char *不应该隐式转换为std :: string吗？

Question

Here is my code:

#include <iostream>
using namespace std;

struct ST {};

bool operator==(const struct ST *s1, const string &s2) {
    return true;
}

int main() {
    struct ST *st = new ST();
    const char *p = "abc";

    if (st == p) {
        return 0;
    }

    return 1;
}

I get compile error:

prog.cpp:14:12: error: comparison between distinct pointer types ‘ST*’ and ‘const char*’ lacks a cast [-fpermissive]
  if (st == p) {
            ^

I wonder why the implicit conversion from char* to string does not work here?

UPDATE Anton's answer makes sense, I updated the code:

#include <string>
using namespace std;

struct ST {};

bool operator==(const struct ST s1, const string &s2) {
    return true;
}

int main() {
    struct ST st;
    const char *p = "abc";

    if (st == p) {
        return 0;
    }

    return 1;
}

Now it compiles.

Answer 1

§13.3.1.2 Operators in expressions [over.match.oper] states:

If no operand of an operator in an expression has a type that is a class or an enumeration, the operator is assumed to be a built-in operator and interpreted according to Clause 5.

This is exactly your case: operator== 's arguments are pointers, so it's considered to be built-in and compiler doesn't look for possible overloads.

Answer 2

Absolutely not.

First of all, you are trying to use a pointer in place of a reference. Any similarity between a pointer and a reference is a implementation detail. Remember the motto: "Implementation Is Irrelevant!"

Next, and more directly to your question, a std::string and a char* are quite different, even if they are used to represent the same things. Conversion between them was deliberately made difficult to prevent using them interchangably.