Ceil function: how can we implement it ourselves?

Question

I know that C++ provides us with a ceil function. For practice, I was wondering how can we implement the ceil function in C++. The signature of the method is public static int ceil(float num)

Please provide some insight.

I thought of a simple way: Convert num to a string, find the index of the decimal point, check if the decimal part is greater than 0. If yes, return num+1 else return num. But I want to avoid using the string conversion

Answer 1

You can take apart the ingredients of an IEEE754 floating point number and implement the logic yourself:

#include <cstring>  float my_ceil(float f) {     unsigned input;     memcpy(&input, &f, 4);     int exponent = ((input >> 23) & 255) - 127;     if (exponent < 0) return (f > 0);     // small numbers get rounded to 0 or 1, depending on their sign      int fractional_bits = 23 - exponent;     if (fractional_bits <= 0) return f;     // numbers without fractional bits are mapped to themselves      unsigned integral_mask = 0xffffffff << fractional_bits;     unsigned output = input & integral_mask;     // round the number down by masking out the fractional bits      memcpy(&f, &output, 4);     if (f > 0 && output != input) ++f;     // positive numbers need to be rounded up, not down      return f; } 

(Insert the usual "not portable" disclaimer here.)

Answer 2

Here is a naive implementation for positive numbers (this uses the fact that casting to (int) truncates toward zero):

int ceil(float num) {     int inum = (int)num;     if (num == (float)inum) {         return inum;     }     return inum + 1; } 

It is easy to extend this to work with negative numbers too.

Your question asked for a function returning int, but normally the ceil() function returns the same type as its argument so there are no problems with the range (that is, float ceil(float num)). For example, the above function will fail if num is 1e20.