cin.get problems

**robid1** · 04-22-2004

I have searched the board but cant seem to find my problem sorry if im repeating old problems.

I can not seem to get the first cin.get(numerator); to get its input, seems to like.. skip it.. i tried clearing it even though the input before hand to me doesnt seem to be getting in the way. The function doesnt have any parameters.

Code:

int enterFraction(void)
{   
    char numerator, denominator;
    
    cout << "\nEnter a fraction in the format A/B: ";
    
    cin.get(numerator);
    cin.ignore(50,'/');
    cin.get(denominator);
    
    // testing variables
    cout << numerator << "/" << denominator << endl;
}

thanks for your help.

**neandrake** · 04-22-2004

From what I remember, you can do this:

Code:

int num, den;
cin.getline(num, '/');
cin.getline(den);

And that will get the numerator and denominator. Of course it's been a while, so I'm not 100%.

**robid1** · 04-22-2004

Yeah i cant use getline

This is homework given to me.
Thanks for your help though, i know how the board feels about homework i just need a kick in the right direction this function in relation to the entire homework is very minimal at best.

**neandrake** · 04-22-2004

Ok, I tried your code. It seems to work fine, except you are only getting in one char for each number. Try your code with 4/5. Try this:

Code:

int enterFraction(void)
{   
    char numerator[12], denominator[12];
    
    cout << "\nEnter a fraction in the format A/B: ";
    
    cin.get(numerator, 12, '/');
    cin.ignore(50,'/');
    cin.get(denominator, 12);
    
    // testing variables
    cout << numerator << endl << "/"<< endl << denominator << endl;
}

You might have problems using them as numbers because the numbers are in string form. I think you can use the function atoi() to get the number from it. Look the function up.

**robid1** · 04-22-2004

Yep that works thank you.

**WaltP** · 04-23-2004

Originally Posted by neandrake

You might have problems using them as numbers because the numbers are in string form. I think you can use the function atoi() to get the number from it. Look the function up.

Or change the values to

Code:

int numerator, denominator;

**Codeplug** · 04-23-2004

Originally Posted by WaltP

Or change the values to

Code:

int numerator, denominator;

If the extractor operators can be used. Otherwise, get() only works with a char, char*, or streambuf.

gg

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