long foo(int h, int l)
{
    return (h << 32) + l;
}

------解决方案--------------------
但是，上面这个函数是有陷阱的，陷阱在哪里呢？

在MSDN里面，有详细的解释：

<< Operator (C# Reference)
http://msdn.microsoft.com/en-us/library/a1sway8w.aspx

If the first operand is an int or uint (32-bit quantity), the shift count is given by the low-order five bits of the second operand. That is, the actual shift count is 0 to 31 bits.

If the first operand is a long or ulong (64-bit quantity), the shift count is given by the low-order six bits of the second operand. That is, the actual shift count is 0 to 63 bits.


因此，上面的函数应该修改为：

C# code

public static long Merge(int high, int low)
{
    return ((long)high << 32) + low;
}