In this code given in Julia documentation i got following results. Is there some discrepancy? julia> function hypot(x, y) x = abs(x) y = abs(y) if x > y r = y/x return x*sqrt(1 + r*r) end if y == 0 r…

I think you meant to refer to Functions · The Julia Language There is no discrepancy, at least not in the code. For the first x=2 and y=0 we have x > y so the first case kicks in If you do hypot(0,0) the second case is reached and you indeed get an integer 0.

Of course, in a purely linear function body like g, the usage of return is pointless since the expression x + y is never evaluated and we could simply make x * y the last expression in the function and omit the return. In conjunction with other control flow, however, return is of real use. Who det…

As was already explained in theprevious responses, in your example, you called hypot(2, 0). This means that x > y, and thus it hits the if x > y branch and before it ever reaches the if y == 0 branch. In the if x > y branch, there’s a return there, so the function terminates when it hits that branc…

Confusion in The return Keyword

New to Julia

Oscar_Smith July 6, 2025, 6:34pm 2

This code runs in order. Since x>y this hits the first return.

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Confusion in The return Keyword

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