sh outline

Spec Outline

sh / std-runner

TODO: sh-baseline

concept:

while read varspec do while read cmdspec do eval $varspec$cmdspec done done

need indented reader first..

when combining varspec one edge case appears; does the std-runner deal with var unset internally? Or should we avail it of a proper 'clean' env. I think it should manage that; would've used '' but that requires escaping with readline; find '/' more fitting, use as prefix to unset varname

/LOG LOG='' LOG=val

trying table; columns varspec, cmdspec; rules:

1. varspec is not required, empty cell no varspec; however cmdspec can also be empty, see 3.

2. cmdspec is required, empty cell uses previous non-empty cell above

3. varspec indents to add to previous value; need a metavalue keyword to indicate empty varspec, using '=' now.

4. skip empty lines or unixy comment/preproc lines

outline rules

• each outline node represents a varspec part;

• the leafs; the final nodes or lists are all cmdspec(s).

• During denormalization each outline node is read until a leaf is found; the result is output as $varspec$cmdspec.

• Leaf nodes should not be terminated, ie. no trailing ';'.

• Denormalization encomasses merging all previous (or root-ward) varspecs.

• Unterminated varspecs generate two testcases: one with and one without.

• Unterminated varspecs are concatenated after concatening all terminated varspecs.

This should help in building test matrices.

Iow. the level at and above an unterminated varspec node is not listed by itself during denormalization. But at and below two cases are generated for each true varspec node in the outline. "non-true" or pure varspec nodes are script segments prepended to

• Var nodes are terminated automatically, unless they match ^[A-Za-z_][A-Za-z0-9_]=.*.

So varspecs can contain unset <var> or any other shell script part. But only variable assignments are aggregated and concatenated as the local-env prefix for the command line.

Order does not matter.

If they are explicitly terminated, they are treated the same. But a second case with no value will be added to the cartesian set as well.

unset x; y=b ..

y=a ..