问为什么C++表达式会涉及重载运算符和隐式转换？

struct Foo {
    Foo() {}
    Foo(int x) {}

    operator bool() const { return false; }

    friend bool operator<(const Foo& a, const Foo& b) {
        return true;
    }
};

int main() {
    Foo a, b;
    if (a < 0) {
        a = 0;
    }
    return 1;
}

g++ foo.cpp
foo.cpp: In function 'int main()':
foo.cpp:18:11: error: ambiguous overload for 'operator<' (operand types are 'Foo' and 'int')
     if (a < 0) {
           ^
foo.cpp:18:11: note: candidate: operator<(int, int) <built-in>
foo.cpp:8:17: note: candidate: bool operator<(const Foo&, const Foo&)
     friend bool operator<(const Foo& a, const Foo& b)