c++ - 保持平等和稳定有什么区别？

Question

根据draft :

An expression is equality-preserving if, given equal inputs, the expression results in equal outputs.[...]

和

[...]stable: two evaluations of such an expression with the same input objects are required to have equal outputs absent any explicit intervening modification of those input objects.[...]

强调我的。

• 这些有什么区别？

• 什么时候表达式可以保持相等但不是稳定和反之亦然？

Answer 1

对于修改其输入的操作，两者会有所不同。

// stable and equality preserving
int foo(int a, int b) { 
    return a + b;
}

// equality preserving, but not stable:
int bar(int a, int &b) { 
    auto ret = a + b;
    ++b;
    return ret;
}

例如:

int x = 1, y = 2;
int z = foo(x, y); // produces 3

int z2 = foo(x, y);  // still produces 3

int zz = bar(x, y); // produces 3
int zz2 = bar(x, y); // produces 4

至于稳定但不保持相等的东西，是的，这也是可能的(对于“相等”的某些定义)。

举一个简单的例子，考虑这样的事情:

struct foo { 
    int bar;

    // they're all equal, just some are more equal than others
    bool operator==(foo const &) const { return true; }
};

int operator+(foo a, foo b) { return a.bar + b.bar; }

foo a{1};
foo b{2};
foo c{3};

// obviously both true
assert(a == b);
assert(b == c);

int x = a + b;
int y = b + c;

assert(x != y); // of course 1 + 2 != 2 + 3;