由于其他答案都没有谈到c++ 11中的规则，这里有一个。来自c++ 11标准(草案n3337)§5/9(强调了差异):

This pattern is called the usual arithmetic conversions, which are defined as follows: — If either operand is of scoped enumeration type, no conversions are performed; if the other operand does not have the same type, the expression is ill-formed. — If either operand is of type long double, the other shall be converted to long double. — Otherwise, if either operand is double, the other shall be converted to double. — Otherwise, if either operand is float, the other shall be converted to float. — Otherwise, the integral promotions shall be performed on both operands. Then the following rules shall be applied to the promoted operands: — If both operands have the same type, no further conversion is needed. — Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank shall be converted to the type of the operand with greater rank. — Otherwise, if the operand that has unsigned integer type has rank greater than or equal to the rank of the type of the other operand, the operand with signed integer type shall be converted to the type of the operand with unsigned integer type. — Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, the operand with unsigned integer type shall be converted to the type of the operand with signed integer type. — Otherwise, both operands shall be converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

这里有一个经常更新的列表。

涉及浮点数的算术运算结果为浮点数。

int + float = float
int * float = float
float * int = float
int / float = float
float / int = float
int / int = int

更多细节请回答。看看c++标准的§5/9节是怎么说的

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows: — If either operand is of type long double, the other shall be converted to long double. — Otherwise, if either operand is double, the other shall be converted to double. — Otherwise, if either operand is float, the other shall be converted to float. — Otherwise, the integral promotions (4.5) shall be performed on both operands.54) — Then, if either operand is unsigned long the other shall be converted to unsigned long. — Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int. — Otherwise, if either operand is long, the other shall be converted to long. — Otherwise, if either operand is unsigned, the other shall be converted to unsigned. [Note: otherwise, the only remaining case is that both operands are int ]

警告!

转换从左到右进行。

试试这个:

int i = 3, j = 2;
double k = 33;
cout << k * j / i << endl; // prints 22
cout << j / i * k << endl; // prints 0

表达式的类型，当不是两个部分是相同类型时，将转换为两者中最大的。这里的问题是理解哪一个比另一个大(它与字节大小无关)。

在包含实数和整数的表达式中，整数将升格为实数。例如，在int + float中，表达式的类型是float。

另一个区别与类型的能力有关。例如，包含int型和long int型的表达式的结果为long int型。

由于其他答案都没有谈到c++ 11中的规则，这里有一个。来自c++ 11标准(草案n3337)§5/9(强调了差异):

This pattern is called the usual arithmetic conversions, which are defined as follows: — If either operand is of scoped enumeration type, no conversions are performed; if the other operand does not have the same type, the expression is ill-formed. — If either operand is of type long double, the other shall be converted to long double. — Otherwise, if either operand is double, the other shall be converted to double. — Otherwise, if either operand is float, the other shall be converted to float. — Otherwise, the integral promotions shall be performed on both operands. Then the following rules shall be applied to the promoted operands: — If both operands have the same type, no further conversion is needed. — Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank shall be converted to the type of the operand with greater rank. — Otherwise, if the operand that has unsigned integer type has rank greater than or equal to the rank of the type of the other operand, the operand with signed integer type shall be converted to the type of the operand with unsigned integer type. — Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, the operand with unsigned integer type shall be converted to the type of the operand with signed integer type. — Otherwise, both operands shall be converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

这里有一个经常更新的列表。

我对这个问题的解决方案得到了WA(错误的答案)，然后我将int中的一个改为long long int，它给出了AC(接受)。之前，我试图做long long int += int * int，在我将其纠正为long long int += long long int * int后。我在谷歌上搜了一下，

1. 算术转换

类型转换条件:

条件满足—>转换

Either operand is of type long double. ---> Other operand is converted to type long double. Preceding condition not met and either operand is of type double. ---> Other operand is converted to type double. Preceding conditions not met and either operand is of type float. ---> Other operand is converted to type float. Preceding conditions not met (none of the operands are of floating types). ---> Integral promotions are performed on the operands as follows: If either operand is of type unsigned long, the other operand is converted to type unsigned long. If preceding condition not met, and if either operand is of type long and the other of type unsigned int, both operands are converted to type unsigned long. If the preceding two conditions are not met, and if either operand is of type long, t he other operand is converted to type long. If the preceding three conditions are not met, and if either operand is of type unsigned int, the other operand is converted to type unsigned int. If none of the preceding conditions are met, both operands are converted to type int.

2。整数转换规则

整数的促销活动:

小于int的整数类型在对其执行操作时被提升。如果原始类型的所有值都可以表示为int型，则较小类型的值转换为int型;否则，它将被转换为无符号整型。整数提升作为对某些参数表达式的常规算术转换的一部分应用;一元+、-和~操作符的操作数;和移位操作符的操作数。

Integer Conversion Rank: No two signed integer types shall have the same rank, even if they have the same representation. The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision. The rank of long long int shall be greater than the rank of long int, which shall be greater than the rank of int, which shall be greater than the rank of short int, which shall be greater than the rank of signed char. The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any. The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width. The rank of char shall equal the rank of signed char and unsigned char. The rank of any extended signed integer type relative to another extended signed integer type with the same precision is implementation-defined but still subject to the other rules for determining the integer conversion rank. For all integer types T1, T2, and T3, if T1 has greater rank than T2 and T2 has greater rank than T3, then T1 has greater rank than T3. Usual Arithmetic Conversions: If both operands have the same type, no further conversion is needed. If both operands are of the same integer type (signed or unsigned), the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank. If the operand that has unsigned integer type has rank greater than or equal to the rank of the type of the other operand, the operand with signed integer type is converted to the type of the operand with unsigned integer type. If the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, the operand with unsigned integer type is converted to the type of the operand with signed integer type. Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type. Specific operations can add to or modify the semantics of the usual arithmetic operations.