With test-case gdb.base/ptype.exp and gcc 15 I run into:
...
(gdb) ptype old_fptr^M
type = double (*)(void)^M
(gdb) FAIL: $exp: ptype old_fptr (compiler doesn't emit unprototyped types)
...
Since C23, non-prototype function declarations are no longer supported, so
"double (*old_fptr) ()" is interpreted as "double (*old_fptr) (void)".
We could try to fix this by detecting the language dialect used, and accepting
the output in that case, but that feels fragile.
We could try to fix this by hard-coding the language dialect, but that doesn't
work for all compilers.
So instead, we opt for the simplest solution: just accept this output, and
produce a pass.
Tested on aarch64-linux.
Approved-By: Tom Tromey <tom@tromey.com>
PR testsuite/32756
Bug: https://sourceware.org/bugzilla/show_bug.cgi?id=32756
fail "ptype $id (compiler doesn't emit prototyped types)"
}
-re "type = $overprototyped\[\r\n\]+$gdb_prompt $" {
- if { [test_compiler_info "armcc-*"] } {
- setup_xfail "*-*-*"
- }
- fail "ptype $id (compiler doesn't emit unprototyped types)"
+ # C23 no longer supports non-prototype function declaration, in which
+ # case the overprototyped regexp is the expected one. Simply pass
+ # in all cases.
+ pass "ptype $id (overprototyped)"
}
}
}