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I was reading Setting an int to Infinity in C++. I understand that when one needs true infinity, one is supposed to use numeric_limits<float>::infinity(); I guess the rationale behind it is that usually integral types have no values designated for representing special states like NaN, Inf, etc. like IEEE 754 floats do (again C++ doesn't mandate neither - int & float used are left to the implementation); but still it's misleading that max > infinity for a given type. I'm trying to understand the rationale behind this call in the standard. If having infinity doesn't make sense for a type, then shouldn't it be disallowed instead of having a flag to be checked for its validity?
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The std::numeric_limits ::digits function is used to find the number of radix digits that the data type can represent without loss of precision.
Setting an int Infinity: The nearby value that we can get is by initializing an “int” to its extreme value. The closest we can get by setting a variable to the maximum value that is double “a = std: numeric_limits<int>:: max();”. Which would be 231-1 if it is 32 bits wide on our implementation.
numeric_limits::minReturns the minimum finite value representable by the numeric type T . For floating-point types with denormalization, min returns the minimum positive normalized value. Note that this behavior may be unexpected, especially when compared to the behavior of min for integral types.
The function numeric_limits<T>::infinity() makes sense for those T for which numeric_limits<T>::has_infinity returns true.
In case of T=int, it returns false. So that comparison doesn't make sense, because numeric_limits<int>::infinity() does not return any meaningful value to compare with.
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