铸造是在C标准定义的负向双unsigned int类型的行为？ [英] Is the behaviour of casting a negative double to unsigned int defined in the C standard?

问题描述

我有上似乎得到不同的结果，不同的平台上运行code。我要寻找一个合适的解释。

Windows系统：

 双DBL = -123.45;
INT d_cast =（无符号整数）DBL;
// d_cast == -123
 

的WinCE（ARM）：

 双DBL = -123.45;
INT d_cast =（无符号整数）DBL;
// d_cast == 0
 

编辑：的

感谢您的指向正确的方向。

修复

 双DBL = -123.45;
INT d_cast =（符号）（INT）DBL;
// d_cast == -123
//适用于两者。
 

解决方案

否

---

这种转换是不确定的，因此不便于携带。

据C99§6.3.1.4脚注50：

  

当真正的浮动类型的值转换为无类型无需执行时整数类型的值转换为无符号类型进行remaindering操作。因此，便携式实浮动值的范围是（-1，Utype_MAX + 1）。

I have code that runs on different platforms that seems to get different results. I am looking for a proper explanation.

Windows :

double dbl = -123.45; 
int d_cast = (unsigned int)dbl; 
// d_cast == -123

WinCE (ARM):

double dbl = -123.45; 
int d_cast = (unsigned int)dbl; 
// d_cast == 0

EDIT:

Thanks for pointing in the right direction.

fix

double dbl = -123.45; 
int d_cast = (unsigned)(int)dbl; 
// d_cast == -123
// works on both. 

解决方案

No

---

This conversion is undefined and therefore not portable.

According to C99 §6.3.1.4 footnote 50:

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).

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