c - 负数问题

Question

我有这个“简单”代码。

union
{
    unsigned int res;
    char         bytes[2];
} ADC;

char ADC_num[5];
float temprature;

void vis_temp (void)        //  Show temp..
{ 
    signed int l, length;
    unsigned int rem;

    GO_nDONE=1;             // initiate conversion on the channel 0   

    while (GO_nDONE) continue;

    ADC.bytes[0]=ADRESL;
    ADC.bytes[1]=ADRESH;

    utoa(ADC_num, ADC.res, 10);

    temprature = (float) ADC.res * 478.1 / 1024;
    temprature = temprature - 50.0;

    l = (signed int) temprature;
    temprature -= (float) l;  
    rem = (unsigned int)(temprature* 1e1);

    sprintf(&ADC_num, "%i.%u", l, rem);

当读取ADC_res(引脚上的电压，温度传感器)时，温度为 0 度或以下，程序将写入“0.65500”而不是“-3.5”或类似内容。我应该将权利声明为有符号和无符号整数。任何修复它的提示，或者有其他转换它的方法。

Answer 1

temprature = (float) ADC.res * 478.1 / 1024;
temprature = temprature - 50.0;

假设现在temprature的值为负值-x.yz。

l = (signed int) temprature;

现在l = -x，并且

temprature -= (float) l;

温度 = -x.yz - (-x) = -0.yz。

rem = (unsigned int)(temprature* 1e1);

乘以 10，并转换为 unsigned int。通常，这会导致未定义的行为 (6.3.1.4 (1)):

When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero). If the value of the integral part cannot be represented by the integer type, the behavior is undefined.⁶¹⁾

⁶¹⁾ The remaindering operation performed when a value of integer type is converted to unsigned type need not be performed when a value of real floating type is converted to unsigned type. Thus, the range of portable real floating values is (−1, Utype_MAX+1).

但是将负值转换为unsigned int无论如何都会产生错误的结果，即使进行了余数运算，你想要的是绝对值，所以你应该转换

rem = (unsigned int)fabsf(temprature * 1e1);

那里。