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Re: [PATCH] gdb: Ensure compiler doesn't optimise variable out in test


On Wed, 29 Aug 2018 11:02:59 PDT (-0700), andrew.burgess@embecosm.com wrote:
In the test gdb.base/funcargs.exp, there's this function:

    void recurse (SVAL a, int depth)
    {
      a.s = a.i = a.l = --depth;
      if (depth == 0)
        hitbottom ();
      else
        recurse (a, depth);
    }

The test script places a breakpoint in hitbottom, and runs the
executable which calls recurse with an initial depth of 4.

When GDB hits the breakpoint in hitbottom the testscript performs a
backtrace, and examines 'a' at each level.

The problem is that 'a' is not live after either the call to
'hitbottom' or the call to 'recurse', and as a result the test fails.

In the particular case I was looking at GCC for RISC-V 32-bit, the
variable 'a' is on the stack and GCC selects the register $ra (the
return address register) to hold the pointer to 'a'.  This is fine,
because, by the time the $ra register is needed to hold a return
address (calling hitbottom or recurse) then 'a' is dead.

In this patch I propose that a use of 'a' is added after the calls to
hitbottom and recurse, this should cause the compiler to keep 'a'
around, which should ensure GDB can find it.

gdb/testsuite/ChangeLog:

	* gdb.base/funcargs.c (use_a): New function.
	(recurse): Call use_a.
---
 gdb/testsuite/ChangeLog           | 5 +++++
 gdb/testsuite/gdb.base/funcargs.c | 7 +++++++
 2 files changed, 12 insertions(+)

diff --git a/gdb/testsuite/gdb.base/funcargs.c b/gdb/testsuite/gdb.base/funcargs.c
index 600792f0a7e..515631f5491 100644
--- a/gdb/testsuite/gdb.base/funcargs.c
+++ b/gdb/testsuite/gdb.base/funcargs.c
@@ -424,6 +424,10 @@ void hitbottom ()
 {
 }

+void use_a (SVAL a)
+{
+}
+
 void recurse (SVAL a, int depth)
 {
   a.s = a.i = a.l = --depth;
@@ -431,6 +435,9 @@ void recurse (SVAL a, int depth)
     hitbottom ();
   else
     recurse (a, depth);
+
+  /* Ensure A is not discarded after the above calls.  */
+  use_a (a);
 }

 void test_struct_args ()

Isn't the compiler still free to kill "a" here because it can see into use_a() and therefor inline it? I'd expected it to choose to inline use_a(), as doing nothing is always cheaper than calling a function.


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