在一个C程序中，我尝试了以下操作(只是为了检查行为)
x = 5 % (-3);
y = (-5) % (3);
z = (-5) % (-3);
printf("%d ,%d ,%d", x, y, z);
在gcc中输出为(2，-2，-2)我以为每次都会有积极的结果。模量可以是负的吗?有人能解释一下这种行为吗?
当前回答
根据C99标准，第6.5.5节 乘法运算符，需要以下条件:
(a / b) * b + a % b = a
结论
余数运算结果的符号 到C99，和红利是一样的。
让我们看一些例子(除数/除数):
只有股息是负的
(-3 / 2) * 2 + -3 % 2 = -3
(-3 / 2) * 2 = -2
(-3 % 2) must be -1
当只有除数为负时
(3 / -2) * -2 + 3 % -2 = 3
(3 / -2) * -2 = 2
(3 % -2) must be 1
除数和被除数都为负
(-3 / -2) * -2 + -3 % -2 = -3
(-3 / -2) * -2 = -2
(-3 % -2) must be -1
6.5.5 Multiplicative operators Syntax multiplicative-expression: cast-expression multiplicative-expression * cast-expression multiplicative-expression / cast-expression multiplicative-expression % cast-expression Constraints Each of the operands shall have arithmetic type. The operands of the % operator shall have integer type. Semantics The usual arithmetic conversions are performed on the operands. The result of the binary * operator is the product of the operands. The result of the / operator is the quotient from the division of the first operand by the second; the result of the % operator is the remainder. In both operations, if the value of the second operand is zero, the behavior is undefined. When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded [1]. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a. [1]: This is often called "truncation toward zero".
其他回答
根据C99标准，第6.5.5节 乘法运算符，需要以下条件:
(a / b) * b + a % b = a
结论
余数运算结果的符号 到C99，和红利是一样的。
让我们看一些例子(除数/除数):
只有股息是负的
(-3 / 2) * 2 + -3 % 2 = -3
(-3 / 2) * 2 = -2
(-3 % 2) must be -1
当只有除数为负时
(3 / -2) * -2 + 3 % -2 = 3
(3 / -2) * -2 = 2
(3 % -2) must be 1
除数和被除数都为负
(-3 / -2) * -2 + -3 % -2 = -3
(-3 / -2) * -2 = -2
(-3 % -2) must be -1
6.5.5 Multiplicative operators Syntax multiplicative-expression: cast-expression multiplicative-expression * cast-expression multiplicative-expression / cast-expression multiplicative-expression % cast-expression Constraints Each of the operands shall have arithmetic type. The operands of the % operator shall have integer type. Semantics The usual arithmetic conversions are performed on the operands. The result of the binary * operator is the product of the operands. The result of the / operator is the quotient from the division of the first operand by the second; the result of the % operator is the remainder. In both operations, if the value of the second operand is zero, the behavior is undefined. When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded [1]. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a. [1]: This is often called "truncation toward zero".
模运算的结果取决于分子的符号，因此y和z都是-2
这是参考资料
http://www.chemie.fu-berlin.de/chemnet/use/info/libc/libc_14.html
整数的除法 介绍整数除法的函数。 这些函数在GNU C库中是多余的，因为在GNU C中 '/'运算符总是四舍五入到零。但是在其他C中 实现中，'/'可以用不同的负参数四舍五入。 Div和ldiv很有用，因为它们指定了如何舍入 商:趋于零。余数的符号和 分子。
C99要求当a/b是可表示的时:
(a/b) * b + a%b等于a
从逻辑上讲，这是有道理的。对吧?
让我们看看这会导致什么:
例A. 5/(-3) = -1
=> (-1) * (-3) + 5%(-3) = 5
这只能在5%(-3)= 2时发生。
例b (-5)/3 = -1
=> (-1) * 3 + (-5)%3 = -5
只有当(-5)%3为-2时才会发生这种情况
我认为没有必要检查数字是否为负。
求正模的一个简单函数是这个-
编辑:假设N > 0和N + N - 1 <= INT_MAX
int modulo(int x,int N){
return (x % N + N) %N;
}
这对x的正负都成立。
原始p.s.:也正如@chux指出的那样，如果你的x和N可能分别达到INT_MAX-1和INT_MAX，只需将int替换为long long int。
如果它们也越过了long long的限制(即在LLONG_MAX附近)，那么你应该分别处理正的和负的情况，在这里的其他答案中描述。
其他答案已经在C99或更高版本中解释过，涉及负操作数的整数除法总是截断为零。
注意，在C89中，结果向上舍入还是向下舍入是由实现定义的。因为(a/b) * b + a%b在所有标准中都等于a，包含负操作数的%的结果也是在C89中实现定义的。
