Number only

I want the use be able to use only number in : cin>>input; How can I do that?

Just use it like CornedBee said. If the user enters a string the var will recieve the value 0.

Actually, If the user enters a string, your value will be assigned the ascii value of the first character. The next cin you try will grab the next character, and so on.

Z.

I did that but the program freak out,
This is the whole code:
If you want to check you will see than all work until you wrote a letter. I can not do something like :
If keyascii = 30 to 39 then
//do the code
else
//do nothing
???

Code:

---

#include <iostream.h>

//'Declaration
double addit(double x,double y);
double soust(double x,double y);
double multi(double x,double y);
double divide(double x,double y);
double x=0;
double y=0;
double msgexit=1;
//'Code
int main()
{
        do{
                int input;
                cout<<"\n:::Menu:::"<<endl<<endl;
                cout<<"1. Add function"<<endl;
                cout<<"2. Soustract function"<<endl;
                cout<<"3. Multi function"<<endl;
                cout<<"4. Divide function"<<endl;
                cout<<"5. Exit"<<endl;
                cout<<"Your choice is : ";
                cin>>input;
                if (input==5)
                        return 0;
                cout<<"First number:  ";
                cin>>x;
                cout<<"Second number:  ";
                cin>>y;

                switch (input)
                {
                case 1:
                        cout<<"The answer is  --->  ";
                        cout<<+addit(x,y);
                        cout<<"  <---"<<endl;
                        break;
                case 2:
                        cout<<"The answer is  --->  ";
                        cout<<soust(x,y);
                        cout<<"  <---"<<endl;
                        break;
                case 3:
                        cout<<"The answer is  --->  ";
                        cout<<multi(x,y);
                        cout<<"  <---"<<endl;
                        break;
                case 4:
                        cout<<"The answer is  --->  ";
                        cout<<divide(x,y);
                        cout<<"  <---"<<endl;
                        break;
                case 5:
                        msgexit = 0;
                        //return 0;
                        break;
                default:
                        cout<<"Error, bad input, quitting"<<endl;
                        return  0;
                }
        }while (msgexit == 1);
}

//'Function
double addit(double x,double y)
{
        return x+y;
}
double soust(double x,double y)
{
        return x-y;
}
double multi(double x,double y)
{
        return x*y;
}
double divide(double x,double y)
{
        return x/y;
}

---

Thx in advance

Quote:

---

Actually, If the user enters a string, your value will be assigned the ascii value of the first character. The next cin you try will grab the next character, and so on.

---

No. The value is still 0.

Code:

---

int input;
cin >> input;
cout<<input<<endl;
//printed that input is 0

---

Vlatko you tryed the code ? if yes you saw then when I enter a letter then you press enter the code freak?

I once created a user input library. It's attached; replace lvp\string.h with apstring.h (get it at http://www.collegeboard.com and search for the Computer Science AP test). I think it needs a Borland compiler.

Thx turtle

Ok, Vlatko, you were right, but if you define input as char, you will get the behavior i described (and oh what a pain it is, sometimes).

Z.

Quote:

---

Ok, Vlatko, you were right, but if you define input as char, you will get the behavior i described (and oh what a pain it is, sometimes).

---

Well, that is right, but if you want a string then you will define a char array, not a single char.

Nevermind, we understand eachother now. ;) ;)

this is because cin>>char asks for a single character, not a number.

Replace String with apstring after you download and copy the AP classes from http://www.collegeboard.org/ap/compu...l/classes.html :

Code:

---

int UserProofInt(const String &Prompt)
// Inputs an integer from the user using Prompt as the prompt. Different
// from cin is that this function uses a slew of debugging techniques to make
// sure that the user really did enter a valid C++ integer.
{
GetNewInput:
        cout << Prompt << ' ';
        String Input;
        cin >> Input;
        for (int pos=0; pos<Input.length(); pos++) {
                // isdigit(char) function returns true if char is a digit
                if (!isdigit(Input[pos]) && Input[pos]!='-') {
                        cout << '\"' << Input << "\" is not an integer!" << endl;
                        goto GetNewInput;
                }
                if (pos!=0 && Input[pos]=='-') {
                        cout << "Invalid placement of negation sign!" << endl;
                        goto GetNewInput;
                }
        }
        if (Input[0]!='-') {
                if (Input.length()>5) {
                        cout << "That number exceeds integer bounds!" << endl;
                        goto GetNewInput;
                }
        }
        else {
                if (Input.length()>6) {
                        cout << "That number exceeds integer bounds!" << endl;
                        goto GetNewInput;
                }
        }
        long ToReturn=atol(Input.c_str());

        if (ToReturn<(-32767) || ToReturn>32767) {
                cout << "That number exceeds integer bounds!" << endl;
                goto GetNewInput;
        }
        // If execution reaches here, the user entered an integer
        // Type-cast it into an int before returning!
        return int(ToReturn);
}

---

I know it uses the evil goto, but I wrote it two years ago. :)

I do not understand a piece of that code. :confused:

One piece or the entire thing?

!isdigit?
goto GetNewInput? That restart the function?
&& Input[pos]!='-' ? whoa....i understand nothing there

I colored the code; do you have trouble with one specific color or should I walk you though each color?

[code]
int UserProofInt(const String &Prompt) // WHY the : &

{
GetNewInput:
cout << Prompt << ' '; // WHY the ' '
String Input;
cin >> Input;
for (int pos=0; pos<Input.length(); pos++) { /*You are checking if the string input have any charater on it, if not he add one ? */
// isdigit(char) function returns true if char is a digit
if (!isdigit(Input[pos]) && Input[pos]!='-')
//I undestand nothing here
{
cout << '\"' << Input << "\" is not an integer!" << endl;
goto GetNewInput; //go to in the function who you are already?
}
if (pos!=0 && Input[pos]=='-') { //Why ' ' and why [pos]
cout << "Invalid placement of negation sign!" << endl;
goto GetNewInput;
}
}
if (Input[0]!='-') {
if (Input.length()>5) {
cout << "That number exceeds integer bounds!" << endl;
goto GetNewInput;
}
}
else {
if (Input.length()>6) {
cout << "That number exceeds integer bounds!" << endl;
goto GetNewInput;
}
}
long ToReturn=atol(Input.c_str()); //WHOA what is that

if (ToReturn<(-32767) || ToReturn>32767) {
cout << "That number exceeds integer bounds!" << endl;
goto GetNewInput;
}
// If execution reaches here, the user entered an integer
// Type-cast it into an int before returning!
return int(ToReturn);
}
[/code¸]

all in bold is what I do not understand.

The first comment (why the &): const type variable can be used to pass a reference pointer to an object if the object doesn't need to be modified. This just saves memory and doesn't make a local copy of the object.

The red block parses through the string and makes sure that the user only entered digits or a - sign. The goto statement re-asks the user for an integer if they violated these rules.

The "why '' and []" comment: stringname[x] is like Mid(stringname, x, 1) in Visual Basic

The atol thing is a standard C function that converts a C-string to a long.

Any other questions? :)

Turtle i want to tell you than you are the man! and you take all the time the time I need to understand . Thank you! I appreciate it a lots.

I bough Sams Teach Yourself C++ today and I have read 40 pages for the moment. I hope helping person, a day, like you are doing so well.

Daok

Me good. :D :D