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();
}

它给出12，就像它应该考虑20 * 0.6 = 12.但是，如果我将所有浮点数改为双打：

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

它给了我11个。并且为了使事情变得更加奇怪，如果我更改代码以使值首先存储在变量中然后再存储在变量中，则会再次给出正确的答案（12）。

#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不能完全以任何二进制浮点格式表示。有时它稍大一些，有点小，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。

“内存中存储”版本不同，因为x87 FPU内部使用80位浮点寄存器。

编辑：详细计算：

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

单精度和双精度铸造提供不同的答案

1 个答案: