小小笔试题（二）

1.9月21日，小米，电子科大笔试题：

void fun()  
{  
    unsigned int a = 2013;  
    int b = -2;  
    int c = 0;  
    while (a + b > 0)  
    {  
        a = a + b;  
        c++;  
    }  
    printf("%d", c);  
}  

　　问：输出是什么？

     此题略有陷进。

     a+b相加会自动转换为unsigned int类型，以前一直以为int和unsigned int相加会转换为int！

     因此当a=1时，a+b不是-1，因此死循环，木有输出。

 2.相关知识  

    C++算数运算类型转换 

    Arithmetic conversion proceeds in the following order:

Operand Type	Conversion
One operand has  type	The other operand is converted to long double.
One operand has double type	The other operand is converted to double.
One operand has float type	The other operand is converted to float.
One operand has unsigned long long int type	The other operand is converted to unsigned long long int
One operand has long long type.	The other operand is converted to long long.
One operand has unsigned long int type	The other operand is converted to unsigned long int.
One operand has unsigned int type and the other operand has long inttype and the value of the unsigned int can be represented in a long int	The operand with unsigned int type is converted to long int.
One operand has unsigned int type and the other operand has long inttype and the value of the unsigned int cannot be represented in a long int	Both operands are converted to unsigned long int.
One operand has long int type	The other operand is converted to long int.
One operand has unsigned int type	The other operand is converted to unsigned int.
Both operands have int type	The result is type int.

3. 参考

    由int与unsigned int 相加看c++类型转换