单精度和双精度铸造给出了不同的答案

Question

所以我有以下代码：

#include <iostream> 
int main(){ 
    float number = 0.6; 
    int anotherNumber = 20; 
    int value = (int) ((float) anotherNumber * number); 
    std::cout << value; 
    std::cin.get(); 
} 

它提供了12，像它应该考虑20 * 0.6 = 12。但是，如果我改变所有彩车双打：

#include <iostream> 
int main(){ 
    double number = 0.6; 
    int anotherNumber = 20; 
    int value = (int) ((double) anotherNumber * number); 
    std::cout << value; 
    std::cin.get(); 
} 

它给我11代替。而为了让事情变得怪异，如果我更改了代码，以便值存储在一个变量，然后再案件后，它给出了正确的答案（12）一次。

#include <iostream> 
int main(){ 
    double number = 0.6; 
    int anotherNumber = 20; 
    double intermediate = (double) anotherNumber * number; 
    int value = (int) intermediate; 
    std::cout << value; 
    std::cin.get(); 
} 

上帝的名字在这里发生了什么？我正在使用g ++ 4.5.3进行编译。

Answer 1

0.6不能在任何二进制浮点格式准确表示。有时候它会稍大一点，但有点小一点，这取决于数据类型。有关详细说明，请参阅What Every Programmer Should Know About Floating-Point Arithmetic。

在“在存储器中存储”版本不同，因为的x87 FPU使用80位的浮点寄存器内部。

编辑：详细的计算：

float 0.6 in memory; 
.100110011001100110011010 
loaded to register: 
.10011001100110011001101000000000000000000000000000000000000000000 
multiplied by 20: 
1100.0000000000000000000010000000000000000000000000000000000000000 
rounded down: 
1100 

double 0.6 in memory 
.10011001100110011001100110011001100110011001100110011 
loaded to register: 
.10011001100110011001100110011001100110011001100110011000000000000 
multiplied by 20: 
1011.1111111111111111111111111111111111111111111111111110000000000 
rounded down: 
1011 

double 0.6 in memory 
.10011001100110011001100110011001100110011001100110011 
loaded to register: 
.10011001100110011001100110011001100110011001100110011000000000000 
multiplied by 20: 
1011.1111111111111111111111111111111111111111111111111110000000000 
converted to double-precision and stored to memory: 
1100.0000000000000000000000000000000000000000000000000 
loaded to register: 
1100.0000000000000000000000000000000000000000000000000000000000000 
rounded down: 
1100