Reference · Math

math.ceil

math.ceil(num) -> int

What it does

The smallest int that is not below the number, as Python's math.ceil. A float gives an int: math.ceil(3.2) is 4 and math.ceil(-3.2) is -3, since rounding goes toward plus infinity. An int is itself. For a decimal, or to keep a float a float, use ceil.

Examples

print(math.ceil(3.2), math.ceil(-3.2), math.ceil(2.0))print(math.ceil(-5), type_of(math.ceil(3.2)))print(math.ceil(7 / 2))

4 -3 2-5 int4

Edge cases

• A negative number rounds toward zero: math.ceil(-0.5) is 0, the same as math.trunc gives there.
• Any positive fraction, however small, rounds up: math.ceil(0.3) is 1.
• A float past what an int holds (2^63, about 9.22 × 10¹⁸) is a runtime error, integer overflow; up to it, an int.
• Pages of ten for 21 items: math.ceil(21 / 10) is 3. Of two ints written in the call, the division is done in ints, so it is exact past 2^53 too; a quotient kept in a variable first is a float.

print(math.ceil(-0.5), math.ceil(0.3), math.ceil(21 / 10))print(math.ceil(9007199254740995 / 2))

0 1 34503599627370498

print(math.ceil(1e19))

line 1: integer overflow

Errors

Infinity and NaN cannot be an int and are a runtime error, not an error value; so is an argument that is not a number, text included. A decimal is refused with a hint to convert it with float, or use ceil.

print(math.ceil("3"))

line 1: `math.ceil` expects a number as argument 1, got str

print(math.ceil(float("nan")))

line 1: `math.ceil`: cannot convert NaN to an int

Chaining

In a pipeline the number comes first:

print(21 / 10 | math.ceil)print([0.1, 1.5, 2.0] | map(math.ceil))

3[1, 2, 2]

See also

math.floor, math.trunc, ceil, round, int

Edit this page on GitHub · every example's output was made by running it (tools/refcheck.mjs) · questions on Discord or r/quillo