c++ - C/C++中原始指针和函数指针支持的操作有哪些？

Question

函数指针支持的所有操作与原始指针有哪些不同？> ， < ， <= ， >= 原始指针支持的运算符如果是这样有什么用？

Answer 1

对于函数指针和对象指针，它们都可以编译，但它们的结果只能保证同一完整对象的子对象的地址一致(您可以比较类或数组的两个成员的地址)并且如果您将函数或对象与自身进行比较。

使用 std::less<> , std::greater<>等等将适用于任何指针类型，并会给出一致的结果，即使相应内置运算符的结果未指定也是如此:

void f() { }
void g() { }

int main() {
  int a, b;

  ///// not guaranteed to pass
  assert((&a < &b) == (&a < &b));

  ///// guaranteed to pass
  std::less<int*> lss1;
  assert(lss1(&a, &b) == lss1(&a, &b));
  // note: we don't know whether lss1(&a, &b) is true or false. 
  //       But it's either always true or always false. 

  ////// guaranteed to pass
  int c[2];
  assert((&c[0] < &c[1]) == (&c[0] < &c[1]));
  // in addition, the smaller index compares less:
  assert(&c[0] < &c[1]);

  ///// not guaranteed to pass
  assert((&f < &g) == (&f < &g));

  ///// guaranteed to pass
  assert((&g < &g) == (&g < &g));
  // in addition, a function compares not less against itself. 
  assert(!(&g < &g));

  ///// guaranteed to pass
  std::less<void(*)()> lss2;
  assert(lss2(&f, &g) == lss2(&f, &g));
  // note: same, we don't know whether lss2(&f, &g) is true or false.

  ///// guaranteed to pass
  struct test {
    int a;
  // no "access:" thing may be between these!
    int b;

    int c[1];
  // likewise here
    int d[1];

    test() {
      assert((&a < &b) == (&a < &b));
      assert((&c[0] < &d[0]) == (&c[0] < &d[0]));

      // in addition, the previous member compares less:
      assert((&a < &b) && (&c[0] < &d[0]));
    }
  } t;
}

虽然所有这些都应该编译(尽管编译器可以自由警告它想要的任何代码片段)。

---

因为函数类型没有 sizeof值，根据 sizeof 定义的操作pointee 类型的将不起作用，这些包括:

void(*p)() = ...;
// all won't work, since `sizeof (void())` won't work.
// GCC has an extension that treats it as 1 byte, though.
p++; p--; p + n; p - n; 

一元 +适用于任何指针类型，并且只会返回它的值，函数指针没有什么特别之处。

+ p; // works. the result is the address stored in p.

最后请注意，指向函数的指针 pointer 不再是函数指针:

void (**pp)() = &p;
// all do work, because `sizeof (void(*)())` is defined.
pp++; pp--; pp + n; pp - n;