CODEWITHSUNDEEP
double
Andrew Wang
Andrew Wang

Reputation: 5

My double is being rounded

Okay, so I have a this equation here that's suppose to get close to the value of pi, but for some reason, my double value is always being converted to a whole number. For example, one of my number is 1/3 but when I print it out, it becomes 0.0. Is it because you have to add d to the end of double values? If so, how would I add it to the end of something like this:

fractionpart = fractionpart + (1/i)

Upvotes: 0

Views: 72

Answers (1)

rodrigo
rodrigo

Reputation: 98338

A D is not used for double constants in C, they are double by default, but a . is. That is, 1 is an integer, while 1. is a double (1.F is a float, BTW, and 1.L is a long double, if it is supported by your compiler).

I personally prefer to add a 0 after the . because 1. looks like the end of an English sentence, while 1.0 looks like a proper number. The leading 0 is also optional, so you can write for example .1, but I prefer 0.1.

So your calculation can be written as:

fractionpart = fractionpart + 1.0 / i;

The alternative would be to add a cast to make the type explicit:

fractionpart = fractionpart + 1 / (double)i;

You can cast any of the two operands of the division.

Upvotes: 1

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