CODEWITHSUNDEEP
c++conditional-statements
CTZStef
CTZStef

Reputation: 1715

Weird conditional statement (same result)

Going through some code, I found this :

#ifdef trunc
# undef trunc
#endif
inline float trunc(float x)
{
    return (x < 0.0f) ? float(int(x)) : float(int(x));
}
inline double trunc(double x)
{
    return (x < 0.0f) ? double(int(x)) : double(int(x));
}
inline long double trunc(long double x)
{
    return (x < 0.0f) ? (long double)(int(x)) : (long double)(int(x));
}
#endif // _WIN32

Of course, the ?: operator always returns one and the same value in each case, regardless of its conditional expression. On the other hand, I guess the author had his reasons to write these functions this way; I can't find one though. Any idea ? Is this just an error (typo) ?

[EDIT] Reply from the author :

Good point - this is just overzealous cut-and-paste from the definition of round(). The following should be just fine (other than the limitation on the range of int):

inline float trunc(float x)
{
    return float(int(x));
}
inline double trunc(double x)
{
    return double(int(x));
}
inline long double trunc(long double x)
{
    return (long double)(int(x));
}

Upvotes: 2

Views: 221

Answers (2)

Joe Z
Joe Z

Reputation: 17956

This code looks wrong.

My guess is that they meant something more like this:

inline float trunc(float x)
{
    return (x < 0.0f) ? -float(int(-x)) : float(int(x));
}

But even that's dubious. I believe int(x) always performs truncation, so even then the two branches of ?: should yield the same result.

In case rounding mode does matter to the typecast (and after a moment's thought, I'm not sure it does), you may really want to use a function like modf, modff or modfl to break the number into integer and fractional portions, and discard the fractional portion.

For example:

inline float trunc(float x)
{
    float int_part;
    modf(x, &int_part);
    return int_part;
}

Edit: One other observation. The original code will fail for values that do not fit in an int. Yet another strike against it.

Upvotes: 5

Aswin Murugesh
Aswin Murugesh

Reputation: 11080

The code returns the same output for both the conditions. This is just a redundancy and moreover, the float(int(x)) doesn't make a point. Because int(x) converts the number to an integer, again converting it to float or double makes no difference but only the type of the variable returned.

Upvotes: 0

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