GDB输出的解释

Question

我没怎么用过GDB。通常我检查简单变量但从不类。这次我正在检查类因为我开始一个新项目 (LLVM)。

这是

的输出

disp CachedTable

CachedTable = {<llvm::DenseMapBase<llvm::DenseMap<std::pair<unsigned int, unsigned int>, unsigned int, llvm::DenseMapInfo<std::pair<unsigned int, unsigned int> > >, std::pair<unsigned int, unsigned int>, unsigned int, llvm::DenseMapInfo<std::pair<unsigned int, unsigned int> > >> = {<No data fields>}, Buckets = 0x1a9e190, NumEntries = 2, NumTombstones = 0, NumBuckets = 64}

如何解释此输出？

感谢任何帮助。

Answer 1

这是 llvm::DenseMap class.
的数据成员的输出 (http://llvm.org/docs/doxygen/html/DenseMap_8h_source.html)

输出的第一部分表示它的基classllvm::DenseMapBase没有数据成员：

<llvm::DenseMapBase<llvm::DenseMap<std::pair<unsigned int, unsigned int>, unsigned int, llvm::DenseMapInfo<std::pair<unsigned int, unsigned int> > >, std::pair<unsigned int, unsigned int>, unsigned int, llvm::DenseMapInfo<std::pair<unsigned int, unsigned int> > >> = {<No data fields>}

其余输出打印 llvm::DenseMap 数据成员值：

Buckets = 0x1a9e190, NumEntries = 2, NumTombstones = 0, NumBuckets = 64

通常很难像这样解释一行输出，尤其是对于来自未知代码的长 structs/classes，因此值得打开 gdb 的漂亮打印模式：

(gdb) set print pretty on

参见Print Settings：

set print pretty on
    Cause gdb to print structures in an indented format with one member per line, like this:

               = {
                next = 0x0,
                flags = {
                  sweet = 1,
                  sour = 1
                },
                meat = 0x54 "Pork"
              }

GDB输出的解释

Interpretation of GDB output

c++

gdb