如何转换为模板类型? [英] How do I cast to a templated type?

问题描述

在gdb中，如果您有指向某物的指针，则可以在打印它之前对其进行转换.

In gdb, if you have a pointer to something, you can cast it before printing it.

例如，这有效:

print *(int*) 0xDEADBEEF

但是，如何打印std::vector<T>?具体是std::vector<std::string>?

However, how do I print a std::vector<T>? Specifically a std::vector<std::string>?

如果它是std::string，我可以使用std::__cxx11::string(它是whatis std::string输出)来实现，但是我不能说服gdb喜欢std::vector<int>(作为示例).引用它无济于事，就像它说的那样，No symbol "std::vector<int>" in current context.

If it's std::string, I can do it with std::__cxx11::string, which whatis std::string outputs, but I can't convince gdb to like std::vector<int> (as an example). Quoting it doesn't help, as it says, No symbol "std::vector<int>" in current context.

推荐答案

一种做到这一点的方法是使用类型混乱的名称.例如，当前gcc和libstdc ++上std::vector<int>的错误名称是_ZSt6vectorIiSaIiEE，我通过在

One way you can do this is by using the mangled name of the type. For example, the mangled name of std::vector<int> on current gcc and libstdc++ is _ZSt6vectorIiSaIiEE, which I found by compiling the following code on the Compiler Explorer:

#include <vector>

void foo(std::vector<int>) {}
// Mangled symbol name: _Z3fooSt6vectorIiSaIiEE
// _Z means "this is C++".
// 3foo means "identifier 3 chars long, which is `foo`"
// Strip those off and you're left with: St6vectorIiSaIiEE
// Add the _Z back: _ZSt6vectorIiSaIiEE

std::vector<std::string>的错误名称是:_ZSt6vectorINSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEESaIS5_EE，可以用whatis进行验证.

The mangled name of std::vector<std::string> is: _ZSt6vectorINSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEESaIS5_EE, which can be verified with whatis.

实际执行演员表:

print *(_Z3fooSt6vectorINSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEESaIS5_EE*) 0xDEADBEEF

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