cast variable to int vs round() function

Question

I have seen it in several places where (int)someValue has been inaccurate and instead the problem called for the round() function. What is the diff

Answer 1

In case of casting a float/double value to int, you generally loose the fractional part due to integer truncation.

This is quite different from rounding as we would usually expect, so for instance 2.8 ends up as 2 with integer truncation, just as 2.1 would end up as 2.

Update:

Another source of potential (gross) inaccuracy with casting is due to the limited range of values being able to be represented with integers as opposed to with floating point types (thanks to @R reminding us about this in the comment below)

Answer 2

1. Casting to int truncates a floating-point number, that is, it drops the fractional part.
2. The round function returns the nearest integer. Halfway cases are rounded away from zero, for example, round(-1.5) is -2 and round(1.5) is 2.

7.12.9.6 The round functions

Synopsis

#include <math.h>
double round(double x);
float roundf(float x);
long double roundl(long double x);

Description

The round functions round their argument to the nearest integer value in floating-point format, rounding halfway cases away from zero, regardless of the current rounding direction.

Returns

The round functions return the rounded integer value.

Source: the C99 standard (ISO/IEC 9899:1999). This section did not change in the C11 standard (ISO/IEC 9899:2011).

(For those who are interested, here is a clear introduction to rounding algorithms.)