sequencer: stop pretending that an assignment is a condition

In 3e81bccdf3 (sequencer: factor out todo command name parsing,
2019-06-27), a `return` statement was introduced that basically was a
long sequence of conditions, combined with `&&`, except for the last
condition which is not really a condition but an assignment.

The point of this construct was to return 1 (i.e. `true`) from the
function if all of those conditions held true, and also assign the `bol`
pointer to the end of the parsed command.

Some static analyzers are really unhappy about such constructs. And
human readers are at least puzzled, if not confused, by seeing a single
`=` inside a chain of conditions where they would have expected to see
`==` instead and, based on experience, immediately suspect a typo.

Let's help all of this by turning this into the more verbose, more
readable form of an `if` construct that both assigns the pointer as well
as returns 1 if all of the conditions hold true.

Signed-off-by: Johannes Schindelin <johannes.schindelin@gmx.de>
Signed-off-by: Junio C Hamano <gitster@pobox.com>

diff --git a/sequencer.c b/sequencer.c
index b5c4043757e948aedaad0782dff947373e2f45ea..e5e3bc6fa5ea5d6096e0df245a07ead59424ff4c 100644 (file)
--- a/sequencer.c
+++ b/sequencer.c
@@ -2600,9 +2600,12 @@ static int is_command(enum todo_command command, const char **bol)
        const char nick = todo_command_info[command].c;
        const char *p = *bol;
 
-       return (skip_prefix(p, str, &p) || (nick && *p++ == nick)) &&
-               (*p == ' ' || *p == '\t' || *p == '\n' || *p == '\r' || !*p) &&
-               (*bol = p);
+       if ((skip_prefix(p, str, &p) || (nick && *p++ == nick)) &&
+           (*p == ' ' || *p == '\t' || *p == '\n' || *p == '\r' || !*p)) {
+               *bol = p;
+               return 1;
+       }
+       return 0;
 }
 
 static int check_label_or_ref_arg(enum todo_command command, const char *arg)