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SharedBuffer: Fix bug in return value of release() 

Since the equality operator '==' has higher precedence than the
assignment operator '=', we were assigning 'prev' to the result of
our comparison and not the result of mRefs.fetch_sub().

This means that 'prev' would only receive the values 0 or 1.  In
the cases where fetch_sub() returned 0 or 1, we were happening to
get the correct value.  But if fetch_sub() was greator than 1,
we would return to the user 0, instead of the previous reference
count.

We fix this by properly adding parentheses.  We also adjust the
whitespace a little to hopefully make the groupings of the logic
easier to see.

Change-Id: Ib129798a7076854b9ca4f6385c42edbf4fb75e57
This commit is contained in:
Greg Kaiser 2016-08-01 12:46:22 -07:00
parent 23a5b3921d
commit dd55734dda
1 changed files with 3 additions and 2 deletions
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3
libutils/SharedBuffer.cpp
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@ -112,7 +112,8 @@ void SharedBuffer::acquire() const {
int32_t SharedBuffer::release(uint32_t flags) const
{
int32_t prev = 1;
if (onlyOwner() || ((prev = mRefs.fetch_sub(1, std::memory_order_release) == 1)
if (onlyOwner()
|| (((prev = mRefs.fetch_sub(1, std::memory_order_release)) == 1)
&& (atomic_thread_fence(std::memory_order_acquire), true))) {
mRefs.store(0, std::memory_order_relaxed);
if ((flags & eKeepStorage) == 0) {