Why is the output 0.0000?

Question

I would like to know why the output in this program is 0.0000:

{
    int a = 65;
    char c = (char)a;
    int m = 3.0/2;
    printf("%f\n", 5 / 2);
    return (0);
}

Answer 1

5 / 2 is integer division which yields an int result. However the "%f" printf flag interprets it as a double value, which will cause undefined behaviour.
In your case, it prints 0.0000, but it could be anything, really.

Use floating point operands in order to get a floating point return value.

printf("%f\n", 5.0 / 2.0);

Answer 2

You are printing a integer with %f specifier. Keep in mind that when you divide int by another int the result is again int. Thus 5/2 is the integer 2, which printed with the %f specifier prints 0.0. To avoid this use 5.0/2 this time dividing a double by an integer and thus getting a double result.

Answer 3

You are most likey getting a parameter mismatch in:

printf("%f\n", 5 / 2);  

This should be written as:

printf("%d\n", 5 / 2);  // use "%d" for integer.  

Although not completely correct, the rest of your assignment statements do not appear to contribute to your print failure.

Answer 4

As Elias has pointed out in his comment every number is int if stated otherwise. That is the key to solution.

In C/C++ 5/2 is int number and equals 2 That is why your :

printf("%f\n", 5 / 2);

is the same as:

printf("%f\n", 2);

But why it prints 0.0 you ask - here is why. When printf is given '%f' flag it will treat a parameter as float.

Now your code disassembly looks / may look like that:

printf("%f\n", 5 / 2);
push        2    
call        dword ptr [__imp__printf (0B3740Ch)]

push 2 moves (usually 32bit 0x2) on the stack

And now the most important how 32 bit value=2 looks (in binary):

(MSB) 00000000 00000000 00000000 00000010 (LSB)

and what this pattern means if treated as 32 bit floating point (according to this):

sign(bit #31) = 0

exponent(next 8 bits ie #30..#23) all = 0 too

fraction(the remaining bits) contains 2(dec)

Now the value should in theory be calculated as:

(-1)^0 * (1+1*2^-22)^(0-127) = 1 * (1+2^-22)^-127

BUT exponent=0x0 which is special case (read link) and is treated as 0.0 - period.

Answer 5

No idea what compiler you are using or what options, but try this:

printf("%f\n",(double)(5 / 2));

or this:

printf("%f\n",5.0 / 2.0));

Answer 6

The result of the expression you pass to printf is an integer 0 ('5 / 2' is an integer division). To correct it:

printf("%f\n", (float) 5 / 2);

or

printf("%f\n", 5.0 / 2.0);