使用LLVM构建项目时，某些函数调用将被内部函数替换。替换是由前端（例如clang）还是LLVM后端完成的？
通过互联网进行的讨论表明，内在功能替换与优化选项有关。那么这是否意味着如果没有优化选项，那么就不会发生内在替代？或者实际上，有一些默认的内部函数替换无法禁用？
如果有任何方法可以禁用所有内部函数，我应该怎么做？
答案 0 :(得分：13)
这取决于。用代码编写的内在函数直接通过前端发出。在IR级别的优化期间（如前端和后端执行此优化），将像llvm.memset这样的内在函数引入代码中。
这是一个（非常愚蠢的）例子：
int main(int argc, char** argv)
{
int a[8];
for (int i = 0; i != 8; ++i)
a[i] = 0;
for (int i = 7; i >= 0; --i)
a[i] = a[i+1] + argc;
return a[0];
}
使用clang 3.5（clang -S -emit-llvm）进行编译，您将获得以下IR而无需任何内在函数：
; Function Attrs: nounwind uwtable
define i32 @main(i32 %argc, i8** %argv) #0 {
%1 = alloca i32, align 4
%2 = alloca i32, align 4
%3 = alloca i8**, align 8
%a = alloca [8 x i32], align 16
%i = alloca i32, align 4
%i1 = alloca i32, align 4
store i32 0, i32* %1
store i32 %argc, i32* %2, align 4
store i8** %argv, i8*** %3, align 8
store i32 0, i32* %i, align 4
br label %4
; <label>:4 ; preds = %11, %0
%5 = load i32* %i, align 4
%6 = icmp ne i32 %5, 8
br i1 %6, label %7, label %14
; <label>:7 ; preds = %4
%8 = load i32* %i, align 4
%9 = sext i32 %8 to i64
%10 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 %9
store i32 0, i32* %10, align 4
br label %11
; <label>:11 ; preds = %7
%12 = load i32* %i, align 4
%13 = add nsw i32 %12, 1
store i32 %13, i32* %i, align 4
br label %4
; <label>:14 ; preds = %4
store i32 7, i32* %i1, align 4
br label %15
; <label>:15 ; preds = %29, %14
%16 = load i32* %i1, align 4
%17 = icmp sge i32 %16, 0
br i1 %17, label %18, label %32
; <label>:18 ; preds = %15
%19 = load i32* %i1, align 4
%20 = add nsw i32 %19, 1
%21 = sext i32 %20 to i64
%22 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 %21
%23 = load i32* %22, align 4
%24 = load i32* %2, align 4
%25 = add nsw i32 %23, %24
%26 = load i32* %i1, align 4
%27 = sext i32 %26 to i64
%28 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 %27
store i32 %25, i32* %28, align 4
br label %29
; <label>:29 ; preds = %18
%30 = load i32* %i1, align 4
%31 = add nsw i32 %30, -1
store i32 %31, i32* %i1, align 4
br label %15
; <label>:32 ; preds = %15
%33 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 0
%34 = load i32* %33, align 4
ret i32 %34
}
再次使用clang -emit-llvm -O1进行编译，您会看到：
; Function Attrs: nounwind readnone uwtable
define i32 @main(i32 %argc, i8** nocapture readnone %argv) #0 {
.preheader:
%a = alloca [8 x i32], align 16
%a6 = bitcast [8 x i32]* %a to i8*
call void @llvm.memset.p0i8.i64(i8* %a6, i8 0, i64 32, i32 4, i1 false)
br label %0
; <label>:0 ; preds = %.preheader, %0
%indvars.iv = phi i64 [ 7, %.preheader ], [ %indvars.iv.next, %0 ]
%1 = add nsw i64 %indvars.iv, 1
%2 = getelementptr inbounds [8 x i32]* %a, i64 0, i64 %1
%3 = load i32* %2, align 4, !tbaa !1
%4 = add nsw i32 %3, %argc
%5 = getelementptr inbounds [8 x i32]* %a, i64 0, i64 %indvars.iv
store i32 %4, i32* %5, align 4, !tbaa !1
%indvars.iv.next = add nsw i64 %indvars.iv, -1
%6 = trunc i64 %indvars.iv to i32
%7 = icmp sgt i32 %6, 0
br i1 %7, label %0, label %8
; <label>:8 ; preds = %0
%9 = getelementptr inbounds [8 x i32]* %a, i64 0, i64 0
%10 = load i32* %9, align 16, !tbaa !1
ret i32 %10
}
初始化循环由llvm.memset内在替换。后端可以根据需要自由处理内在函数，但通常将llvm.memset降低到libc库调用。
回答您的第一个问题：是的，如果您没有优化代码，那么您的IR就不会获得内在函数。
要防止在代码中引入内在函数，您只需在IR上找到优化传递，然后再运行它。以下是一个相关问题：如何找出在IR上完成的传递：Where to find the optimization sequence for clang -OX?
我们得到-O1的：
prune-eh -inline-cost -always-inline -functionattrs -sroa -domtree -early-cse -lazy-value-info -jump-threading -correlated-propagation -simplifycfg -instcombine -tailcallelim -simplifycfg -reassociate -domtree -loops -loop-simplify -lcssa -loop-rotate -licm -loop-unswitch -instcombine -scalar-evolution -lcssa -indvars -loop-idiom -loop-deletion -loop-unroll -memdep -memcpyopt -sccp -instcombine -lazy-value-info -jump-threading -correlated-propagation -domtree -memdep -dse -adce -simplifycfg -instcombine -barrier -domtree -loops -loop-simplify -lcssa -branch-prob -block-freq -scalar-evolution -loop-vectorize -instcombine -simplifycfg -strip-dead-prototypes -verify
一个疯狂的猜测：instcombine正在引入llvm.memset。我在没有instcombine的情况下运行pass并选择未经优化的IR并得到这个：
; Function Attrs: nounwind readnone uwtable
define i32 @main(i32 %argc, i8** %argv) #0 {
%a = alloca [8 x i32], align 16
%1 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 8
%2 = load i32* %1, align 4
%3 = add nsw i32 %2, %argc
%4 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 7
store i32 %3, i32* %4, align 4
%5 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 7
%6 = load i32* %5, align 4
%7 = add nsw i32 %6, %argc
%8 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 6
store i32 %7, i32* %8, align 4
%9 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 6
%10 = load i32* %9, align 4
%11 = add nsw i32 %10, %argc
%12 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 5
store i32 %11, i32* %12, align 4
%13 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 5
%14 = load i32* %13, align 4
%15 = add nsw i32 %14, %argc
%16 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 4
store i32 %15, i32* %16, align 4
%17 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 4
%18 = load i32* %17, align 4
%19 = add nsw i32 %18, %argc
%20 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 3
store i32 %19, i32* %20, align 4
%21 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 3
%22 = load i32* %21, align 4
%23 = add nsw i32 %22, %argc
%24 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 2
store i32 %23, i32* %24, align 4
%25 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 2
%26 = load i32* %25, align 4
%27 = add nsw i32 %26, %argc
%28 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 1
store i32 %27, i32* %28, align 4
%29 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 1
%30 = load i32* %29, align 4
%31 = add nsw i32 %30, %argc
%32 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 0
store i32 %31, i32* %32, align 4
%33 = getelementptr inbounds [8 x i32]* %a, i32 0, i64 0
%34 = load i32* %33, align 4
ret i32 %34
}
没有说明。因此，为了防止（至少memset）你的代码中的内在函数不要在你的IR上运行instcombine。然而，instcombine是一个强大的选择通行证，真正缩短了代码。
现在您有两个选择：
我希望这能以某种方式帮助你。干杯！