I'm struggling to apply a "While" loop to the following problem: Design the logic for a program that allows a user to enter a number. Display the sum of every number from one through the entered number.
Start
int userNumber;
Declarations
int number = 1
while number <= userNumber
++number
endwhile
output number
Stop
I know my code isn't correct as it is just adding one to the initial value until the user's number is reached, thus making the output the user's number. How would I go about adding each subsequent value without writing them out e.g. user's number is 10, so the program would add 1+2+3+4+5+6+7+8+9+10 and output the total of 55?
Thank you!
Here's a tip. You'll want to start at the users number and count down to 0. Like this:
int finalNum = 0;
int userNum;
//This is where you need to get the user's number....
while(userNum > 0)
{
finalNum += userNum;
userNum--;
}
//Do whatever you need to finalNum....
EDIT: It appears you've posted pseudocode; usually a big no-no here unless stated otherwise. It's better to post the actual code as it's easier to tell what exactly is going on.
The function you need could look like this for c++:
#include <iostream>
using namespace std;
void calc(unsigned x)
{
unsigned t = 0; // Assume the result to be 0 (zero)
for (i = 1; i <= x; i++) // Continue until i is greater than x
{
t += i; // Accumulate, i.e. t = t +i
}
cout << "sum=" << t << endl; // Print the result
}
int main()
{
calc(10);
return 0;
}
An alternative is:
#include <iostream>
using namespace std;
void calc(unsigned x)
{
cout << "sum=" << (x*(x+1)/2) << endl; // Print the result
}
int main()
{
calc(10);
return 0;
}
This works because the sum of all integers from 1 to n is n*(n+1)/2
Related
I am trying to write a code to print values using function and the program doesn't print anything and I don't know why. I know void function don't return any value so I put cout inside the function and I call is correctly
#include <iostream>
using namespace std;
void Print(int n){
for (int i = 0; i <= n; i++){
if (n%i == 0){
cout << i << ' ';
}
}
cout << endl;
}
int main(){
int x;
cin >> x;
Print(x);
return 0;
}
Start your loop from 1 in Print function. You can not mod a number with 0. Division by zero is not allowed.
When you run your program, it wont print anything. You have to enter a number first, then press Enter/Return, and then it should print something.
Assuming you do this and it still does not print, that will be because your Print has a loop that starts from i = 0, and you then do n % i. Doing the modulo % operation with a right-hand-side operand of 0 is not going to work. Start the loop from i = 1 or something else >0.
Lastly, your program will terminate in main at the return 0; right after the first Print. This means that most of your main won't actually be executed. This is probably intentional, but keep that in mind.
Fix these issues and make sure you enter a number into your program when it runs that is valid, like 12, and it should print something.
I'm trying to understand something. I'm still a beginner to c++ and I just created this little program where you input a value and it tells you whether it's even or odd. To do this, I made an integer called "result" which takes value, and then does % 2 operation.
However, my first mistake was that I put int result above "cin >> value" so for some reason that messed up the program and the number would always be even no matter what. Then when I put int result below "cin >> value" the program worked like it should. Why is it doing this?
Any help would be appreciated thank you. I apologize if this is a duplicate but I don't know what to search for.
#include <iostream>
#include <string>
#include "Human.h"
#include <ctime>
using namespace std;
int main() {
int value = 0; // where I input
cin >> value;
// if you put int result above cin program changes.
int result = value % 2;
if (result == 0) {
cout << "Even number." << endl;
}
else {
cout << "Odd number." << endl;
}
return 0;
}
Any code whichever programming language you use runs from top to bottom.
You need to first declare the variable, give it a value and then check for being even or odd.
When you used cin after setting the value of result = value%2; the compiler used the originally initialized value for value which is 0 to compute the value of result which will be 0%2.
That's why you need to use cin>>value; before setting result = value%2;.
C++ read the code top to bottom , line by line. So you will have to int your variable first.I made a much more simpler version of the program if you want to read it:
#include <iostream>
using namespace std;
int main() {
int a;
cout << "a=";
cin >> a ;
if(a%2==0)
{cout<<"a is even";}
else
{cout<<"a is uneven";}
}
When you put int result = value % 2; before cin >> value;, your program will calculate the result before you put a value inside int value via your input.
So your program does calculate int result = 0 % 2;
I want to write a simple program where I ask the user to input a set of numbers to sum. The user can input an unknown number of numbers.
Here's my code:
#include <iostream>
using namespace std;
//ask the user to input a set of numbers to sum (unknown number of input)
int main (){
int sum = 0, value = 0;
while (cin >> value){
sum += value;
cout << "sum is " << sum << endl;
return 0;
}
}
However, when I input several numbers, the result always equal to the first number, not the sum of all numbers entered. As in:
5 6 7 8
sum is 5
What I am doing wrong?
The problem is return 0. Put it outside of the while block.
You are doing it wrong. You must have to put the print method outside of the loop and the return too.If you want user to terminate at any time he wants then you need to take input at an specific keyword or any other keyword other than the data type(like char at place of integer) , so that it will terminate the loop.
#include <iostream>
using namespace std;
int main (){
int sum = 0, value = 0;
/*press any key other than number to terminnate*/
while (cin >> value){
sum += value;
}
cout << "sum is " << sum << endl;
return 0;
}
Like for example:
#include <iostream>
using namespace std;
int main()
{
for (int n=10; n>0; n--){
cout<< n <<", ";}
}
This will output the numbers 10,9,8,7,6,5,4,3,2,1
So is there a new way so I just get the last instance of the loop, the 1?
I new at this and google isn't giving me any answers.
There is no direct way to detect whether the current iteration of a for loop is the last one. But if the behavior of the loop is predictable, you can usually write code that can detect when you're on the last iteration.
In this case, you could do something like:
if (n == 1) {
cout << n << "\n";
}
in the body of the loop. (Of course it would be simpler in this case to replace the entire loop with cout << "1\n";, but I presume this is an example of something more complex.)
In more complicated cases, you can save whatever information you need in the body of the loop:
int value_to_print:
for ( ... ) {
value_to_print = i;
}
std::cout << value_to_print << "\n";
On each iteration, value_to_print is replaced by the current value of i. The final value is the value of i on the last iteration.
You could create a variable (outside the loop) to hold the "current" value of n; whatever happens to the loop (exit condition reached, break, an exception is thrown...) the value will stay there:
int last_n;
for (int n=10; n>0; n--) {
last_n = n;
cout<< n <<", ";
if (something) {
break; // works in this case
} else if (something else) {
throw some_random_error; // works in this case too
}
}
cout << "The last value of 'n' was " << last_n << endl;
You can use a simple if statement for that.
int main()
{
for (int n=10; n>0; n--) {
cout << n << ", ";
if( n == 1 ) {
return n;
}
}
}
The simplest way to accomplish this is: -
#include <iostream>
using namespace std;
int main()
{
int x;
for (int n = 10; n > 0; n--){
x = n;
}
cout << x;
return 0;
}
I'm new to programming too and was trying to figure out something which will allow me to get the last instance of my loop as output.
I tried something and got the output, see if it can help you (if there's a mistake please let me know).
Here user input string is being replaced by "*" and instead of giving output of every instance i have made so only last instance is given as output.
#include <iostream>
#include <string>
using namespace std;
int main()
{
string str;
int string_length;//string length
cout<<"Enter your Email-ID: ";
cin>>str;
string_length = str.length(); //to give the length of input string and use it for the loop
cout<<"lentgh of the string: "<<string_length <<endl;
for(int x = 0; x <= string_length; x++){
str[x] = '*';
while(x==string_length) //string_length is the last instance of the loop
{
cout<<"Here's your Encrypted Email-ID: " <<str<<endl;
break;
}
}
return 0;
}
Hello I am having trouble with a c++ program. Basically its a loop that iterates the amount of times the user wants it to. Now when it reaches a number divisible by 4 it keeps track of that number and finally then outputs how many times the number entered was divisible by 4.
#include<iostream>
using namespace std;
int num;
int count;
int test = 0;
int main()
{
cin>> num;
for (int count = 0; count < num; count++)
if (count % 4 == 0)
(test++);
else
cout<<"";
return 0;
}
Well - if you use return in main, your program will just exit, because that's what return does - ends the function and returns some value. If you want to actually print the value of test, do it before you return:
cout << test;
getch(); // use this so the console won't close automatically
return 0;
Also, the whole program could be written much better:
int main()
{
cin>> num;
cout << num/4;
getch(); // use this so the console won't close automatically
return 0;
}
Do you need to use a loop? If you just need "How many times is a given number divisible by 4" and are not required to loop
#include<iostream>
using namespace std;
int main()
{
int num;
cin>> num;
cout<< num<<" is divisible by 4 "<< (num>>2) <<" time"<<(num>>2>1?"s":"") <<endl;
return 0;
}
num>>2 is bit shifting to teh right twice, which is the same as doing an integer divide by 4. It could be replaced by num/4 if you wanted. Integer division always truncates, so for all positive numbers, it's like rounding down: the same behavior your loop gives you.