[LLVMdev] LLVM assembly without basic block

Seung Jae Lee lee225 at uiuc.edu
Fri Jun 29 12:08:23 PDT 2007


Hello, guys.

I just wonder if there is any way to spit out LLVM assembly without any basic block division.
E.g., 
If I emit LLVM assembly for the following simple code:
------------------------------------------------------------
void f_loop(long* c, long sz) {

   long i;
   for (i = 0; i < sz; i++) {
      long offset = i * sz;
      long* out = c + offset;
      out[i] = 0;
   }
}
------------------------------------------------------------
You know the LLVM assembly is printed out as follows for this code.
------------------------------------------------------------
void %f_loop(int* %c, int %sz) {
entry:
        %sz = cast int %sz to uint              ; <uint> [#uses=1]
        %tmp18 = setgt int %sz, 0               ; <bool> [#uses=1]
        br bool %tmp18, label %bb, label %return

bb:             ; preds = %bb, %entry
        %indvar = phi uint [ 0, %entry ], [ %indvar.next, %bb ]         ; <uint> [#uses=2]
        %i.0.0 = cast uint %indvar to int               ; <int> [#uses=2]
        %tmp2 = mul int %i.0.0, %sz             ; <int> [#uses=1]
        %tmp4.sum = add int %tmp2, %i.0.0               ; <int> [#uses=1]
        %tmp7 = getelementptr int* %c, int %tmp4.sum            ; <int*> [#uses=1]
        store int 0, int* %tmp7
        %indvar.next = add uint %indvar, 1              ; <uint> [#uses=2]
        %exitcond = seteq uint %indvar.next, %sz                ; <bool> [#uses=1]
        br bool %exitcond, label %return, label %bb

return:         ; preds = %bb, %entry
        ret void
}
------------------------------------------------------------
As you can see, there are three basic blocks composing this code.
Is there any way to get the assembly for this function without any basic block division?
I know this may be absurd but curious.

Thanks,
Seung Jae Lee



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