什么是由原始的指针和函数指针在C / C ++支持的操作？ [英] What are the operations supported by raw pointer and function pointer in C/C++?

问题描述

什么是由函数指针支持的所有操作从原始指针会有所不同？
是>，＆LT; ，＆LT; =，> =由原始指针支持若然是利用运营商？

What are all operations supported by function pointer differs from raw pointer? Is > , < , <= , >=operators supported by raw pointers if so what is the use?

推荐答案

对于这两个函数和对象指针，它们编译，但其结果只能保证是相同的完整对象的子对象的地址是一致的（你可以比较一类或数组的两个成员），如果你比较函数或对象对本身。

For both function and object pointers, they compile but their result is only guaranteed to be consistent for addresses to sub-objects of the same complete object (you may compare the addresses of two members of a class or array) and if you compare a function or object against itself.

使用的std ::以下＆LT;＆GT; ，的std ::更大＆LT;＆GT; 等会工作与任何指针类型，并且将提供一致的结果，即使各自内置的操作的结果是不确定的：

Using std::less<>, std::greater<> and so on will work with any pointer type, and will give consistent results, even if the result of the respective built-in operator is unspecified:

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;
}

那一切都应该编译虽然（尽管编译器是免费的警告它想要的任何code段）。

Everything of that should compile though (although the compiler is free to warn about any code snippet it wants).

---

因为函数类型具有指针对象类型没有的sizeof 价值，业务遍及的术语定义的sizeof 即会没有工作，其中包括：

Since function types have no sizeof value, operations that are defined in terms of sizeof of the pointee type will not work, these include:

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;

单目 + 适用于任何指针类型，并且将只返回它的价值，没有什么特别之处函数指针。

The unary + works on any pointer type, and will just return the value of it, there is nothing special about it for function pointers.

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

最后请注意，一个指向一个函数的指针的不是一个函数指针了：

Finally note that a pointer to a function pointer is not a function pointer anymore:

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

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