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gdb: make interp_add static

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Since this commit:

  commit 8322445e0584be846f5873b9aab257dc9fbda05d
  Date:   Tue Jun 21 01:11:45 2016 +0100

      Introduce interpreter factories

Interpreters should be registered with GDB, not by calling interp_add,
but with a call to interp_factory_register.  I've checked the insight
source, and it too has moved over to using interp_factory_register.

In this commit I make interp_add static within interps.c.

There should be no user visible change after this commit.
This commit is contained in:
Andrew Burgess 2021-11-24 17:18:34 +00:00
parent 19c26da69d
commit 9716aa0a27
2 changed files with 1 additions and 3 deletions
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2
gdb/interps.c
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@ -127,7 +127,7 @@ interp_factory_register (const char *name, interp_factory_func func)
/* Add interpreter INTERP to the gdb interpreter list. The
interpreter must not have previously been added. */
void
static void
interp_add (struct ui *ui, struct interp *interp)
{
struct ui_interp_info *ui_interp = get_interp_info (ui);

2
gdb/interps.h
View File

@ -94,8 +94,6 @@ public:
bool inited;
};
extern void interp_add (struct ui *ui, struct interp *interp);
/* Look up the interpreter for NAME, creating one if none exists yet.
If NAME is not a interpreter type previously registered with
interp_factory_register, return NULL; otherwise return a pointer to