Why it returns 0?
print(70 * (50 / 100));
70 * 0,5 = 35
I don't know why this is happening. Or did I make some stupid mistake either ... I don't know
Why it returns 0?
print(70 * (50 / 100));
70 * 0,5 = 35
I don't know why this is happening. Or did I make some stupid mistake either ... I don't know
The 50/100
is a division between integers. Since 50<100
, the result is 0 and consequently 70*(50/100)
results in 0.
If you want to avoid this, you have to cast one of them as a double
.
70 * (50 / (double)100)
or
70 * ((double) 50/ 100)
When you divide two integers, the result is always an integer. For example, the result of 7 / 3 is 2
.
So on 50/100 is equal to 0.35
and because you are diving two Integers, it will ignore decimal places so it's gonna be a zero - 0, so computer see it as : (ignoring .35)
70 * 0 = 0
P.S
Maybe you could explore little bit about invoking Decimal.Divide, your int arguments get implicitly converted to Decimals so .35 won't be ignored.
You can also enforce non-integer division on int arguments by explicitly casting at least one of the arguments to a floating-point type, e.g.:
int a = 42;
int b = 23;
double result = (double)a / b;
That's how it comes.. :)
Because /
will output int
and not double
value. The result should be 0.5 and 0 will be taken as int
then it will be multiplied by 70 and the result will be 0.
You need to make a cast as follows :
double x = 50/(double)100 ;
Then:
print(70 * x);
50 / 100
is 0
and 70 * 0
is 0
.