binutils-gdb/gdb/sparc-pinsn.c
John Gilmore 32c298e08f * sparc-pinsn.c: Remove kludge for preferring architectures.
Remove #ifdef's for SORT_NEEDED.  We need to sort the table, now
and forever.  Add `add' instruction to the set that get checked
for a preceding `sethi' in order to print an absolute address.
Corresponding changes in ../include/opcode/sparc.h needed
to eliminate garbage instructions.
1992-01-30 09:06:49 +00:00

481 lines
12 KiB
C

/* Print SPARC instructions for GDB, the GNU Debugger.
Copyright 1989, 1991 Free Software Foundation, Inc.
This file is part of GDB, the GNU disassembler.
This program is free software; you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation; either version 2 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program; if not, write to the Free Software
Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA. */
#include <stdio.h>
#include "defs.h"
#include "symtab.h"
#include "opcode/sparc.h"
#include "gdbcore.h"
#include "string.h"
#include "target.h"
extern char *reg_names[];
#define freg_names (&reg_names[4 * 8])
union sparc_insn
{
unsigned long int code;
struct
{
unsigned int anop:2;
#define op ldst.anop
unsigned int anrd:5;
#define rd ldst.anrd
unsigned int op3:6;
unsigned int anrs1:5;
#define rs1 ldst.anrs1
unsigned int i:1;
unsigned int anasi:8;
#define asi ldst.anasi
unsigned int anrs2:5;
#define rs2 ldst.anrs2
#define shcnt rs2
} ldst;
struct
{
unsigned int anop:2, anrd:5, op3:6, anrs1:5, i:1;
unsigned int IMM13:13;
#define imm13 IMM13.IMM13
} IMM13;
struct
{
unsigned int anop:2;
unsigned int a:1;
unsigned int cond:4;
unsigned int op2:3;
unsigned int DISP22:22;
#define disp22 branch.DISP22
} branch;
#define imm22 disp22
struct
{
unsigned int anop:2;
unsigned int adisp30:30;
#define disp30 call.adisp30
} call;
};
/* Nonzero if INSN is the opcode for a delayed branch. */
static int
is_delayed_branch (insn)
union sparc_insn insn;
{
unsigned int i;
for (i = 0; i < NUMOPCODES; ++i)
{
const struct sparc_opcode *opcode = &sparc_opcodes[i];
if ((opcode->match & insn.code) == opcode->match
&& (opcode->lose & insn.code) == 0)
return (opcode->flags & F_DELAYED);
}
return 0;
}
static int opcodes_sorted = 0;
extern void qsort ();
/* Print one instruction from MEMADDR on STREAM.
We suffix the instruction with a comment that gives the absolute
address involved, as well as its symbolic form, if the instruction
is preceded by a findable `sethi' and it either adds an immediate
displacement to that register, or it is an `add' or `or' instruction
on that register. */
int
print_insn (memaddr, stream)
CORE_ADDR memaddr;
FILE *stream;
{
union sparc_insn insn;
register unsigned int i;
if (!opcodes_sorted)
{
static int compare_opcodes ();
qsort ((char *) sparc_opcodes, NUMOPCODES,
sizeof (sparc_opcodes[0]), compare_opcodes);
opcodes_sorted = 1;
}
read_memory (memaddr, &insn, sizeof (insn));
for (i = 0; i < NUMOPCODES; ++i)
{
const struct sparc_opcode *opcode = &sparc_opcodes[i];
if ((opcode->match & insn.code) == opcode->match
&& (opcode->lose & insn.code) == 0)
{
/* Nonzero means that we have found an instruction which has
the effect of adding or or'ing the imm13 field to rs1. */
int imm_added_to_rs1 = 0;
/* Nonzero means that we have found a plus sign in the args
field of the opcode table. */
int found_plus = 0;
/* Do we have an `add' or `or' instruction where rs1 is the same
as rsd, and which has the i bit set? */
if ((opcode->match == 0x80102000 || opcode->match == 0x80002000)
/* (or) (add) */
&& insn.rs1 == insn.rd)
imm_added_to_rs1 = 1;
if (insn.rs1 != insn.rd
&& strchr (opcode->args, 'r') != 0)
/* Can't do simple format if source and dest are different. */
continue;
fputs_filtered (opcode->name, stream);
{
register const char *s;
if (opcode->args[0] != ',')
fputs_filtered (" ", stream);
for (s = opcode->args; *s != '\0'; ++s)
{
if (*s == ',')
{
fputs_filtered (",", stream);
++s;
if (*s == 'a')
{
fputs_filtered ("a", stream);
++s;
}
fputs_filtered (" ", stream);
}
switch (*s)
{
case '+':
found_plus = 1;
/* note fall-through */
default:
fprintf_filtered (stream, "%c", *s);
break;
case '#':
fputs_filtered ("0", stream);
break;
#define reg(n) fprintf_filtered (stream, "%%%s", reg_names[n])
case '1':
case 'r':
reg (insn.rs1);
break;
case '2':
reg (insn.rs2);
break;
case 'd':
reg (insn.rd);
break;
#undef reg
#define freg(n) fprintf_filtered (stream, "%%%s", freg_names[n])
case 'e':
case 'v': /* double/even */
case 'V': /* quad/multiple of 4 */
freg (insn.rs1);
break;
case 'f':
case 'B': /* double/even */
case 'R': /* quad/multiple of 4 */
freg (insn.rs2);
break;
case 'g':
case 'H': /* double/even */
case 'J': /* quad/multiple of 4 */
freg (insn.rd);
break;
#undef freg
#define creg(n) fprintf_filtered (stream, "%%c%u", (unsigned int) (n))
case 'b':
creg (insn.rs1);
break;
case 'c':
creg (insn.rs2);
break;
case 'D':
creg (insn.rd);
break;
#undef creg
case 'h':
fprintf_filtered (stream, "%%hi(%#x)",
(int) insn.imm22 << 10);
break;
case 'i':
{
/* We cannot trust the compiler to sign-extend
when extracting the bitfield, hence the shifts. */
int imm = ((int) insn.imm13 << 19) >> 19;
/* Check to see whether we have a 1+i, and take
note of that fact.
FIXME: No longer true/relavant ???
Note: because of the way we sort the table,
we will be matching 1+i rather than i+1,
so it is OK to assume that i is after +,
not before it. */
if (found_plus)
imm_added_to_rs1 = 1;
if (imm <= 9)
fprintf_filtered (stream, "%d", imm);
else
fprintf_filtered (stream, "%#x", imm);
}
break;
case 'L':
print_address ((CORE_ADDR) memaddr + insn.disp30 * 4,
stream);
break;
case 'l':
if ((insn.code >> 22) == 0)
/* Special case for `unimp'. Don't try to turn
it's operand into a function offset. */
fprintf_filtered (stream, "%#x",
(int) (((int) insn.disp22 << 10) >> 10));
else
/* We cannot trust the compiler to sign-extend
when extracting the bitfield, hence the shifts. */
print_address ((CORE_ADDR)
(memaddr
+ (((int) insn.disp22 << 10) >> 10) * 4),
stream);
break;
case 'A':
fprintf_filtered (stream, "(%d)", (int) insn.asi);
break;
case 'C':
fputs_filtered ("%csr", stream);
break;
case 'F':
fputs_filtered ("%fsr", stream);
break;
case 'p':
fputs_filtered ("%psr", stream);
break;
case 'q':
fputs_filtered ("%fq", stream);
break;
case 'Q':
fputs_filtered ("%cq", stream);
break;
case 't':
fputs_filtered ("%tbr", stream);
break;
case 'w':
fputs_filtered ("%wim", stream);
break;
case 'y':
fputs_filtered ("%y", stream);
break;
}
}
}
/* If we are adding or or'ing something to rs1, then
check to see whether the previous instruction was
a sethi to the same register as in the sethi.
If so, attempt to print the result of the add or
or (in this context add and or do the same thing)
and its symbolic value. */
if (imm_added_to_rs1)
{
union sparc_insn prev_insn;
int errcode;
errcode = target_read_memory (memaddr - 4,
(char *)&prev_insn, sizeof (prev_insn));
if (errcode == 0)
{
/* If it is a delayed branch, we need to look at the
instruction before the delayed branch. This handles
sequences such as
sethi %o1, %hi(_foo), %o1
call _printf
or %o1, %lo(_foo), %o1
*/
if (is_delayed_branch (prev_insn))
errcode = target_read_memory
(memaddr - 8, (char *)&prev_insn, sizeof (prev_insn));
}
/* If there was a problem reading memory, then assume
the previous instruction was not sethi. */
if (errcode == 0)
{
/* Is it sethi to the same register? */
if ((prev_insn.code & 0xc1c00000) == 0x01000000
&& prev_insn.rd == insn.rs1)
{
fprintf_filtered (stream, "\t! ");
/* We cannot trust the compiler to sign-extend
when extracting the bitfield, hence the shifts. */
print_address (((int) prev_insn.imm22 << 10)
| (insn.imm13 << 19) >> 19, stream);
}
}
}
return sizeof (insn);
}
}
printf_filtered ("%#8x", insn.code);
return sizeof (insn);
}
/* Compare opcodes A and B. */
static int
compare_opcodes (a, b)
char *a, *b;
{
struct sparc_opcode *op0 = (struct sparc_opcode *) a;
struct sparc_opcode *op1 = (struct sparc_opcode *) b;
unsigned long int match0 = op0->match, match1 = op1->match;
unsigned long int lose0 = op0->lose, lose1 = op1->lose;
register unsigned int i;
/* If a bit is set in both match and lose, there is something
wrong with the opcode table. */
if (match0 & lose0)
{
fprintf (stderr, "Internal error: bad sparc-opcode.h: \"%s\", %#.8lx, %#.8lx\n",
op0->name, match0, lose0);
op0->lose &= ~op0->match;
lose0 = op0->lose;
}
if (match1 & lose1)
{
fprintf (stderr, "Internal error: bad sparc-opcode.h: \"%s\", %#.8lx, %#.8lx\n",
op1->name, match1, lose1);
op1->lose &= ~op1->match;
lose1 = op1->lose;
}
/* Because the bits that are variable in one opcode are constant in
another, it is important to order the opcodes in the right order. */
for (i = 0; i < 32; ++i)
{
unsigned long int x = 1 << i;
int x0 = (match0 & x) != 0;
int x1 = (match1 & x) != 0;
if (x0 != x1)
return x1 - x0;
}
for (i = 0; i < 32; ++i)
{
unsigned long int x = 1 << i;
int x0 = (lose0 & x) != 0;
int x1 = (lose1 & x) != 0;
if (x0 != x1)
return x1 - x0;
}
/* They are functionally equal. So as long as the opcode table is
valid, we can put whichever one first we want, on aesthetic grounds. */
/* Our first aesthetic ground is that aliases defer to real insns. */
{
int alias_diff = (op0->flags & F_ALIAS) - (op1->flags & F_ALIAS);
if (alias_diff != 0)
/* Put the one that isn't an alias first. */
return alias_diff;
}
/* Except for aliases, two "identical" instructions had
better have the same opcode. This is a sanity check on the table. */
i = strcmp (op0->name, op1->name);
if (i)
if (op0->flags & F_ALIAS) /* If they're both aliases, be arbitrary. */
return i;
else
fprintf (stderr,
"Internal error: bad sparc-opcode.h: \"%s\" == \"%s\"\n",
op0->name, op1->name);
/* Fewer arguments are preferred. */
{
int length_diff = strlen (op0->args) - strlen (op1->args);
if (length_diff != 0)
/* Put the one with fewer arguments first. */
return length_diff;
}
/* Put 1+i before i+1. */
{
char *p0 = (char *) strchr(op0->args, '+');
char *p1 = (char *) strchr(op1->args, '+');
if (p0 && p1)
{
/* There is a plus in both operands. Note that a plus
sign cannot be the first character in args,
so the following [-1]'s are valid. */
if (p0[-1] == 'i' && p1[1] == 'i')
/* op0 is i+1 and op1 is 1+i, so op1 goes first. */
return 1;
if (p0[1] == 'i' && p1[-1] == 'i')
/* op0 is 1+i and op1 is i+1, so op0 goes first. */
return -1;
}
}
/* They are, as far as we can tell, identical.
Since qsort may have rearranged the table partially, there is
no way to tell which one was first in the opcode table as
written, so just say there are equal. */
return 0;
}