为什么c＃编译器允许返回值类型和变量类型的错误匹配？

Question

Consider the following code:

static void Main()
{
    dynamic a = 1;
    int b = OneMethod(a);
}

private static string OneMethod(int number)
{
    return "";
}

Please notice that type of b and return type of OneMethod does not match. Nevertheless it builds and throws the exception at runtime.

The reason behind this may be Compiler does not know which OneMethod would be called, because a is dynamic. But why it cannot see that there is only one OneMethod . There will surely be an exception at runtime.

Answer 1

Any expression that has an operand of type dynamic will have a type of dynamic itself.

Thus your expression OneMethod(a) returns an object that's typed dynamically

so the first part of your code is equivalent to

static void Main()
{
    dynamic a = 1;
    dynamic temp = OneMethod(a);
    int b = temp;
}

one way of argue why this is sensible even in your case depends on whether or not you think the compiler should change behavior for that particular line depending when you add the below method

private static T OneMethod<T>(T number)

Now the compiler won't know the type returned until runtime. It won't even know which method is called. The generic or the non generic. Wouldn't you be surprised if it in the first case marked the assignment as a compile error and then by adding a completely different method it got moved to a runtime error?