What is valid in a switch statement for type compatibility?
I would have expected it to follow what appears to be the gimple
"standard" of allowing types that pass the
"useless_type_convserion_p()" test.
I am doing some switch analysis and an triggering a failure in ADA when
the gimple_switch_index() is a signed 64 bit value and the case labels
are only 32 bit integers. I would have expected that these needed to be
the same precision? I don't seem to get this failure anywhere else.
(gdb) p print_gimple_stmt (stderr, sw, 0, 0)
switch (_22) <default: <L9> [67.00%], case 1: <L6> [33.00%]>
(gdb) p gimple_switch_index (sw)
$16 = (tree_node *) 0x7fffee3593f0
(gdb) p print_generic_expr (stderr, $16, 0)
_22
(gdb) p print_generic_expr (stderr, $16->typed.type, 0)
SIGNED_64 << signed 64
bit index
low is the value of CASE_LOW()
(gdb) p print_generic_expr (stderr, low, 0)
1
(gdb) p low->typed.type
$19 = (tree) 0x7fffefacf5e8
(gdb) p print_generic_expr (stderr, $19, 0)
integer
(gdb) p low->typed.type->type_common.precision
$22 = 32 <<< 32 bit case label
(gdb) p type->type_common.precision
$23 = 64
(gdb) p useless_type_conversion_p ($16->typed.type, $19)
$24 = false
(gdb) p useless_type_conversion_p ($19, $16->typed.type)
$25 = false
Is this valid? If so I'll do the promotion myself but this fails the
"is_useless_type_conversion_p ()" test....
Andrew
