# The cltq story

Stare at the C code below and guess what could go wrong. It’s originally from a CPU simulator, implementing a 16×16⇒32 unsigned multiplication. Thanks to Mattias Engdegård at Intel for sharing the story.

To be more specific, let’s assume x86-64. Is it possible that for some a and b, the result of mul(a, b) changes when compiling the code with different optimization levels using gcc?

Spoiler alert: here’s what will happen. When compiling the code with gcc -O0, you’ll get:

$./mul 65535 65535 65535 * 65535 = 4294836225 (0xfffe0001)  Hope you’re not expecting the multiplication to overflow and output 1 here, which would be the case in Go but not in C/C++. We’ll come back to that later. With gcc -O2, you’ll get this: $ ./mul 65535 65535
65535 * 65535 = 18446744073709420545 (0xfffffffffffe0001)


This doesn’t look pretty, does it? And here’s what gcc 4.7.3/4.8.1 actually emits with -O2:

Wait, the C code uses unsigned integers only. What’s this sign-extension instruction cltq (cdqe in Intel/AMD manuals) doing here? Is this a gcc bug?

Actually, even though emitting cltq is sort of creepy, it doesn’t violate the C standard. There’s a very subtle bug in the mul function. I’ll show how gcc emits cltq by exploiting two odd C rules, integer promotions and undefined behavior, and discuss possible ways to avoid/detect such problems.

## How gcc emits cltq

To see what gcc does, invoke it with -fdump-tree-all to dump the IR after each pass. Here’s the (simplified) output after vrp (value range propagation).

First of all, a and b are of type uint16_t, which is smaller than int (assuming 32-bit int). According to the rule of integer promotions (C11, 6.3.1.1/2), they are converted to int for multiplication. In other words, a * b is actually (int)a * (int)b, a signed multiplication.

Second, since signed integer overflow is undefined behavior in C, gcc assumes the signed multiplication doesn’t overflow. This is shown in the range information: the product _5 is considered non-negative. Therefore, c and _5 are indistinguishable.

Now the sign conversion c = (uint32_t) _5 becomes dead code, and gcc’s next pass dce (dead code elimination) removes it. The return value _7 is basically (uint64_t)((int)a * (int)b). Note that this is a sign extension from int to uint64_t, which will be lowered to cltq. This is how cltq pops out.

## Fixes and workarounds

One way to fix the C code is to convert a and b to uint32_t before multiplication (assuming int is 32-bits).

Instead of fixing the code, another approach is to add a workaround compiler option. Here’s what gcc 4.7.3/4.8.1 emits with -O2 -fwrapv:

Everything looks good now. However, if you use -fno-strict-overflow instead of -fwrapv, congratulations, you’ll get bitten by cltq again:

I never really understand the difference between -fwrapv and -fno-strict-overflow, and which option to choose. Looks like it’s safer to just fix the C code.

As a comparison, here’s what clang 3.3 emits with -O2:

Don’t worry, no cltq here.

## Possible detection methods

In our STACK paper, we use the term unstable code to refer to program fragments being optimized away by the compiler due to undefined behavior. STACK is a static checker for detecting unstable code. While it works well for other cases, unfortunately, STACK doesn’t work here. The main reason is that STACK accepts LLVM IR, which has no sign conversion instruction; the code being optimized away, the sign conversion c = (uint32_t) _5 as in gcc, doesn’t exist in LLVM IR:

Since in LLVM IR there’s no instruction to be optimized away, STACK doesn’t report any warning. One could port STACK to gcc to catch this case though.

Another possible way is to extend STACK with an oracle that rewrites zext (zero extension) to sext (sign extension). The idea is that if the code after rewriting is equivalent to the original only under the assumption of undefined behavior, then this rewriting exploits undefined behavior, and the code may be unstable.

I coded up a prototype and it did work for this mul case (see the sign branch in STACK’s git repository if interested). But it reported too many false positives. After all, the compiler being able to remove code often indicates a bug, while being able to flip between zext and sext doesn’t. Maybe we need better heuristics to reduce false positives.

There are a few handy tools for finding undefined behavior. For example, run the C code with Frama-C’s value analysis and you’ll see the following warning:

$frama-c -val mul.c [value] Analyzing a complete application starting at main ... mul.c:3:[kernel] warning: signed overflow. assert (int)a*(int)b ≤ 2147483647; mul.c:3:[value] assigning non deterministic value for the first time ...  You can also compile it with clang’s -fsanitize=undefined (based on IOC). You need an input to trigger the undefined behavior, such as: $ ./mul 65535 65535
mul.c:3:17: runtime error: signed integer overflow
: 65535 * 65535 cannot be represented in type 'int'


Though these tools won’t tell you what code may be “miscompiled” as STACK does, fixing the warnings will reduce the chances of being bitten by the compiler.

## Summary

This is one of my favorite unstable code examples. The code is very short, uses unsigned integers only, and yet confusing. Fixing the problem, especially by using gcc’s workaround options, is tricky.

Update from Sep 18th: Add results from Frama-C and clang’s -fsanitize=undefined. Thanks to David Mentré and John Regehr.