c++/8935: Failure to apply trivial peephole optimizations

[martin at xemacs dot org]

>Number:         8935
>Category:       c++
>Synopsis:       Failure to apply trivial peephole optimizations
>Confidential:   no
>Severity:       non-critical
>Priority:       medium
>Responsible:    unassigned
>State:          open
>Class:          pessimizes-code
>Submitter-Id:   net
>Arrival-Date:   Sat Dec 14 02:46:01 PST 2002
>Closed-Date:
>Last-Modified:
>Originator:     Martin Buchholz
>Release:        gcc-3.2.1
>Organization:
>Environment:
Linux x86
>Description:
Some very easy optimizations are being missed on x86 at -O3.

The x86 asm for a function that effectively does { return 0; } is

	pushl	%ebp
	xorl	%eax, %eax
	movl	%esp, %ebp
	subl	$40, %esp
	movl	%ebp, %esp
	popl	%ebp
	ret

A trivial peephole optimizer with a two instruction window can convert
the sequence

	subl	$40, %esp
	movl	%ebp, %esp
to 
	movl	%ebp, %esp

Subsequently, we can convert the sequence

	movl	%esp, %ebp
	movl	%ebp, %esp
to
	movl	%esp, %ebp

leaving us with 

	pushl	%ebp
	xorl	%eax, %eax
	movl	%esp, %ebp
	subl	$40, %esp
	movl	%ebp, %esp
	popl	%ebp
	ret

which is the asm generated for other functions.

Detail: gcc 3.2.1, x86 Linux, g++ -O3

Entire source files follows:

    struct S
    {
    public:
      enum E { E1, E2 } e;
      S (enum E ee) : e (ee) {}
      S (const S&other);
      operator E () { return e; }
    };

    bool foo () { return S::E1 == S::E2; }
    bool bar () { return S::E(S(S::E1)) == S::E(S(S::E2)); }


>How-To-Repeat:

>Fix:

>Release-Note:
>Audit-Trail:
>Unformatted: