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c++ - clang AST 匹配 ostream <<

转载 作者:塔克拉玛干 更新时间:2023-11-03 07:41:58 27 4
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我正在尝试为涉及流式传输的某些场景编写一个 clang-tidy 检查。考虑这个简单的函数:

#include <iostream>
#include <stdint.h>

void foo(std::ostream& os, uint8_t i) {
os << i;
os << 4 << i;
}

如果我运行 clang-query测试一些匹配器:

$ clang-query foo.cxx --
clang-query> match cxxOperatorCallExpr(hasOverloadedOperatorName("<<"))
// [ ... snip some std library matches ... ]
Match #7:

foo.cxx:5:5: note: "root" binds here
os << i;
^~~~~~~

Match #8:

foo.cxx:6:5: note: "root" binds here
os << 4 << i;
^~~~~~~~~~~~

Match #9:

foo.cxx:6:5: note: "root" binds here
os << 4 << i;
^~~~~~~

这匹配 operator<< 的所有 3 种用法我在这个程序中有。伟大的。但是,如果我尝试为第一个参数添加一个过滤器:

clang-query> match cxxOperatorCallExpr(hasOverloadedOperatorName("<<"), hasArgument(0, expr(hasType(asString("std::ostream")))))

Match #1:

foo.cxx:5:5: note: "root" binds here
os << i;
^~~~~~~

Match #2:

foo.cxx:6:5: note: "root" binds here
os << 4 << i;
^~~~~~~

就是这样。它不匹配整个表达式 os << 4 << i .为什么不?该表达式确实具有类型 std::ostream .如果我直接转储 AST,我会看到:

|-CXXOperatorCallExpr 0x953f618 <line:6:5, col:16> 'basic_ostream<char, struct std::char_traits<char> >':'class std::basic_ostream<char>' lvalue
| |-ImplicitCastExpr 0x953f600 <col:13> 'basic_ostream<char, struct std::char_traits<char> > &(*)(basic_ostream<char, struct std::char_traits<char> > &, unsigned char)' <FunctionToPointerDecay>
| | `-DeclRefExpr 0x953f5d8 <col:13> 'basic_ostream<char, struct std::char_traits<char> > &(basic_ostream<char, struct std::char_traits<char> > &, unsigned char)' lvalue Function 0x94bcd40 'operator<<' 'basic_ostream<char, struct std::char_traits<char> > &(basic_ostream<char, struct std::char_traits<char> > &, unsigned char)'
| |-CXXOperatorCallExpr 0x953f160 <col:5, col:11> 'std::basic_ostream<char, struct std::char_traits<char> >::__ostream_type':'class std::basic_ostream<char>' lvalue
| | |-ImplicitCastExpr 0x953f148 <col:8> 'std::basic_ostream<char, struct std::char_traits<char> >::__ostream_type &(*)(int)' <FunctionToPointerDecay>
| | | `-DeclRefExpr 0x953f0c0 <col:8> 'std::basic_ostream<char, struct std::char_traits<char> >::__ostream_type &(int)' lvalue CXXMethod 0x94b7740 'operator<<' 'std::basic_ostream<char, struct std::char_traits<char> >::__ostream_type &(int)'
| | |-DeclRefExpr 0x953ec88 <col:5> 'std::ostream':'class std::basic_ostream<char>' lvalue ParmVar 0x953e470 'os' 'std::ostream &'
| | `-IntegerLiteral 0x953ecb0 <col:11> 'int' 4
| `-ImplicitCastExpr 0x953f5c0 <col:16> 'uint8_t':'unsigned char' <LValueToRValue>
| `-DeclRefExpr 0x953f1a8 <col:16> 'uint8_t':'unsigned char' lvalue ParmVar 0x953e500 'i' 'uint8_t':'unsigned char'

内外兼修CXXOperatorCallExpr说他们的表达是class std::basic_ostream<char> lvalue .实际正确匹配此参数的正确方法是什么?

最佳答案

肯定是第一个子调用<<的返回类型有问题,请参阅以下匹配器,它也与 "class std::__1::basic_ostream<char>" 匹配以及这个丢失的电话:

clang-query> match callExpr(hasArgument(0, expr(hasType(anyOf(asString("class std::__1::basic_ostream<char>"), asString("std::ostream"))))))

Match #1:

/Users/ug/clangtest/main.cc:5:5: note: "root" binds here
os << i;
^~~~~~~

Match #2:

/Users/ug/clangtest/main.cc:6:5: note: "root" binds here
os << 4 << i;
^~~~~~~~~~~~

Match #3:

/Users/ug/clangtest/main.cc:6:5: note: "root" binds here
os << 4 << i;
^~~~~~~
3 matches.

关于c++ - clang AST 匹配 ostream <<,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48068716/

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