c++ - 指针算术递增后缀/前缀

Question

我无法理解下面一段特定代码的逻辑。

int i[] = { 21, 4, -17, 45 };

int* i_ptr = i;

std::cout << (*i_ptr)++ << std::endl;   // 21

std::cout << *i_ptr << std::endl;       // 22

std::cout << *i_ptr++ << std::endl;     // 22

std::cout << *(i_ptr - 1) << std::endl; // 22

std::cout << *i_ptr << std::endl;       // 4

std::cout << ++*i_ptr << std::endl;     // 5

std::cout << *++i_ptr << std::endl;     // -17

system("pause");

我的问题是这段代码是如何从 22...开始的

std::cout << *(i_ptr - 1) << std::endl; // 22

到 4。

std::cout << *i_ptr << std::endl;       // 4

然后到 5.

std::cout << ++*i_ptr << std::endl;     // 5

当我第一次浏览这段代码时，我以为 22 只是从 22 到 21。我知道它与 C++ 运算符优先级有关，但这对我来说毫无意义。

Answer 1

std::cout << (*i_ptr)++ << std::endl;   // 21
//i_ptr points to i[0], which is increased from 21 to 22

std::cout << *i_ptr << std::endl;       // 22
//prints i[0], which is 22

std::cout << *i_ptr++ << std::endl;     // 22
//prints i[0] and increments i_ptr to point to i[1]

std::cout << *(i_ptr - 1) << std::endl; // 22
//prints i[0], i_ptr points to i[1], so i_ptr - 1 points to i[0]

std::cout << *i_ptr << std::endl;       // 4
//prints i[1], which is 4

std::cout << ++*i_ptr << std::endl;     // 5
//prints the incremented i[1], which was 4 and is 5 now

std::cout << *++i_ptr << std::endl;     // -17
//increment i_ptr to point to i[2] and prints the value