Am 16.03.2020 um 21:24 schrieb Artur Tarassow: > Hi, > > this is a point I don't understand and frequently stumble over. > > Function main() calls checker(). When checker() is called first, it > correctly prints the message "Some error message" and correctly returns 1. > > When checker() is called within the conditional-statement,  <if > checker(1)>, it also returns correctly 1 which triggers printing the > message "Check returned some other error". However, printing the > checker's message "Some error message" 'fails' or is not done. Why is > this the case? Hi Artur, I think your example could be much shorter: <hansl> function scalar printsth(void) print "hey" return 1 end function if printsth() # does not spit out "hey" endif </hansl> Frankly, don't know if it's by design, but could be I guess.

On Mon, 16 Mar 2020, Allin Cottrell wrote: > On Mon, 16 Mar 2020, Sven Schreiber wrote: > >> Hi Artur, I think your example could be much shorter: >> >> <hansl> >> function scalar printsth(void) >> print "hey" >> return 1 >> end function >> >> if printsth() # does not spit out "hey" >> endif >> </hansl> >> >> Frankly, don't know if it's by design, but could be I guess. > > The idea is that if printsth() is called in its own right we want to see > anything it might print, but if it's called under "if" then all we're > interested in is its return value. From that point of view, while the > "if-block" above is entered OK (which can be verified by putting a print > statement within it) it's empty and so nothing is printed. > > Admittedly one could make a case for showing printed output, if any, from a > function call regardless of whether it serves as a boolean switch. So far > as I know that's not an issue that has been raised to date, but it's open > for discussion. FWIW, the C translation of Sven's script <C> #include<stdio.h> int printsth() { printf("Hey\n"); return 1; } int main() { if(printsth()){ } } </C> does indeed print "Hey".

On Mon, 16 Mar 2020, Riccardo (Jack) Lucchetti wrote: > On Mon, 16 Mar 2020, Allin Cottrell wrote: > >> On Mon, 16 Mar 2020, Sven Schreiber wrote: >> >>> Hi Artur, I think your example could be much shorter: >>> >>> <hansl> >>> function scalar printsth(void) >>>  print "hey" >>>  return 1 >>> end function >>> >>> if printsth()    # does not spit out "hey" >>> endif >>> </hansl> >>> >>> Frankly, don't know if it's by design, but could be I guess. >> >> The idea is that if printsth() is called in its own right we want to >> see anything it might print, but if it's called under "if" then all >> we're interested in is its return value. From that point of view, >> while the "if-block" above is entered OK (which can be verified by >> putting a print statement within it) it's empty and so nothing is >> printed. >> >> Admittedly one could make a case for showing printed output, if any, >> from a function call regardless of whether it serves as a boolean >> switch. So far as I know that's not an issue that has been raised to >> date, but it's open for discussion. > > FWIW, the C translation of Sven's script > > <C> > #include<stdio.h> > int printsth() { printf("Hey\n"); return 1; } > int main() { if(printsth()){ } } > </C> > > does indeed print "Hey". And now in git hansl works that way too.