A "cout" question

How do I make it so that this statement shows .8 instead of 0:
cout << 4/5;

It keeps displaying 0 instead of .8. Thanks for your help.

cout<<4.0/5.0;

Now how would I do that if it were a variable?

If it were a variable, the << operator would know how. If both top/bottom are floats, it will output 0.8. But if they are ints, then you will get the result of integer division, which drops the remainder and the << will output an int, or 0.

Type cast it. I believe C++ likes you to use "static_cast". IIRC, it's something like:

Code:

int var1 = 4, var2 = 5;
cout << static_cast<float>(var1) / static_cast<float>(var2);

Quzah.

Here is my line of code:

cout << "PI= " << 4 * static_cast<float>(c) / static_cast<float>(n) << endl;

Both of the variables, c & n, were defined as integers. I still can't get the line to print out any decimals. I'm going to fiddle around with it, but maybe you know what I am doing wrong? Thanks for all your help already quzah.

Well, if you look at that line, you'll see that 4 isn't a floating point number. Cast it also, or just ad a .0 to it.

Quzah.

Thanks for your help, it worked!

Originally Posted by scrub05

Both of the variables, c & n, were defined as integers. I still can't get the line to print out any decimals.

Even casting and performing a floating point division operation might result in no decimals being output if the result of that operation results in a value such as X.0. The decimals will not be shown if there isn't anything to show or unless you tell cout to format its output differently:

Code:

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    float f = 5.0f;
    cout << "f is: " << f << endl;
    cout << setiosflags(ios::fixed) << setprecision(4)
         << "f is: " << f << endl;

    int c = 8, n = 4;
    float result = 4 * static_cast<float>(c) / static_cast<float>(n);

    cout.unsetf(ios::fixed);
    cout << "Result is: " << result << endl;
    cout << "Result is: " << setiosflags(ios::fixed)
         << setprecision(2) << result << endl;

    return 0;
}

My output:

Code:

f is: 5
f is: 5.0000
Result is: 8
Result is: 8.00

So you see it can work but depending on what your values for c and n are you might not be seeing the decimals.

I'd sooner write a function which prototypes the arguments as Double, as well as returns a Double (regardless what type you passed to it). Then allow C++'s Promotion Rules to do the work for you. Much more elegant IMHO. Here's a sample proggie...

Code:

#include <iostream>
using namespace std;

double divide ( double, double );

int main()
{
    int x, y;
    
    x = 4;
    y = 5;
    
    cout << "4 / 5 = " << divide ( x, y );
    
    while ( cin.get() != '\n' );
    
    return 0;
    
}

double divide ( double a, double b )
{
    return a / b;
    
}

I woudl just use a solution like what XSquared Said; define your int as a float,

ie- Intstead of INT X call it Float X. That should get you squared away.