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The precision of float is only 7 digits Overflow occurs only if the dividend is already infinity (so a test would be isinf(candidate)). Adding number 1 to a float larger than 2^24 gives a wrong result
With using double types instead of float you will get a correct result. If the magnitude exceeds 1, the division never creates a new overflow Les types flottants permettent de manipuler des nombres décimaux en c++
Contrairement aux entiers, ils incluent des fonctionnalités supplémentaires telles que la représentation des infinis, de nan (« not a number »), et une précision variable selon la norme ieee 754.
Ce que tu fais, c'est une fonction qui sépare le signe, l'exposant et la mantisse (la, il s'agit de lire le float ou le double bit a bit), et une fois cela fait, se référant au format officiel des flottants, tu crées ta fonction not a number. If an arithmetic operation that yields a floating point type produces a value that is not in the range of representable values of the result type, the behavior is undefined according to the c++ standard, but may be defined by other standards the machine might conform to, such as ieee 754. We need to include the division code in the try block and catch any exception in the corresponding catch block. When dealing with floating point numbers the term underflow means that the number is 'too small to represent', which usually just results in 0.0
Note that i have also heard the term underflow being used for overflow to a very large negative number, but this is not the best term for it. But do keep in mind when an overflow actually occurs, the result is undefined behavior So while this probably works on most machines, it isn't guaranteed You're better of catching it before it happens
Because this isn't trivial to read at first, i'd wrap it into a function
Bool will_overflow(const t& px, const t& pvalue,.
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