3* ( *p)++ 为什么先算3 * ( *p)而不先算自增？++的优先级不是优先于乘法吗
++写在前面会先进行自增，再按优先级顺序进行运算。
++写在后面会先按优先级顺序完成除了自增以外的所有运算，最后进行自增。
( *p)++ 是先计算求值，但是返回的是*p在自增之前的值。
3* (++(*p))，这才是返回自增之后的值
自增运算符写在后面都是当前语句结束后在执行
//C++ Operators
// Operators specify an evaluation to be performed on one of the following:
// One operand (unary operator)
// Two operands (binary operator)
// Three operands (ternary operator)
// The C++ language includes all C operators and adds several new operators.
// Table 1.1 lists the operators available in Microsoft C++.
// Operators follow a strict precedence which defines the evaluation order of
//expressions containing these operators. Operators associate with either the
//expression on their left or the expression on their right; this is called
//“associativity.” Operators in the same group have equal precedence and are
//evaluated left to right in an expression unless explicitly forced by a pair of
//parentheses, ( ).
// Table 1.1 shows the precedence and associativity of C++ operators
// (from highest to lowest precedence).
//
//Table 1.1 C++ Operator Precedence and Associativity
// The highest precedence level is at the top of the table.
//+------------------+-----------------------------------------+---------------+
//| Operator | Name or Meaning | Associativity |
//+------------------+-----------------------------------------+---------------+
//| :: | Scope resolution | None |
//| :: | Global | None |
//| [ ] | Array subscript | Left to right |
//| ( ) | Function call | Left to right |
//| ( ) | Conversion | None |
//| . | Member selection (object) | Left to right |
//| -> | Member selection (pointer) | Left to right |
//| ++ | Postfix increment | None |
//| -- | Postfix decrement | None |
//| new | Allocate object | None |
//| delete | Deallocate object | None |
//| delete[ ] | Deallocate object | None |
//| ++ | Prefix increment | None |
//| -- | Prefix decrement | None |
//| * | Dereference | None |
//| & | Address-of | None |
//| + | Unary plus | None |
//| - | Arithmetic negation (unary) | None |
//| ! | Logical NOT | None |
//| ~ | Bitwise complement | None |
//| sizeof | Size of object | None |
//| sizeof ( ) | Size of type | None |
//| typeid( ) | type name | None |
//| (type) | Type cast (conversion) | Right to left |
//| const_cast | Type cast (conversion) | None |
//| dynamic_cast | Type cast (conversion) | None |
//| reinterpret_cast | Type cast (conversion) | None |
//| static_cast | Type cast (conversion) | None |
//| .* | Apply pointer to class member (objects) | Left to right |
//| ->* | Dereference pointer to class member | Left to right |
//| * | Multiplication | Left to right |
//| / | Division | Left to right |
//| % | Remainder (modulus) | Left to right |
//| + | Addition | Left to right |
//| - | Subtraction | Left to right |
//| << | Left shift | Left to right |
//| >> | Right shift | Left to right |
//| < | Less than | Left to right |
//| > | Greater than | Left to right |
//| <= | Less than or equal to | Left to right |
//| >= | Greater than or equal to | Left to right |
//| == | Equality | Left to right |
//| != | Inequality | Left to right |
//| & | Bitwise AND | Left to right |
//| ^ | Bitwise exclusive OR | Left to right |
//| | | Bitwise OR | Left to right |
//| && | Logical AND | Left to right |
//| || | Logical OR | Left to right |
//| e1?e2:e3 | Conditional | Right to left |
//| = | Assignment | Right to left |
//| *= | Multiplication assignment | Right to left |
//| /= | Division assignment | Right to left |
//| %= | Modulus assignment | Right to left |
//| += | Addition assignment | Right to left |
//| -= | Subtraction assignment | Right to left |
//| <<= | Left-shift assignment | Right to left |
//| >>= | Right-shift assignment | Right to left |
//| &= | Bitwise AND assignment | Right to left |
//| |= | Bitwise inclusive OR assignment | Right to left |
//| ^= | Bitwise exclusive OR assignment | Right to left |
//| , | Comma | Left to right |
//+------------------+-----------------------------------------+---------------+
【以下回答由 GPT 生成】
这个问题是由于表达式的解析顺序引起的。表达式的解析顺序由运算符优先级和结合性决定。
在C/C++中，一元运算符（如自增操作符++）的优先级是最高的，而二元运算符（如乘法操作符*）的优先级较低。然而，表达式的解析顺序不仅仅取决于运算符的优先级，还受到括号的影响。
在这个特定的表达式3(p)中，由于括号的存在，解析器会首先计算*p，然后再将结果与3相乘。这是因为括号具有更高的优先级，它们会改变运算符的解析顺序。
为了更好地理解，让我们来看一个简化的例子：
int x = 2;
int* p = &x;
int result = 3 * (*p);
在这个例子中，我们首先声明一个变量x并将其初始化为2。然后，我们声明一个指针变量p并将其指向x。最后，我们计算表达式3 * (*p)并将结果保存在result变量中。
在计算表达式3 * (*p)时，解析器首先执行括号内的操作，即解引用指针p并获取其指向的值x的内容。然后，它将此值2乘以3，得到最终的结果6。
因此，表达式3(p)中的自增操作符并没有被计算，因为它在这个表达式中并不存在。如果你想使用自增操作符，可以按照以下方式修改代码：
int x = 2;
int* p = &x;
++(*p); // 先执行自增操作
int result = 3 * (*p); // 再执行乘法运算
在这个修改后的版本中，我们先执行自增操作符++来递增指针p指向的值x。然后，我们再计算乘法表达式3 * (*p)并将结果保存在result变量中。
总结起来，表达式的解析顺序取决于运算符的优先级和结合性，并受到括号的影响。在处理表达式时，要注意运算符的优先级和结合性以及括号的使用，以确保表达式得到正确的求值顺序。
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