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@mkczarkowski
Created May 23, 2018 13:47
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// OK, brak bloku pozwala na wykorzystanie niejawnego return.
const sum = (a, b) => a + b;
// :(, każde wywołanie zwróci undefined, wewnątrz bloków wymagane jest jawne return.
const sum = (a, b) => { a + b }
// OK, jawne return.
const sum = (a, b) => { return a + b };
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