Edebug provides rudimentary coverage testing and display of execution frequency.
Coverage testing works by comparing the result of each expression with the previous result; each form in the program is considered covered if it has returned two different values since you began testing coverage in the current Emacs session. Thus, to do coverage testing on your program, execute it under various conditions and note whether it behaves correctly; Edebug will tell you when you have tried enough different conditions that each form has returned two different values.
Coverage testing makes execution slower, so it is only done if
edebug-test-coverage is non-
nil. Frequency counting is
performed for all executions of an instrumented function, even if the
execution mode is Go-nonstop, and regardless of whether coverage testing
Use C-x X = (
edebug-display-freq-count) to display both
the coverage information and the frequency counts for a definition.
Just = (
edebug-temp-display-freq-count) displays the same
information temporarily, only until you type another key.
This command displays the frequency count data for each line of the current definition.
It inserts frequency counts as comment lines after each line of code.
You can undo all insertions with one
undo command. The counts
appear under the ‘(’ before an expression or the ‘)’ after
an expression, or on the last character of a variable. To simplify
the display, a count is not shown if it is equal to the count of an
earlier expression on the same line.
The character ‘=’ following the count for an expression says that the expression has returned the same value each time it was evaluated. In other words, it is not yet covered for coverage testing purposes.
To clear the frequency count and coverage data for a definition,
simply reinstrument it with
For example, after evaluating
(fac 5) with a source
breakpoint, and setting
edebug-test-coverage to t, when
the breakpoint is reached, the frequency data looks like this:
(defun fac (n) (if (= n 0) (edebug)) ;#6 1 = =5 (if (< 0 n) ;#5 = (* n (fac (1- n))) ;# 5 0 1)) ;# 0
The comment lines show that
fac was called 6 times. The
if statement returned 5 times with the same result each
time; the same is true of the condition on the second
The recursive call of
fac did not return at all.