Call function in gdb
J.T. Conklin
jtc@redback.com
Fri Nov 12 14:12:00 GMT 1999
>>>>> "hjl" == H J Lu <hjl@lucon.org> writes:
hjl> Hi, With old gdb, I can do
hjl>
hjl> (gdb) call foo (1)
hjl>
hjl> Now with gdb in CVS, I got "segmentation fault". Any ideas, BTW,
hjl> I am using Linux/i386.
You don't offer much to go on. For what it's worth, calling inferior
functions with a freshly built-from-the-last-snapshot NetBSD/i386 gdb
with the following test code:
int
foo(int x)
{
return x + 1;
}
main()
{
exit (0);
}
appears to work fine.
./gdb -nx a.out
GNU gdb 19991108
Copyright 1998 Free Software Foundation, Inc.
GDB is free software, covered by the GNU General Public License, and you are
welcome to change it and/or distribute copies of it under certain conditions.
Type "show copying" to see the conditions.
There is absolutely no warranty for GDB. Type "show warranty" for details.
This GDB was configured as "i386-unknown-netbsd1.3E"...
(gdb) break main
Breakpoint 1 at 0x1804: file foo.c, line 10.
(gdb) run
Starting program: /sandbox/gdb/gdb-sync3/=i386-netbsd/gdb/a.out
Breakpoint 1, main () at foo.c:10
10 exit (0);
(gdb) call foo(1)
$1 = 2
--
J.T. Conklin
RedBack Networks
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