 Integer count in a closed interval

For real numbers $a\le b$, the integers in the closed interval $[a,b]$ begin at $\lceil a\rceil$ and end at $\lfloor b\rfloor$, if there are any. Their number is thereforeThe extra one includes both ends of the resulting integer interval. The maximum with zero handles an empty intersection: $[0.2,0.8]$ contains no integer, even though it has positive length.

For $[-2.7,1.1]$, the ceiling function selects $-2$ and the floor function selects $1$, giving $1-(-2)+1=4$. The four points are $-2,-1,0,1$. For $[-2,2]$, the count is five, not four: the real length $2-(-2)$ and the number of integer points are different quantities.

The endpoint convention matters. For the half-open interval $[a,b)$ with $a\le b$, the last allowed integer is $\lceil b\rceil -1$, so the count becomes $\max (0,\lceil b\rceil -\lceil a\rceil )$. This handles an integer endpoint $b$ without including it accidentally. For example, $[-2,2)$ has four integer points.