c++ - char* 不应该隐式转换为 std::string 吗？

Question

这是我的代码:

#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;
}

编译错误:

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

我想知道为什么从 char* 到 string 的隐式转换在这里不起作用？

更新 安东的回答有道理，我更新了代码:

#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;
}

现在可以编译了。

Answer 1

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

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.

这正是您的情况:operator== 的参数是指针，因此它被认为是内置的，编译器不会寻找可能的重载。