英文:
math.Mod in Go returns integer part instead of floating-point remainder
问题
Golang的**math.Mod(10, 4)**返回2 - 即除法结果2.5的整数部分 - 但它不应该是“浮点数的余数”，也就是0.5吗？
英文:
Golang's math.Mod(10, 4) returns 2 -- ie. the integer part of division result 2.5 -- but shouldn't it be "the floating point remainder", that is, 0.5?
答案1
得分: 22
结果是正确的。math.Mod返回的是余数，在这个例子中实际上是2。它相当于%运算符，但适用于浮点数。
英文:
The result is correct. math.Mod returns the remainder, which really is 2 in this case. It's equivalent to the % operator, but for floating point numbers.
答案2
得分: -1
如果你想要在浮点数除法后获得小数部分的值，我理解你想要的是这样的代码：
double A = 10.0;
double B = 4.0;
double divResult = A / B;
double fractional = divResult % 1.0;
//打印结果为 "10.0 / 4.0 小数部分余数: 0.5"
System.out.printf("%f / %f 小数部分余数: %f", A , B, fractional);
英文:
if you're looking for the value after the decimal place after doing floating point division, as I understand, you would want something like this:
double A = 10.0;
double B = 4.0;
double divResult = A / B;
double fractional = divResult % 1.0;
//prints "10.0 / 4.0 fractional remainder: 0.5"
System.out.printf("%f / %f fractional remainder: %f", A , B, fractional);
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