C++ C++；运算符优先查询

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前缀++和*的优先级相同。两者的关联性是从右向左的

后缀++的优先级高于*和前缀++。postfix++的关联性是从左到右的

第一个代码示例：

int x[4] = {0, 1, 2, 3};
int *ptr = x;

cout << x[0] << " at " << &x[0] << "\n";
cout << x[1] << " at " << &x[1] << "\n";
cout << x[2] << " at " << &x[2] << "\n";
cout << x[3] << " at " << &x[3] << "\n";
cout << "*ptr = " << *ptr << " at " << ptr << "\n";
cout << "*++ptr = " << *++ptr << " at " << ptr << "\n";
cout << "++*ptr = " << ++*ptr << " at " << ptr << "\n";
cout << "*ptr++ = " << *ptr++ << " at " << ptr << "\n";

int cd = 7;
cout << "cd = " << cd << " at " << &cd<< "\n";
cout << "++cd = " << ++cd << " at " << &cd << "\n";
cout << "cd++ = " << cd++ << " at " << &cd << "\n";

int c = 10;
int d = 1;
cout << c << " at " << &c << "\n";
int e = c+++d;
cout << c << " at " << &c << "\n";
cout << d << " at " << &d << "\n";
cout << e << " at " << &e << "\n";

在最后一个cout语句中，除了增量后指针，它在使用之前首先递增指针“ptr”值

第二个代码示例：

int x[4] = {0, 1, 2, 3};
int *ptr = x;

cout << x[0] << " at " << &x[0] << "\n";
cout << x[1] << " at " << &x[1] << "\n";
cout << x[2] << " at " << &x[2] << "\n";
cout << x[3] << " at " << &x[3] << "\n";
cout << "*ptr = " << *ptr << " at " << ptr << "\n";
cout << "*++ptr = " << *++ptr << " at " << ptr << "\n";
cout << "++*ptr = " << ++*ptr << " at " << ptr << "\n";
cout << "*ptr++ = " << *ptr++ << " at " << ptr << "\n";

int cd = 7;
cout << "cd = " << cd << " at " << &cd<< "\n";
cout << "++cd = " << ++cd << " at " << &cd << "\n";
cout << "cd++ = " << cd++ << " at " << &cd << "\n";

int c = 10;
int d = 1;
cout << c << " at " << &c << "\n";
int e = c+++d;
cout << c << " at " << &c << "\n";
cout << d << " at " << &d << "\n";
cout << e << " at " << &e << "\n";

在这里的最后一个cout语句中，请注意cd递增，然后使用post increment操作符进行访问

第三个代码示例：

int x[4] = {0, 1, 2, 3};
int *ptr = x;

cout << x[0] << " at " << &x[0] << "\n";
cout << x[1] << " at " << &x[1] << "\n";
cout << x[2] << " at " << &x[2] << "\n";
cout << x[3] << " at " << &x[3] << "\n";
cout << "*ptr = " << *ptr << " at " << ptr << "\n";
cout << "*++ptr = " << *++ptr << " at " << ptr << "\n";
cout << "++*ptr = " << ++*ptr << " at " << ptr << "\n";
cout << "*ptr++ = " << *ptr++ << " at " << ptr << "\n";

int cd = 7;
cout << "cd = " << cd << " at " << &cd<< "\n";
cout << "++cd = " << ++cd << " at " << &cd << "\n";
cout << "cd++ = " << cd++ << " at " << &cd << "\n";

int c = 10;
int d = 1;
cout << c << " at " << &c << "\n";
int e = c+++d;
cout << c << " at " << &c << "\n";
cout << d << " at " << &d << "\n";
cout << e << " at " << &e << "\n";

我们看到，++在语句中访问vars值后，会增加vars值

问题是第三个代码示例与前两个代码示例不同，为什么增量后运算符在访问变量“c”之前不增加它的值？ 为什么在上一个代码示例中，变量没有收到12的值

“为什么增量后运算符不增加变量的值 “c”在访问它之前

因为它是一个后增量操作符。它计算其操作数，然后递增

所以

inte=c+++d被解释为“将
c
+
d
分配给
e
，并增加
c
”

在前两个例子中，你编写了一个代码，它改变一个变量的值，并对该变量进行评估，没有任何序列点来限制操作的顺序。
是<代码> CUT @弗兰Cou-OISANDURIX（C++（？）：）的版本]Yakk AdamNevraumont为什么以及如何改变？看起来它不再是UB了，因为c++17是可信的。我想我必须重新学习这一切。编辑：虽然我找不到什么改变了。后续：谢谢蒂姆的回复。这是我们所期望的。但在前两个代码示例中，情况并非如此。为什么在前两个示例代码中它首先递增并计算操作数？谢谢。我补充了答案