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# Basic console project

## 4 posts in this topic

Hey

I'm trying to make a simple calculator, at this point I can't even get the input right.

I know my code is kinda ugly but afaik it should work.

int _tmain(int argc, _TCHAR* argv[])
{
int get1=0, get2=0;
cout << "Binary operation" << endl;

//cin.get();
do
{
if(get1==0)cout << "Enter first operand (integer): >";
if (!(cin >> get1) && get1==0) {
cout << "Did you even try?" << endl;
cin.clear();
cin.ignore(1000, '\n');
}
else {
if(get2==0)cout << "Enter second operand (integer): >";
if (!(cin >> get2)){
cout << "Cmon! You can do it!" << endl;
cin.clear();
cin.ignore(1000, '\n');
}
else cout << "Good boy! Press ENTER to continue" << endl;
}

} while (get1==0 || get2==0);

return 0;
}



This is how the console responds:

Why do I get an blanc line on the end instead of the line "Enter second operand (integer): >" ?

What should I change?

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EDIT: Careful ... read my other post ... I was too quick ... and wrong ;-)

Any reason why the cin >> get1 is part of the condition? I don't know what kind of operator >> is exactly and what the operation should return, but it seems to evaluate as false and that is probably why the else is never executed.

With get1 you have the additional "&& get1==0" ... you could add that to the 2nd condition as well ...

But would't that work?

		cin >> get1;
if (get1==0) {
cout << "Did you even try?" << endl;
cin.clear();
cin.ignore(1000, '\n');
}
else {
if(get2==0)cout << "Enter second operand (integer): >";
cin >> get2;
if (get2==0){
cout << "Cmon! You can do it!" << endl;
cin.clear();
cin.ignore(1000, '\n');
}
else cout << "Good boy! Press ENTER to continue" << endl;
}


That still would not be quite pretty, though ... but you know that :-)

EDIT: Careful ... read my other post ... I was too quick ... and wrong ;-)

Edited by DareDeveloper
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It happens because get1 isn't 0, so it fails the initial if(get1==0) test when you insert something invalid on the second operand.

You could add a get1 = 0 after your cout << "Cmon[...]":

			if (!(cin >> get2)){
cout << "Cmon! You can do it!" << endl;
get1 = 0;


But I don't know if this will break other stuff, I don't really understand what your code is supposed to do.

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No wait ... the 2nd time the loop is performed ... if(get1==0)cout << "Enter first operand (integer): >"; is not shown because get1 is not 0 ... but you read get1 again because cin >> get1 is called again ... so you need to find a different approach I guess. Probably turning

if(get1==0)cout << "Enter first operand (integer): >";

into a grown up If-Block is the first step to doing that ... maybe turning it into a regular while loop instead of do ... while ...

... or ...

int _tmain(int argc, _TCHAR* argv[])
{
int get1=0, get2=0;
cout << "Binary operation" << endl;

do
{
cout << "Enter first operand (integer): >";
cin >> get1;
if (get1==0) {
cout << "Did you even try?" << endl;
cin.clear();
cin.ignore(1000, '\n');
}
} while (get1==0);

do
{
cout << "Enter second operand (integer): >";
cin >> get2;
if (get2==0){
cout << "Cmon! You can do it!" << endl;
cin.clear();
cin.ignore(1000, '\n');
}

} while (get2==0);

cout << "Good boy! Press ENTER to continue" << endl;

return 0;
}


Is that any good?

Edited by DareDeveloper
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The cin >> get1 operation waits on an input and puts the entered number in get1 and checks if the operation was valid.

The problem was indeed that after entering an non integral number this code did not jump back to the "else"  but it waits on  cin >> get1 (again!) before concluding the statement is false anyway, because get1==0 is should be false at this point.

The solution is:

do
{
if(get1==0){
cout << "Enter first operand (integer): >";
if (!(cin >> get1)) {
cout << "Did you even try?" << endl;
cin.clear();
cin.ignore(1000, '\n');
}
}
else {
if(get2==0)cout << "Enter second operand (integer): >";
if (!(cin >> get2)){
cout << "Cmon! You can do it!" << endl;
cin.clear();
cin.ignore(1000, '\n');
}
else cout << "Good boy! Press ENTER to continue" << endl;
}

} while (get1==0 || get2==0);


Thanks a lot!

Edited by Flandria4
0

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