C++中typeid(var).name()输出的PKc具体含义是什么？

Great question! Let’s break down exactly what’s going on here and decode that mysterious PKc output you’re seeing.

First, let’s clarify the chain of calls #

• When you run typeid(check).name(), you’re getting GCC’s mangled shorthand for the type of check: 'c' for char, 'd' for double, etc.
• The critical detail you might have missed: std::type_info::name() always returns a const char*, no matter what type check is. This is a fixed part of the C++ standard.

Decoding PKc #

GCC uses a name mangling system to pack complex type names into short strings. Let’s unpack PKc piece by piece:

• P: Abbreviation for "pointer"
• K: Abbreviation for "const" (it comes from the German word Konstant, which is why it’s not C)
• c: Abbreviation for char

Put it all together, PKc is GCC’s mangled way of writing const char* — which is exactly the type of the value returned by typeid(check).name(). That’s why changing check’s type doesn’t change this output: you’re always asking for the type of the same const char* pointer.

Verify with demangled output #

If you want to see a human-readable version instead of the mangled string, you can use GCC’s built-in demangling function. Here’s a quick example:

#include <bits/stdc++.h>
#include <cxxabi.h>

int main() {
    double check = 69.666;
    const char* mangled_type = typeid(typeid(check).name()).name();
    
    int demangle_status;
    char* demangled = abi::__cxa_demangle(mangled_type, nullptr, nullptr, &demangle_status);
    
    if (demangle_status == 0) {
        std::cout << "Demangled type: " << demangled << "\n";
        free(demangled); // Clean up the allocated memory
    } else {
        std::cout << "Couldn't demangle the type name\n";
    }
    
    return 0;
}

Running this will output:

Demangled type: char const*

Which confirms our explanation perfectly.

内容的提问来源于stack exchange，提问作者pkdyn