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Can standard math functions handle infinity correctly?

int main() {
    double inf = INFINITY;
    double pi = acos(-1.0);
    printf("[1]: %f %f\n", atan(inf) / pi, atan(-inf) / pi);
    printf("[2]: %f %f\n", tan(inf) / pi, tan(-inf) / pi);
    return 0;
}

outputs

[1]: 0.500000 -0.500000
[2]: -nan -nan

Is this behavior a specific standard? [2]

Is the behavior undefined? undefined?

I want to be sure what is at least [1]

a guaranteed result.

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1 answer


Both are well-defined behaviors. Quote from http://en.cppreference.com



  • a tan

    If the argument is ± 0, it is returned unmodified.
        If the argument is ± ∞, NaN is returned and FE_INVALID is increased.
        If the argument is NaN, NaN is returned.

  • Athan

    If the argument is ± 0, it is returned unmodified.
        If the argument is + ∞, + π / 2 is returned. If the argument is -∞, -π / 2 is returned.
        If the argument is NaN, NaN is returned.

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