolve this ?
I would suggest that you encapsulate the functionality of step 3 into a separate function. You could define a function as follows, and then call it at an appropriate location in the main function.
void CheckVector(vector<int> yourlist)
{
.... // Take user input for number to search for
.... // The logic of searching for number.
if (number exists)
{
// cout twice the number
// return CheckVector(yourlist)
}
else
return;
}
The same functionality can be implemented with a goto statement, avoiding the need for a function. However, using goto is considered bad practice and I won't recommend it.
Your issue is that you set isinlist to false as soon as one single value in the list is not equal to the user input.
You should set isinlist to false ay the beginning of your while loop and change it to true if you find a match.
Stepping your code with a debugger should help you understand the issue. I encourage you to try it.
I have used assert function to ensure that the first number input is between 1 and 7 (inclusive). However, when I execute the program and enter an invalid number, it causes the program to crash. So, how is the assert function being of any use here if that's the case?
Please correct my implementation where required.
Code:
#include <iostream>
#include <assert.h>
using namespace std;
int main() {
int num;
int n;
int max;
cout << "Please enter a number between 1 & 7 \n";
cin >> num;
assert(num >= 1 && num <= 7);
for (int i = 0; i < num; i++) {
cout << "Please enter number " << (i + 1) << ": ";
cin >> n;
if (i == 0) {
max = n;
}
max = (n > max) ? n : max;
}
cout << "The maxmum value is: " << max << endl;
system("pause");
return 0;
}
Assert is not what you want here. What you need is validation. Assertions are for debugging, for identifying completely invalid program states. The user entering invalid input is not invalid program state, it's just invalid user input.
To perform validation, you need an if-test. You will also need some code ready to handle the case of invalid user input. There's absolutely no way to prevent the user from providing invalid input (short of insanely aggressive dynamic validation where you capture keyboard events as they occur and prevent those keystrokes from translating into character input to your program, but now we're just getting ridiculous), so you need to react to it when it happens, say, by printing an error message and then asking for more input.
One way of doing this is as follows:
do {
cin >> num;
if (!(num >= 1 && num <= 7)) {
cerr << "invalid number; must be between 1 and 7!" << endl;
num = -1;
}
} while (num == -1);
Just to expand on that point about assertions, they are supposed to make the program crash. An assertion failure means your code is broken and must be fixed before it can be used in real life. Assertion failures should never be triggered in production code; they simply aid in testing and catching bugs.
What does the "crash" does? According to me, assert will abort the program execution, maybe as another value other than 0.
I wrote a program to sum all odd numbers less than or equal to N. It's not the most efficient or eloquent program, but it works in the compiler on Codepad.org and does not work in DevC++. Usually when a program I wrote is stuck in some kind of infinite loop the program crashes in DevC++ and Windows stops it and lets me know.
Here, the program compiles and runs, but just sits with the cursor blinking and does nothing. Windows doesn't stop it, nothing happens, the program doesn't finish, no matter for how long I let it sit. I'm guessing this is a problem with DevC++ unless it's a problem with my code that Codepad overlooks. Will anyone explain to me what is happening here?
Here is my code:
#include <iostream>
using namespace std;
int odd(int N)
{
int i;
int sum = 0;
for(i = 0; i <= N; ++i)
{
while((i % 2) != 0)
{
sum = sum + i;
}
}
return sum;
}
int main()
{
int N;
cout << "Pick a value: ";
cin >> N;
cout << "The sum of all numbers <= to " << N << " is: " << odd(N);
return 0;
}
I've made the suggested change to an if-statement and the same problem is occuring:
#include <iostream>
using namespace std;
int odd(int N)
{
int i;
int sum = 0;
for(i = 0; i <= N; ++i)
{
if ((i % 2) != 0)
{
sum = sum + i;
}
}
return sum;
}
int main()
{
int N;
cout << "Pick a value: ";
cin >> N;
cout << "The sum of all odd numbers <= to " << N << " is: " << odd(N);
return 0;
}
while((i % 2) != 0)
{
sum = sum + i;
}
This is a infinite loop.Because if (i % 2 != 0) is true then the program will increment sum again and again.What you are probably looking to do is have an if statement instead of while
Seems like the edit is working, please try deleting the old output file and rebuilding and re-compile the entire program.
The output seems to be as follows:
Pick a value: 52
The sum of all odd numbers <= to 52 is: 676
Process exited after 1.034 seconds with return value 0
Press any key to continue . . .
make sure the window of the previous run is closed else the compiler will not recompile but just runs the previous version before you changed it.you may see this as an error stated at bottom in debug mode.
the while() is an infinite loop because i is not changed inside the while() or its {} so use if
Sorry about last time for those who saw my previous thread. It was riddled with careless errors and typos. This is my assignment:
"Write a program that will enable the user to enter a series of non-negative numbers via an input statement. At the end of the input process, the program will display: the number of odd numbers and their average; the number of even numbers and their average; the total number of numbers entered. Enable the input process to stop by entering a negative value. Make sure that the user is advised of this ending condition."
And here is my code:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int number, total1=0, total2=0, count1=0, count2=0;
do
{
cout << "Please enter a number. The program will add up the odd and even ones separately, and average them: ";
cin >> number;
if(number % 2 == 0)
{
count1++;
total1+=number;
}
else if (number >= 0)
{
count2++;
total2+=number;
}
}
while (number>=0);
int avg1 = total1/count1;
int avg2 = total2/count2;
cout << "The average of your odd numbers are: " << avg1 << endl;
cout << "The average of your even numbers are " << avg2 << endl;
}
It seems to be working fine, but when I enter a negative number to terminate the program, it includes it with the rest of the averaged numbers. Any advice to get around this? I know it's possible, but the idea escapes me.
Your main loop should be like this:
#include <iostream>
for (int n; std::cout << "Enter a number: " && std::cin >> n && n >= 0; )
{
// process n
}
Or, if you want to emit a diagnostic:
for (int n; ; )
{
std::cout << "Enter a number: ";
if (!(std::cin >> n)) { std::cout << "Goodbye!\n"; break; }
if (n < 0) { std::cout << "Non-positve number!\n"; break; }
// process n
}
After here:
cout << "Please enter a number. The program will add up the odd and even ones seperately, and average them: ";
cin >> number;
Immediately check if the number is negative
if(number < 0) break;
Now you wouldn't need to use your do-while loop in checking if the number is negative. Thus, you can use an infinite loop:
while(true) {
cout << "Please enter a number. The program will add up the odd and even ones seperately, and average them: ";
cin >> number;
if(number < 0) break;
// The rest of the code...
}
ADDITIONAL:
There is something wrong in your code. You aren't showing the user how much the number of even and odd numbers are, and the total number of numbers entered.
ANOTHER ADDITIONAL: You should use more meaningful variable names:
int totalNumEntered = 0, sumEven = 0, sumOdd = 0, numEven = 0, numOdd = 0;
Of course I am not limiting you to these names. You can also use other similar names.
FOR THE INTEGER DIVISION PROBLEM:
You must cast your expression values to the proper type (in this case, it is float). You should also change the averages variables' types to float:
float avg1 = float(total1) / float(count1);
float avg2 = float(total2) / float(count2);
Immediately after cin >> number, check for < 0, and break if so. Try to step through the program line by line to get a feel for the flow of execution. Have fun learning, and good luck!