有谁能解释一下为什么bool
与if (a < 0)
表达式可以作为特定的运算符,是否有解决办法?
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;
}
当编译时,得到:
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)
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