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> Also struggling a bit with
> result *= y++;
> I interpreted that as:
> #local Result = Result * (Y+1);
> (line 102)
The whole statement, when decomposed, can be interpreted in C/C++ as
old_y = y ;
y = y + 1 ;
result = result * old_y ;
'y++' is interpreted as get the value of y, increment y by one then return the
value of y prior to incrementing it.
#local Old_Y = Y ;
#local Y = Y + 1 ;
#local Result = Result * Old_Y ;
In your interpretation, Y will never be incremented ...
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