LLVM从Value获取常量整数* [英] LLVM get constant integer back from Value*

问题描述

我从一个整数常量创建一个llvm :: Value *，如下所示：

  llvm :: Value * constValue = llvm :: ConstantInt :: get（llvmContext，llvm :: APInt（node-> someInt（）））; 
  

现在我想要检索编译时常量值;

  int constIntValue = constValue->？ 
  

LLVM程序员手册似乎暗示cast<>将接受一个指针，当使用类型（而不是类型加指针）模板参数，但是我很确定的失败从2.8：

  llvm :: Value * foo = 0; 
 llvm :: ConstantInt * intValue =& llvm :: cast< llvm :: ConstantInt，llvm :: Value>（foo）; 
 
 //构建错误：
 //错误：没有匹配的函数调用'cast（llvm :: Value *&）'
  
 
 
 
这里正确的方法是什么？
解决方案
给定 llvm :: Value * foo ，你知道 foo 实际上是一个 ConstantInt ，我相信惯用的LLVM代码方法是使用 dyn_cast 如下：
  if（llvm :: ConstantInt * CI = dyn_cast< llvm :: ConstantInt>（foo））{
 // foo确实是一个ConstantInt，我们可以使用CI here 
} 
 else {
 // foo实际上不是ConstantInt 
} 
  
 
 $ b b 
如果您确定  foo 是 ConstantInt 如果不是，则可以使用 cast 而不是 dyn_cast 。 p> 
 
 
 
 
 
 PS请注意， cast 和 dyn_cast 是LLVM自己的RTTI实现的一部分。  dyn_cast 的行为与标准C ++  dynamic_cast 有些类似，尽管在实现和性能方面存在差异href =http://llvm.org/docs/ProgrammersManual.html#isa>在此处阅读）。
 
I create a llvm::Value* from a integer constant like this:
llvm::Value* constValue = llvm::ConstantInt::get( llvmContext , llvm::APInt( node->someInt() ));
now i want to retrieve the compile-time constant value back;
int constIntValue = constValue->???
The examples shown in LLVM Programmer manual seem to imply that cast<> will accept a pointer when using the type (rather than the type plus pointer) template parameter, however i'm pretty sure thats failing as from 2.8:
llvm::Value* foo = 0;
llvm::ConstantInt* intValue = & llvm::cast< llvm::ConstantInt , llvm::Value >(foo );

//build error:
//error: no matching function for call to ‘cast(llvm::Value*&)’
What would be the correct approach here?
解决方案
Given llvm::Value* foo and you know that foo is actually a ConstantInt, I believe that the idiomatic LLVM code approach is to use dyn_cast as follows:
if (llvm::ConstantInt* CI = dyn_cast<llvm::ConstantInt>(foo)) {
  // foo indeed is a ConstantInt, we can use CI here
}
else {
  // foo was not actually a ConstantInt
}
If you're absolutely sure that foo is a ConstantInt and are ready to be hit with an assertion failure if it isn't, you can use cast instead of dyn_cast.



P.S. Do note that cast and dyn_cast are part of LLVM's own implementation of RTTI. dyn_cast acts somewhat similarly to the standard C++ dynamic_cast, though there are differences in implementation and performance (as can be read here).

                        
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