On 11-09-19 5:11 PM, Gene Leynes wrote:
I thought that invisible works like return()
However, it appears that it doesn't exit a function. Is it supposed to work
this way?
Yes, invisible() just marks its argument as non-printing. You still
need to return it. (The man page is a little ambiguous, but what it is
saying is that a call to invisible() returns a non-printing object. It
isn't saying that it triggers a return from the caller.)
Duncan Murdoch
>
funInvisible = function(){
invisible(10)
cat('I was not expecting this to print\n')
cat('because it occurs after the invisible return\n')
}
funInvisible()
funVisible = function(){
cat('start of function\n')
return(10)
cat('This error is safely avoided: \n')
cat(1 / 0)
}
funVisible()
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