Do...While loop

[squirrel27]

Hello! Thank you for your response in advance.. My program is C++ and I use Codeblocks to run my code. My question is what am I doing wrong in my do while loop? I can get the program to "run" but it repeatedly asks me the enter a positive number and does not recognize my negative entry or "y" to do again. Is it necessary to include a call statement? Thank you again.

// add the following lines at the beginning of the source file
#include <iostream>

using namespace std;

// create translation from generic keywords to C++
#define Declare
#define ByRef ref
#define Call
#define Constant const
#define Declare
#define Display cout <<
#define Eof cin.eof()
#define Function
#define Input cin >>
#define InputLine(x) getline(cin, x)
#define Integer int
#define Module
#define Newline endl
#define Real double
#define String string
#define Set


int main()
{
int sum = 0;
int number;
char again = 'y';


do {
cout << "Enter a positive number \n";
cout << "then a negative number after your last entry." << endl;
cin >> number;
sum = sum + number;



}while (again == 'y' || again == 'Y');


cout << "Enter a positive number \n";
cout << "then a negative number after your last entry." << endl;
cin >> number;
cout << "The sum of your numbers are: " << sum << endl;
cout << "Do you want to do this again?";
cin >> again;
return 0;


}

[MikeyBoy]

You don't change the value of again inside your loop. Once the loop enters, again will always have the value 'y', so it will never stop iterating.

[Aceix]

1. Your loop depends on whether again=Y or y. But outside the loop, you made it equal to y(line 29), and do not allow the user to change it, so you loop becomes an "eternal" one.
2.

cout << "Enter a positive number \n";
cout << "then a negative number after your last entry." << endl;
cin >> number;

Here you are intending to accept two integers(if I'm right), but providing one variable for input. You could do something like:

cout << "Enter a positive number \n";
cin >> number;
sum=number;
cout << "then a negative number after your last entry." << endl;
cin>>number;
sum += number;

If you still want to maintain the nymber of variables you're using.

Aceix.

[Nanyo]

hello I need help
I'm trying to do simple calc but I'm faling with all my tryings:

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
float firstnum, secondnum, result;
char sign;
while(true)
{
cout << "first number:\n";
cin >> firstnum;
cout << "What do yo want to do: * / + - ^\n";
cin >> sign;
cout << "second number:\n";
cin >> secondnum;
if(sign == '*')
{
cout << "the result is:" << firstnum * secondnum << "\n\n";
}
if(sign == '+')
{
cout << "the result is:" << firstnum + secondnum << "\n\n";
}
if(sign == '-')
{
cout << "the result is:" << firstnum - secondnum << "\n\n";
}
if(sign == '/')
{
if(secondnum == 0)
{
cout << "invalid denominator" << "\n\n";
}
else
{
cout << "the result is:" << firstnum / secondnum << "\n\n";
}
}
if(sign == '^')
{
result == pow((float) firstnum, secondnum);
cout << result << "\n\n";
}
}
return 0;
}

this is my code (C++code blocks compiler)
and this is the error:
undefined reference to `pow(double, double)'

HELP!!! Thank you!

[AbstractionAnon]

Line 41: You're using the comparison operator (==), not the assignment operator (=).

PLEASE USE CODE TAGS (the <> formatting button) when posting code.
It makes it easier to read your code and also easier to respond to your post.
http://www.cplusplus.com/articles/jEywvCM9/

Please don't post in somebody else's thread.

[squirrel27]

Thank you AbstractionAnon

[Irhcsa]

I'm not 100% sure but wouldn't this be way easier just using a for loop?

[AbstractionAnon]

For loops are normally used where a fixed number of iterations are desired. The OP has coded an infinite loop.

An infinite loop can be coded as either while (true) or for (;;) Either idiom works. I personally prefer the while (true) construct for an infinite loop.