Question

我正在尝试在基本块上编写一个简单传递，代码如下：

struct SimplePass : BasicBlockPass, InstVisitor<SimplePass>
{
   ... some initialisation and some finalization code 
    virtual bool runOnBasicBlock(BasicBlock& B) {
        std::cout << "---This is a block divider---" << B.size() << std::endl;
        visit(B);
        return false;
    }
    void visitInstruction(Instruction& I){
        std::cout << "Visiting every single instruction:" << I.getOpcodeName(I.getOpcode()) << std::endl;
    }
    void visitBranchInst(BranchInst& I) {
        if(I.isUnconditional()) {
            std::cout << "Encountered an unconditional branch!" << std::endl;
        }
    }

}

非常奇怪的是我得到了这样的输出：

...
---This is a block divider---5
Visiting every single instruction:call
Visiting every single instruction:load
Visiting every single instruction:add
Visiting every single instruction:store
Encountered an unconditional branch!
---This is a block divider---7 
Visiting every single instruction:phi 
Visiting every single instruction:load
Visiting every single instruction:sub
Visiting every single instruction:call
Visiting every single instruction:load
Visiting every single instruction:icmp
---This is a block divider---3
......

很容易看出，在上面的两个块中，实际的指令数应该是5和7，但 visitInstrucion 函数有时不会访问基本块的最后一条指令，为什么这样做发生？这应该发生吗？

Answer 1

在第一个区块中：

Visiting every single instruction:call
Visiting every single instruction:load
Visiting every single instruction:add
Visiting every single instruction:store
Encountered an unconditional branch!

它是 5！最后一行来自您的void visitBranchInst(BranchInst& I)，其优先于visitInstruction。更具体的访问者优先于更通用的访问者。如果你想要调用visitInstruction，你必须明确地从更具体的访问者那里做到 - 它不会自动发生。

至于下一个块，也许它以一个有条件的分支结束？那么您的visitBranchInst不会打印任何内容，也不会传播到visitInstruction。

llvm：visitInstruction不访问基本块中的每条指令？

1 个答案: