Rounding numbers in C++ error - expression must have integral or enum type [duplicate]

Question

I have a function which is designed to round a number down to the nearest even number.

double round(double d) { floor(d + 0.5); if(d % 2 == 1) { d = d-1; } return d; }

However, this returns the error "expression must have integral or enum type" when I try to compile the code. The error is reported from the same line as the if statement.

Can anyone point me in the right direction?

Adding to what @interjay said, it does do something - it returns a value, which you are silently throwing away. You need to assign the result to a variable. — Mark Ransom
– Mark Ransom, Commented Jun 6, 2013 at 18:16
What results do you want for inputs 1 (0, 2, or either?), 3 (2, 4, or either?), –1 (0, –2, either, or anything [will not be used]), and –3 (–2, –4, either, or anything?)? — Eric Postpischil
– Eric Postpischil, Commented Jun 6, 2013 at 18:28

Community · Accepted Answer · 2017-05-23 12:31:29Z

5

The % operator is only defined for integers. You wanna use the fmod() function.

Bill is right about a proper implementation of round(double x).

edited May 23, 2017 at 12:31

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answered Jun 6, 2013 at 18:11

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Bill Carey · Accepted Answer · 2013-06-06 18:13:04Z

3

The floor() function returns a double:

double floor (double x);

which is a floating point type, not an 'integral type', like int or char. Instead of calling floor(d + 0.5); which rounds d and discards the result, you'd want something like:

int i = static_cast<int>(floor(d + 0.5));

answered Jun 6, 2013 at 18:13

Bill Carey

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2 Comments

Eric Postpischil Over a year ago

Why is this the accepted answer? The question asks for code to round down to an even number. This neither rounds down nor rounds to an even number (when an odd number is nearer). Was the question written incorrectly?

Bill Carey Over a year ago

The question asked,

However, this returns the error 'expression must have integral or enum type' when I try to compile the code. The error is reported from the same line as the if statement

. I was trying to explain why that particular error arose, and what could be done to fix it.

Joop Eggen · Accepted Answer · 2013-06-06 18:13:21Z

return floor(d/2 + 0.5) * 2;

Of course doubles are an approximation. For 10^50 you won't get even numbers probably.

naumcho · Accepted Answer · 2013-06-06 18:14:40Z

floor doesn't work in place, it returns the floor-ed value. Also % applies for integers so you can't re-use d. What you want is:

 int i = floor(d + 0.5); if(i % 2 == 1) { i = i-1; } return i;

abelenky · Accepted Answer · 2013-06-06 18:20:43Z

This version will do what you ask, returning an int.

If the parameter d is outside the range of an int, it returns 0 instead.
(maybe you want to throw an OutOfRange exception or something)

int roundDown2Even(double d) { return (INT_MIN <= d && d <= INT_MAX)? ((int)d) & ~1 : 0; }

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