Here's a simple C file with an enum definition and a main
function:
enum days {MON, TUE, WED, THU};
int main() {
enum days d;
d = WED;
return 0;
}
It transpiles to the following LLVM IR:
define dso_local i32 @main() #0 {
%1 = alloca i32, align 4
%2 = alloca i32, align 4
store i32 0, i32* %1, align 4
store i32 2, i32* %2, align 4
ret i32 0
}
%2
is evidently the d
variable, which gets 2 assigned to it. What does %1
correspond to if zero is returned directly?
The %1
register was generated by clang to handle multiple return statements in a function. Imagine you were writing a function to compute an integer's factorial. Instead of this
int factorial(int n){
int result;
if(n < 2)
result = 1;
else{
result = n * factorial(n-1);
}
return result;
}
You'd probably do this
int factorial(int n){
if(n < 2)
return 1;
return n * factorial(n-1);
}
Why? Because Clang will insert that result
variable that holds the return value for you. Yay. That's the reason for that %1
variable. Look at the IR for a slightly modified version of your code.
Modified code,
enum days {MON, TUE, WED, THU};
int main() {
enum days d;
d = WED;
if(d) return 1;
return 0;
}
IR,
define dso_local i32 @main() #0 !dbg !15 {
%1 = alloca i32, align 4
%2 = alloca i32, align 4
store i32 0, i32* %1, align 4
store i32 2, i32* %2, align 4, !dbg !22
%3 = load i32, i32* %2, align 4, !dbg !23
%4 = icmp ne i32 %3, 0, !dbg !23
br i1 %4, label %5, label %6, !dbg !25
5: ; preds = %0
store i32 1, i32* %1, align 4, !dbg !26
br label %7, !dbg !26
6: ; preds = %0
store i32 0, i32* %1, align 4, !dbg !27
br label %7, !dbg !27
7: ; preds = %6, %5
%8 = load i32, i32* %1, align 4, !dbg !28
ret i32 %8, !dbg !28
}
Now you see %1
making itself useful huh? Most functions with a single return statement will have this variable stripped by one of LLVM's passes.