Question

我需要确定一组输入的标准偏差。这需要（每个输入 - 平均值）^ 2的总和。我有以下代码来执行此操作（510是用于测试的示例输入）：

int testDiff = 510;       
console.printf("testDiff = %i \r\n",testDiff);

double testDiff_double = static_cast<double>(testDiff);
console.printf("testDiff_double = %d \r\n",testDiff_double);

double result = pow(static_cast<double>(testDiff),2);
console.printf("result = %d \r\n",result);

但是，此代码不会生成预期输出（260100）。如果我打印在每个阶段获得的值，我会得到以下结果：

testDiff = 510
testDiff_double = 0
pow = 0

为什么从整数到双精度的转换失败？我正在使用static_cast，所以如果转换存在逻辑问题，我在编译时会遇到错误。

注意（虽然没关系）：我在MBED微控制器上运行此代码。

Answer 1

因为您使用%d而不是%f来显示浮点值。

Answer 2

在printf函数中，尝试使用％f而不是％d。

Answer 3

您必须使用%f或%e或%E或%g才能显示双/浮点数。

来自printf reference page：

c   Character   
d or i  Signed decimal integer
e   Scientific notation (mantise/exponent) using e character
E   Scientific notation (mantise/exponent) using E character
f   Decimal floating point  
g   Use the shorter of %e or %f
G   Use the shorter of %E or %f
o   Unsigned octal
s   String of characters
u   Unsigned decimal integer
x   Unsigned hexadecimal integer    
X   Unsigned hexadecimal integer (capital letters)
p   Pointer address
n   Nothing printed. The argument must be a pointer to a signed int, where the number of characters written so far is stored.   
%   A % followed by another % character will write % to stdout.

将整数转换为double

3 个答案: