在 C# 中，是否可以在不进行类型转换的情况下将整数转换为无符号整数？

Question

是否可以在不需要显式类型转换的情况下将整数转换为无符号整数？

我有以下代码-

using System;

class MainClass
{
    public static void Main ()
    {
        int num1 = -33;
        uint num2 = Convert.ToUInt32 (num1);
        Console.WriteLine (num2);
    }
}

它抛出错误 -

System.OverflowException has been thrown.

Value was either too large or too small for a UInt32

我不想使用显式类型转换。这样可以吗？

问题与问题 不重复，正如我特别提到的，我想知道是否可以不使用显式类型转换？

Answer 1

I want to know if it is possible without explicit typecasting?

简短的回答是：没有。

I want to know the reason of existence of the function Convert.ToUInt32()

答案：微软的某个人发现它很有帮助并创建了它。其他人发现它 helpful/easy 来使用并使用它。我想我们找不到实现该功能的具体原因。

有关此主题的更多信息：

表示 4 种可能性。 他们每个人都包含一个类型转换。我不知道将 uint 转换为 int 的另一个内置功能。 That answer 很好地解释了 uint->int 的转换及其陷阱。

(UltraShort)Summary: You have to take care of overflows.

在处理无符号类型时，您应该查看 checked and unchecked 关键字。

类型转换

根据this类型转换有4种方法。

• 隐式
• 显式
• 使用 conversion oerators 的用户定义转换。
• helüer 的转化次数 classes（如 Convert class）

Convert.ToInt32() / Convert.ToUInt32()的用例

ToInt32() 的代码示例取自 msdn。

float[] values= { Single.MinValue, -1.38e10f, -1023.299f, -12.98f,
                  0f, 9.113e-16f, 103.919f, 17834.191f, Single.MaxValue };
int result;

foreach (float value in values)
{
   try {
      result = Convert.ToInt32(value);
      Console.WriteLine("Converted the {0} value {1} to the {2} value {3}.",
                        value.GetType().Name, value, result.GetType().Name, result);
   }
   catch (OverflowException) {
      Console.WriteLine("{0} is outside the range of the Int32 type.", value);
   }   
}                                 
// The example displays the following output:
//    -3.40282346638529E+38 is outside the range of the Int32 type.
//    -13799999488 is outside the range of the Int32 type.
//    Converted the Double value -1023.29901123047 to the Int32 value -1023.
//    Converted the Double value -12.9799995422363 to the Int32 value -13.
//    Converted the Double value 0 to the Int32 value 0.
//    Converted the Double value 9.11299983940444E-16 to the Int32 value 0.
//    Converted the Double value 103.918998718262 to the Int32 value 104.
//    Converted the Double value 17834.19140625 to the Int32 value 17834.
//    3.40282346638529E+38 is outside the range of the Int32 type.

转换代码示例。ToUInt(Int32) 也取自 msdn

int[] numbers = { Int32.MinValue, -1203, 0, 121, 1340, Int32.MaxValue };
uint result;
foreach (int number in numbers)
{
   try {
      result = Convert.ToUInt32(number);
      Console.WriteLine("Converted the {0} value {1} to the {2} value {3}.",
                        number.GetType().Name, number,
                        result.GetType().Name, result);
   }
   catch (OverflowException) {
      Console.WriteLine("The {0} value {1} is outside the range of the UInt32 type.",
                        number.GetType().Name, number);
   }
}
// The example displays the following output:
//    The Int32 value -2147483648 is outside the range of the UInt32 type.
//    The Int32 value -1203 is outside the range of the UInt32 type.
//    Converted the Int32 value 0 to the UInt32 value 0.
//    Converted the Int32 value 121 to the UInt32 value 121.
//    Converted the Int32 value 1340 to the UInt32 value 1340.
//    Converted the Int32 value 2147483647 to the UInt32 value 2147483647.

Answer 2

根据 OP 评论的要求：

//Sample inputs
string str = "15";
bool b = false;
int i = 15;
double d = 14.5;
object o = str;
object oi = i;

//These are not valid
//uint ustr = (uint)str; 
//uint ub = (uint)b;

//This is valid at compile time but throw a run time exception
//uint uo = (uint)o;
//uint uoi = (uint)oi;

//These are valid
uint ustr = Convert.ToUInt32(str);
uint ub = Convert.ToUInt32(b);
uint uo = Convert.ToUInt32(o);
uint uoi2 = Convert.ToUInt32(oi);

//both ways are valid
uint ud = (uint)d;
uint ud2 = Convert.ToUInt32(d);
uint ui = (uint)i;
uint ui2 = Convert.ToUInt32(i);

//Some inputs can't be converted by either way (casting may be possible but result in wrong values)
string str2 = "Bananas";
int i2 = -15;