Hi,
I was wondering why I can serialize a call such as in
serialize(call("ripley"), NULL)
[1] 58 0a 00 00 00 02 00 03 00 02 00 02 03 00 00 00 00 06 00 00 00 01 00 04
00 09 00 00 00 06 72 69 70 6c 65 79 00 00 00 fe
but if I try from C++ the call gets evaluated
library("Rcpp")
rcpp.serialize = cppFunction(code="RObject my_serialize(RObject
x){Function r_serialize(\"serialize\"); return r_serialize(x,
R_NilValue);}")
rcpp.serialize(10)
[1] 58 0a 00 00 00 02 00 03 00 02 00 02 03 00 00 00 00 0e 00 00 00 01 40
24 00 00 00 00 00 00
# easy things work
rcpp.serialize(call("ripley"))
Error: could not find function "ripley"
# wants to evaluate it
unserialize(rcpp.serialize(call("sqrt", 2)))
[1] 1.414214
# How is one to serialize the call itself
rcpp.serialize(list(call("ripley")))
[1] 58 0a 00 00 00 02 00 03 00 02 00 02 03 00 00 00 00 13 00 00 00 01 00
00 00 06 00 00 00 01 00 04 00 09 00 00 00 06 72 69 70 6c 65 79 00 00 00 fe
#wrapping in a list is enough to stop evaluation
eval(call("ripley"))
Error in eval(expr, envir, enclos) : could not find function "ripley"
eval(list(call("ripley")))
[[1]]
ripley()
#OK that's consistent
Could anyone enlighten me on the evaluation mechanism? Not saying it's
broken, but certainly my expectations were off. Thanks
Antonio
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