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0296197999
Very simple / crude method of removing 'is_static' from initialization. Why should we require an application tell us whether it is linked statically or dynamically to libLVM? If the application is linked statically, but libraries exist and dlopen() calls succeed, why do we care if it's statically linked?
41 lines
1.2 KiB
C
41 lines
1.2 KiB
C
/*
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* Copyright (C) 2003-2004 Sistina Software, Inc. All rights reserved.
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* Copyright (C) 2004-2007 Red Hat, Inc. All rights reserved.
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*
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* This file is part of LVM2.
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*
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* This copyrighted material is made available to anyone wishing to use,
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* modify, copy, or redistribute it subject to the terms and conditions
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* of the GNU Lesser General Public License v.2.1.
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*
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* You should have received a copy of the GNU Lesser General Public License
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* along with this program; if not, write to the Free Software Foundation,
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* Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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*/
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#ifndef _LVM_CMDLINE_H
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#define _LVM_CMDLINE_H
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struct cmd_context;
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struct cmdline_context {
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struct arg *the_args;
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struct command *commands;
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int num_commands;
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int commands_size;
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int interactive;
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};
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int lvm2_main(int argc, char **argv);
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void *cmdlib_lvm2_init(unsigned is_static);
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void lvm_fin(struct cmd_context *cmd);
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struct cmd_context *init_lvm(void);
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void lvm_register_commands(void);
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int lvm_split(char *str, int *argc, char **argv, int max);
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int lvm_run_command(struct cmd_context *cmd, int argc, char **argv);
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int lvm_shell(struct cmd_context *cmd, struct cmdline_context *cmdline);
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#endif
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