, . clang :
unsigned sl(char const* s) {
return (*s) ? (1+sl(s+1)) : 0;
}
:
clang -emit-llvm -O1 -c sl.c -o sl.o
# ^^ Yes, O1 is already sufficient.
llvm-dis-3.2 sl.o
llvm (sl.o.ll)
define i32 @sl(i8* nocapture %s) nounwind uwtable readonly {
%1 = load i8* %s, align 1, !tbaa !0
%2 = icmp eq i8 %1, 0
br i1 %2, label %tailrecurse._crit_edge, label %tailrecurse
tailrecurse: ; preds = %tailrecurse, %0
%s.tr3 = phi i8* [ %3, %tailrecurse ], [ %s, %0 ]
%accumulator.tr2 = phi i32 [ %4, %tailrecurse ], [ 0, %0 ]
%3 = getelementptr inbounds i8* %s.tr3, i64 1
%4 = add i32 %accumulator.tr2, 1
%5 = load i8* %3, align 1, !tbaa !0
%6 = icmp eq i8 %5, 0
br i1 %6, label %tailrecurse._crit_edge, label %tailrecurse
tailrecurse._crit_edge: ; preds = %tailrecurse, %0
%accumulator.tr.lcssa = phi i32 [ 0, %0 ], [ %4, %tailrecurse ]
ret i32 %accumulator.tr.lcssa
}
I do not see a recursive call. Indeed, clang is called a loop label tailrecurse, which gives us a pointer as to what clang is doing here.
So finally ( tl; dr ) yes, this code is completely safe, and a decent compiler with a decent flag will do recursion.
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