cin doubt

[vbkris]

i am a newbie to c++ i have a doubt

when we give cin the curso appears asking for the use to enter the info

eg:
cout "Enter Number:";
cin >> i;

will display
Enter Number:_(cursor)


i want a default value to be displayed...
like:
Enter Number:95
the use can change it at his option..

is this possible using C++?


i have another doubt
int a,b;
float c;
a=15;
b=16;
c=a/b;
cout << c;
output: &quot;0&quot;

why is it not .975?

if i change a or b to float it gives .975, why?





Known is handfull, Unknown is worldfull

 

[mrscary]

In answer to your first question - using cin it is not possible.

In your other question,

c = a/b - what you have is this

float = int/int

The int/int will evalulate to an int
This is then cast to a float

To get the correct value you will need to cast everything:

c = (float) ( (float)a)/((float)b);


Hollingside Technologies, Making Technology work for you.

http://www.hollingside.com

 

[chipperMDW]

First question: You can't explicitly give a value to be input by default. You can say something to the effect of: &quot;Enter a number (default 6):&quot; and then check to see if they entered anything. If they did, use what they put; if not, use the default.

Second question: mrscary is right, but uses more casting than necessary.

  int/int   = int    (e.g. 7/5 == 1)
  int/float = float  (e.g. 7/(float)5 == 1.4)
float/int   = float  (e.g. (float)7/5 == 1.4)
float/float = float  (e.g. (float)7/(float)5 == 1.4)

In the first example, int/int, (which is what you have,) dividing an int by an int gives you an int. In this case, you get 1. When you assign that to a float, you get 1 as a float, which is just 1.0.

In each of the other examples, you get the correct value because at least one of the operands was treated as a float. It's unnecessary to cast both operands, and it's also unnecessary to cast the final result before assigning it.

 

[vbkris]

thanks guys... but have u come across any program that has satisfied my first question?

Known is handfull, Unknown is worldfull

 

[mrscary]

To do the first part of the question - you would need to write your own input routine.
M.

Hollingside Technologies, Making Technology work for you.

http://www.hollingside.com

 

[vbkris]

i am a newbie mrscary...

Known is handfull, Unknown is worldfull