單精度和雙精度鑄造給出了不同的答案

Question

所以我有以下代碼：

#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，像它應該考慮20 * 0.6 = 12。但是，如果我改變所有彩車雙打：

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

它給我11代替。而爲了讓事情變得怪異，如果我更改了代碼，以便值存儲在一個變量，然後再案件後，它給出了正確的答案（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(); 
} 

上帝的名字在這裏發生了什麼？我正在使用g ++ 4.5.3進行編譯。

Answer 1

0.6不能在任何二進制浮點格式準確表示。有時候它會稍大一點，但有點小一點，這取決於數據類型。有關詳細說明，請參閱What Every Programmer Should Know About Floating-Point Arithmetic。

在「在存儲器中存儲」版本不同，因爲的x87 FPU使用80位的浮點寄存器內部。

編輯：詳細的計算：

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