encryption – How to set another encrypted boot volume without unlocking it?

I have two MacOS installations as part of a dual boot configuration on my MBP. Both happen to be Mojave 10.14. They are installed on two APFS volumes encrypted in the same APFS container.

I want to be able to set the next boot volume without providing a decryption password or holding down the option key while rebooting. In relation, I want the current "other volume" to be the default volume. The current default is the second MacOS volume and installation that I have made.

The problems I see:

1] Under System Preferences -> Startup Disk, the decryption password is required: "It must be unlocked to determine if it is a valid boot disk."
– that can not technically be true. When I hold down the Option key during startup, the decryption passwords are not known at this time, both options are presented.
Screen capture of the startup window

2] Examples of command-line blessed commands are based on a volume mounted under / Volumes, which of course means that the decryption password is made available.

bitwise – overload c ++ enum class operators and enable the control value set

I usually use c ++ 11 enum class instead of old enums, because many enums may contain the same value
but as we know, they do not have the bit operators out of the box, so we should set them manually

under windows it's already done for us with this macro in windows.h header:

DEFINE_ENUM_FLAG_OPERATORS

but then I encountered another problem

if I use these enumerations as flags, I can put a lot of values ​​in the falg file using the bitwise operator | , but to test if the flag contains a value that I can not check with the bit-level operator like this:

if (flags & value) {...}

I could, however, write a model function like this one to test:

template 
constexpr bool has_value(T flags, T value)
{
    return (std::underlying_type_t)flags & (std::underlying_type_t)value;
}

but I would really like to use the old format to test the values ​​in the flags
I already found a solution on github which included all the emums chosen in a class called flags and which was used like the old ones but with the safety type of enum classes

Since I do not want to use templates for this purpose and I just want the & operator to give me a bool value in if expressions, I've used another class containing the enum value and can be converted to bool. That's what I wrote.

#define OVERLOAD_ENUM_OPERATORS(x) 
class EnumBool##x 
{ 
    x enum_value; 
public: 
    constexpr EnumBool##x(const x e_val) : enum_value(e_val) {} 
    constexpr operator x() const { return enum_value; } 
    explicit operator bool() const { return (std::underlying_type_t)enum_value != 0; } 
}; 
inline constexpr EnumBool##x operator&(const x lhs, const x rhs) 
{ 
    return EnumBool##x((x)((std::underlying_type_t)lhs & (std::underlying_type_t)rhs)); 
} 
inline constexpr x operator|(const x lhs, const x rhs) 
{ 
    return (x)((std::underlying_type_t)lhs | (std::underlying_type_t)rhs); 
} 
inline constexpr x operator^(const x lhs, const x rhs) 
{ 
    return (x)((std::underlying_type_t)lhs ^ (std::underlying_type_t)rhs);
} 
inline constexpr x operator~(const x lhs) 
{ 
    return (x)(~(std::underlying_type_t)lhs);
} 
inline constexpr x& operator|=(x& lhs, const x rhs) 
{ 
    lhs = (x)((std::underlying_type_t)lhs | (std::underlying_type_t)rhs); 
    return lhs; 
} 
inline constexpr x& operator&=(x& lhs, const x rhs) 
{ 
    lhs = (x)((std::underlying_type_t)lhs & (std::underlying_type_t)rhs); 
    return lhs; 
} 
inline constexpr x& operator^=(x& lhs, const x rhs) 
{ 
    lhs = (x)((std::underlying_type_t)lhs ^ (std::underlying_type_t)rhs); 
    return lhs; 
} 
inline constexpr bool operator==(const x lhs, const x rhs) 
{ 
    return (std::underlying_type_t)lhs == (std::underlying_type_t)rhs; 
} 
inline constexpr bool operator!=(const x lhs, const x rhs) 
{ 
    return (std::underlying_type_t)lhs != (std::underlying_type_t)rhs; 
} 
inline constexpr bool operator>(const x lhs, const x rhs) 
{ 
    return (std::underlying_type_t)lhs > (std::underlying_type_t)rhs; 
} 
inline constexpr bool operator<(const x lhs, const x rhs) 
{ 
    return (std::underlying_type_t)lhs < (std::underlying_type_t)rhs; 
} 
inline constexpr bool operator>=(const x lhs, const x rhs) 
{ 
    return (std::underlying_type_t)lhs >= (std::underlying_type_t)rhs; 
} 
inline constexpr bool operator<=(const x lhs, const x rhs) 
{ 
    return (std::underlying_type_t)lhs <= (std::underlying_type_t)rhs; 
}

Set screaming frog?

What's a screaming frog? why is it used?

.

monsters – Can we be part of a multitude of Covens and derive the benefits? Can only one Coven of Hags form several conventions using the same set of Hags?

I'm joking with Hag Covens and not that I do not think they're weak or anything (because they're anything but), I just wanted to add an extra touch to a particular Hag that I'm I have in my world.

I'm curious to know if it's possible that a Hag is separated from several Covens and its rules, or that three Hags form more than one Coven using the same three Hags, so 39, get more Coven spells, etc.

Why do people still add me who are not friends of friends? I've already set up my settings for friend-only friends

Why do people still add me that I do not know?
I've set my settings only as friend friends and when people add me, this does not show up.
It's annoying

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How to update the emoji set for standard email applications?

I have a basic compatibility problem with emoji: the emoji available on an Android device (Android 8.1) do not match those available on a second (Android 6.0?) In applications such as the application of SMS messaging or Google Hangouts. Some emoji appear as empty boxes on receipt.

Is there a way to update the old Android device to include new emoji? Suppose email applications such as Google Hangouts have the latest updates for their respective platforms and the receiving device is a normal end user device that is not rooted.

Relationship between continued dependence of varieties and set topology

This proposition is found in Shub's book: The Global Stability of Dynamical Systems. Right here $ R $ is a rectangle. I wondered why in this marked part the continued dependence on stable varieties implies that given sets are open and closed? Is it a property of multiples? or something else?
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fpga – Vivado: TCL command to set synchronization paths between clock1 and clock2 as false paths (TIMING-6 and TIMING-7)

Suppose that CDC synchronization is already in place on each path between a clock domain and another clock domain. In this case, what is the Vivado TCL command to set the synchronization path between clock1 and clock2 so as to be a false path for each synchronization path between clock1 and clock2?

Compiler warning example:

WARNING: (TIMING-6) The clocks clk_1 and clk_2 are related (timed together) but they have no common primary clock. The design could fail in hardware. To find a timing path between these clocks, run the following command: report_timing -from (get_clocks clk_fpga_0) -to (get_clocks clk_out1_design_zynq_zyboz720_clk_wiz_0_0) 

WARNING: (TIMING-7) The clocks clk_1 and clk_2 are related (timed together) but they have no common node. The design could fail in hardware. To find a timing path between these clocks, run the following command: report_timing -from (get_clocks clk_1) -to (get_clocks clk_2)

set theory – Partition of n elements in k subsets

Given a set of n identical elements. Find the number of possible ways to form k subassemblies
Let's say that A = {0,0,0,0,0,0,0,0} and k = 3 like {{0,0}, {0,0,0}, {0,0}}
The answer given is 36, but I am unable to understand the math behind that. Can someone explain it?