I'm trying to figure out how the byte counting works in "Sum the numbers on standard in". In one answer using the jq
util counts the non-space bytes after the util name, but not the util or the spaces. Using similar reckoning, (i.e. don't count the util name), one could get a byte count of 0 from numsum
.
Possibly the distinction is between programs which are also languages, (i.e. jq
describes itself as a language, therefore don't count the language binary name), and utils that are not, (so do count the util name). If so, perhaps the question is what distinguishes a language from a util with a lot of switches.
Supposing a clear distinction exists between language executables and utils, there's then the question as to language executables used as utils, as in this answer where the pipe |bc
is counted even though bc
describes itself as a language.
add
is ajq
program, then explain why+
is not abc
program. To simplify, assume the input to stdin is strictly two numbers. \$\endgroup\$+
would be a perfectly valid answer. When a challenge is very easy (such as adding two numbers), it will often have a very short answer in many languages; that's not really a problem (except possibly with the challenge itself). \$\endgroup\$|bc
as three bytes. It's not about what's passed to|bc
. \$\endgroup\$