Is this a bug in the Intel C++ Compiler (ICC)?

Question

The following testing code does correctly in VS either with debug or release, so does it in gcc. It also does correctly for ICC with debug, but not when optimization enabled (-O2).

---

#include <cstdio>

class tClassA{
public:
  int m_first, m_last;

  bool isEmpty() const {return (m_first == m_last);}
  void updateFirst() {m_first = m_first + 1;}
  void updateLast() {m_last = m_last + 1;}

  tClassA() : m_first(0), m_last(0) {}

  ~tClassA() {}

  void doSomething() {printf("should not reach here\r\n");}

};


int main() {

 tClassA q;

 while(true) {

    while(q.isEmpty()) ;

    q.doSomething();
  }

  return 1;

}

---

It is supposed to stoped at while(q.isEmpty()). When -O2 enabled under ICC (release), however, it starts to "doSomething" infintely.

Since this is single-threaded program AND isEmpty() should be evaluated as "true", I can find no reason the ICC should behave in this way? Do I miss anything?

Answer 1

Your best bet is to take the resulting binary step into it and dis-assemble the main function and see what assembly was generated. Not saying you will be able to see a bug, but you can see if something was optimized out.

Answer 2

It sure sounds like a bug. Here's a (pretty wild) guess about what reasoning might have lead to it...

After inlining, it sees:

while (q.m_first == q.m_last) /* do nothing */ ;
do_something();

and any sequence of do nothing repeatedly ; do something can be translated to simply "do something". This falls down if the repeated part is endless (as in this case). But perhaps they don't test their compiling on examples that intentionally have endless looping ;-).

Answer 3

I think it may have been your version of gcc. I compiled your prog under 4.4.2 and it worked exactly as it should have.

Answer 4

Any chance the actual code that you built and ran was missing the semicolon after while(q.isEmpty()) ? That would certainly result in the next line being called infinitely.

Answer 5

Because the while (q.isEmpty()) ; loop contains no statements that can ever cause an externally-visible side-effect, the entire loop is being optimised out of existence. It's the same reason that:

for (int i = 0; i < 10; i++)
    ;

could be optimised out of existence, as long as i was not volatile (stores to volatile objects are part of the "externally visible" effects of a program).

In the C language, it is actually an outstanding bone of contention as to whether an infinite loop is allowed to be optimised away in this manner (I do not know what the situation is with C++). As far as I know, a consensus has never been reached on this issue - smart and knowledgeable people have taken both sides.

Answer 6

As a slight aside, this version of icc does what you want. That is, it never calls doSomething().

[9:41am][wlynch@computer /tmp] icc --version
icc (ICC) 11.0 20081105

Answer 7

The C++ standard allows loops without side-effects to be removed, even if they don't terminate:

It is generally felt that it is important to allow the transformation of potentially non-terminating loops (e.g. by merging two loops that iterate over the same potentially infinite set, or by eliminating a side-effect-free loop), even when that may not otherwise be justified in the case in which the first loop never terminates. http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2429.htm

See the discussion here: http://blog.regehr.org/archives/161