Abakaki abasontekile bayasetshenziswa ekuchazweni kwezincazelo zomsebenzi ngokuhlukaniswa kwamacala. Silandela umbuzo olula wokuthi ngabe lokhu kumelwa nakho kungasuswa futhi umsebenzi ungancishiselwa kwisaziso esenza ngaphandle kwaso. Isibonelo, umsebenzi
$$f: \mathbb{R} \to \mathbb{R}, f(x) = \left\{\begin{matrix} 42, & \text{falls } x = 0 \\ x, & \text{sonst} \end{matrix}\right.$$
ngosizo lwemisebenzi eyisisekelo yezibalo kusetshenziswa igama lomugqa owodwa?
Lokho akunakwenzeka futhi sikufakazisa ngosizo lokuqhubeka.
Sibheka ukulandelana \((x_n)\) ne- \(x_n = \frac{1}{n}\) . Kulokhu kulandelana \( \lim_{ n \to \infty } x_n = \lim_{ n \to \infty } \frac{1}{n} = 0\) . Ngaphezu kwalokho, \(\lim_{ n \to \infty } f(x_n) = \lim_{ n \to \infty } \frac{1}{n} = 0 \neq 42 = f(0)\) . Ngakho-ke \(f\) ayisebenzi lapho iphuzu \(x=0\) isb.
Njengoba isamba nomkhiqizo wemisebenzi eqhubekayo kuqhubeka futhi ngenxa yezigaba eziboshwe ngamaketanga, umuntu angenza kuphela imisebenzi eqhubekayo (isb.ngakaze i- \(f\) ) ngosizo lwemisebenzi eyisisekelo yezibalo.
Kodwa-ke, uma sivumela ukusebenza kwe- signum engasebenzi , ngokwesibonelo, singakuthola kalula ukubhalwa okunjalo. Ngemuva kwalokho okungukuthi
$$f: \mathbb{R} \to \mathbb{R}, f(x) = sgn^2(x-42)+42.$$
Ngomsebenzi ojwayelekile \(f\) umahluko wecala uyasebenza
$$f,g,h,a: \mathbb{R} \to \mathbb{R}, f(x) = \begin{Bmatrix} g(x), & \text{falls } a(x) = 0 \\ h(x), & \text{falls } a(x) \neq 0 \end{Bmatrix} = sgn^2 \left(a(x)\right)\cdot h(x) + \left(1-sgn^2\left(a(x)\right)\right)\cdot g(x).$$
Ngakolunye uhlangothi, uma ubheka imisebenzi ngezilimi zohlelo, amagatsha angaxazululwa. Isibonelo, ku-PHP umsebenzi wesiginali ungenziwa imephu nge:
e367d0ca10c4f0ac43640ad7fd1b3f0d
\(f\) nayo ingakhonjiswa ngaphandle kwanoma yiziphi if / else control structures with:
e367d0ca10c4f0ac43640ad7fd1b3f0d
Uma ufuna ukwenza ngaphandle ukuqhathanisa opharetha, ungaya isinyathelo esisodwa okunye futhi ukuzigcina wena emhlabeni elihle bitwise opharetha:
e367d0ca10c4f0ac43640ad7fd1b3f0d