I used the following syntax
if (uf > 0x7f800000)
But this is not working. Any help will be appreciated!
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Ancient_Bison
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4https://en.cppreference.com/w/c/numeric/math/isnan – n. m. could be an AI Oct 16 '22 at 10:49
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Assuming that `uf` is a `float` variable, then the comparison `uf > 0x7f800000` will convert the integer value into a `float` and do a straight numeric comparison. – Some programmer dude Oct 16 '22 at 10:51
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By the way, there's almost no need to use `float` in your programs, and haven't been this side of the millennium. Use `double` instead. – Some programmer dude Oct 16 '22 at 10:51
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NaNs always result in false when compared using ==, > or any other relational operator. And your integer 0x7f800000 will be converted to the common type of it and uf.
You should use the standard isnan() function instead.
Toby Speight
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Not in a portable way. What's you're objection to using the function? (Actually, if you look closely, it's a function-like macro, so might be less objectionable to you). – Toby Speight Oct 16 '22 at 13:31