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On 1/21/11 2:50 PM, Chuck Zhao wrote:
<blockquote cite="mid:4D39F197.6090306@eecg.toronto.edu" type="cite">
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I need to figure out all LLVM Instructions that may write to
memory.<br>
<br>
In <a moz-do-not-send="true"
href="http://llvm.org/docs/tutorial/OCamlLangImpl7.html">http://llvm.org/docs/tutorial/OCamlLangImpl7.html</a>,
it mentions that<br>
"<span class="Apple-style-span" style="border-collapse: separate;
color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-style:
normal; font-variant: normal; font-weight: normal;
letter-spacing: normal; line-height: normal; orphans: 2;
text-indent: 0px; text-transform: none; white-space: normal;
widows: 2; word-spacing: 0px; font-size: medium;"><span
class="Apple-style-span" style="text-align: left;">In LLVM,
all memory accesses are explicit with load/store instructions,
and it is carefully designed not to have (or need) an
"address-of" operator.</span></span>"<br>
<br>
I take this as "StoreInst is the only one that writes to memory".
<br>
</blockquote>
<br>
There are intrinsic functions which write to memory also, such as
memcpy.<br>
<blockquote cite="mid:4D39F197.6090306@eecg.toronto.edu" type="cite">
<br>
However, this doesn't seem to be enough.<br>
</blockquote>
<br>
Your observation is correct. Strictly speaking, any instruction can
write to memory after code generation because it may access a stack
spill slot or a function parameter which the ABI places on the
stack.<br>
<br>
When the Language Reference Manual talks about writing to memory, it
is talking about writing to memory that is visible at the LLVM IR
level. The stack frame is invisible at the LLVM IR level. Put
another way, "memory" is a set of memory locations which can be
explicitly accessed by LLVM load and store instructions and are not
in SSA form; it is not all of the memory within the computer.<br>
<br>
If you're interested in finding instructions that write to RAM
(including writes to stack spill slots), it may be better to work on
Machine Instructions within the code generator framework.<br>
<br>
-- John T.<br>
<br>
<br>
<blockquote cite="mid:4D39F197.6090306@eecg.toronto.edu" type="cite">
<br>
Consider: <br>
...<br>
int a, b, d;<br>
d = a + b;<br>
...<br>
<br>
The above code is turned into LLVM IR:<br>
<pre wrap=""> %0 = load i32* @a, align 4
%1 = load i32* @b, align 4
%2 = add nsw i32 %1, %0
store i32 %2, i32* @d, align 4
Is it possible that temps such as %0, %1 and/or %2 will NOT being register allocated later in the compilation stage, and thus left in memory?
The above code, when converted back to C level, looks like this:
...
unsigned int llvm_cbe_tmp__6;
unsigned int llvm_cbe_tmp__7;
unsigned int llvm_cbe_tmp__8;
unsigned int llvm_cbe_tmp__9;
llvm_cbe_tmp__6 = *(&a);
llvm_cbe_tmp__7 = *(&b);
llvm_cbe_tmp__8 = ((unsigned int )(((unsigned int )llvm_cbe_tmp__7) + ((unsigned int )llvm_cbe_tmp__6)));
*(&d) = llvm_cbe_tmp__8;
llvm_cbe_tmp__9 = /<b class="moz-txt-star"><span class="moz-txt-tag">*</span>tail<span class="moz-txt-tag">*</span></b>/ printf(((&_OC_str.array[((signed int )0u)])), llvm_cbe_tmp__8);
...
It seems the compiler-generated temps are _actually_ left on stack, and writes to them are actually writes to stack memory (via load, add, ...).
I am confused here.
Could somebody help to clarify it?
Thank you
Chuck
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</pre>
</blockquote>
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