如何确定函数参数是否注释？

Question

我使用标签bar来标注功能参数，如下所示。

int foo (char* s __attribute__((annotate("bar")))) { 
    ... 
} 

接下来，我正在运行函数传递。如何确定给定的函数参数是否使用标签bar注释？

Answer 1

您将不得不阅读llvm.var.annotation和llvm.dbg.declare内部函数。

更具体地说，这里是LLVM-IR通过上面的代码生成：

@.str = private unnamed_addr constant [4 x i8] c"bar\00", section "llvm.metadata" 
@.str.1 = private unnamed_addr constant [75 x i8] c"/tmp/compiler-explorer-compiler117030-12962-1rhu4lb.ojfaiz4cxr/example.cpp\00", section "llvm.metadata" 

; Function Attrs: nounwind uwtable 
define i32 @foo(char*)(i8*) #0 !dbg !6 { 
    %2 = alloca i8*, align 8 
    store i8* %0, i8** %2, align 8 
    call void @llvm.dbg.declare(metadata i8** %2, metadata !12, metadata !13), !dbg !14 
    %3 = bitcast i8** %2 to i8* 
    call void @llvm.var.annotation(i8* %3, i8* getelementptr inbounds ([4 x i8], [4 x i8]* @.str, i32 0, i32 0), i8* getelementptr inbounds ([75 x i8], [75 x i8]* @.str.1, i32 0, i32 0), i32 1) 
    ret i32 0, !dbg !15 
} 

!6 = distinct !DISubprogram(name: "foo", linkageName: "foo(char*)", scope: !1, file: !1, line: 1, type: !7, isLocal: false, isDefinition: true, scopeLine: 1, flags: DIFlagPrototyped, isOptimized: false, unit: !0, variables: !2) 
!7 = !DISubroutineType(types: !8) 
!8 = !{!9, !10} 
!9 = !DIBasicType(name: "int", size: 32, align: 32, encoding: DW_ATE_signed) 
!10 = !DIDerivedType(tag: DW_TAG_pointer_type, baseType: !11, size: 64, align: 64) 
!11 = !DIBasicType(name: "char", size: 8, align: 8, encoding: DW_ATE_signed_char) 
!12 = !DILocalVariable(name: "s", arg: 1, scope: !6, file: !1, line: 1, type: !10) 

的dbg.declare指令告诉你，2％实际上是函数（命名为S）的第一个参数。

％3是％2的bitcast，所以基本上是一个别名。

而且，llvm.var.annotation指令告诉你％2用常量字符串@str注解，该值为“bar”。