C++ 将 int 转换为 double
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C++ casting int to double
提问by nenito
I'm not a C++ developer, but today I've found a C++ code and try to understand it. So I've stacked on this piece of code:
我不是 C++ 开发人员，但今天我找到了 C++ 代码并尝试理解它。所以我堆积了这段代码：
int m = 2, n = 3, i = 1;
double mid = (double)m / n * i;
int d = (int)mid + 1;
printf("%d %d\n", mid, d);
The result which is going to be printed to the console is: 1431655765 1071994197. It seems to be related with the casting of variable m to double, but I have no idea how it is happening. I need someone to help me understand it. Thanks in advance!
将要打印到控制台的结果是：1431655765 1071994197。这似乎与将变量 m 强制转换为双倍有关，但我不知道它是如何发生的。我需要有人帮助我理解它。提前致谢！
回答by Ivaylo Strandjev
You should print a double(mid) with the %lfformat specifier in printf.
您应该打印双（mid）与%lf格式符printf。
回答by PoP
Changing the printf to
将 printf 更改为
printf("%f %i\n", mid, d);
will actually print what you expect i.e., 0.666667 1
实际上会打印您期望的内容，即 0.666667 1
回答by Emeche Cash
A simpler way to solve it would be
一种更简单的解决方法是
double m_Doubled;
m_Doubled = static_cast(m);
