如何从LLVM指令获取文件名和目录? [英] How to get the filename and directory from a LLVM Instruction?

问题描述

我需要在llvm传递过程中提取目录和文件名. 当前版本的llvm将getFilename和getDirectory从DebugLoc移到了DebugInfoMetadata.我无法直接在DebugLoc标头中找到类成员getFilename.因此，如何从指令转到源代码的文件名和目录?

I need to extract the directory and filename during a llvm pass. The current version of llvm moved getFilename and getDirectory from DebugLoc to DebugInfoMetadata. I can't find a class member getFilename directly in the DebugLoc header. Thus, how to do I go from an instruction to source code filename and directory?

http://llvm.org/docs/doxygen/html/classllvm_1_1DebugLoc.html

此外，还有一个打印功能可能会有所帮助，但它仅需使用llvm::raw_ostream且不能重定向到std::string.

Additionally, there is a print function that might help but it only takes a llvm::raw_ostream and can't be redirected to a std::string.

void print (raw_ostream &OS) const
// prints source location /path/to/file.exe:line:col @[inlined at]

下面的代码是导致错误的原因

The code below is what gives the error

const DebugLoc &location = an_instruction_iter->getDebugLoc()
StringRef File = location->getFilename() // Gives an error

---我几分钟前想出的解决方案----

---solution I figured out a few minutes ago----

const DebugLoc &location = i_iter->getDebugLoc();
const DILocation *test =location.get();
test->getFilename();`

推荐答案

1)

std::string dbgInfo;
llvm::raw_string_ostream rso(dbgInfo);
location->print(rso);
std::sting dbgStr = rso.str()

2)

auto *Scope = cast<DIScope>(location->getScope());
std::string fileName = Scope->getFilename();

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