unsign | Remove code signatures from OSX Mach

 by   steakknife C Version: Current License: No License

kandi X-RAY | unsign Summary

kandi X-RAY | unsign Summary

unsign is a C library typically used in macOS applications. unsign has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.

Remove code signatures from OSX Mach-O binaries (note: unsigned binaries cannot currently be re-codesign'ed. Patches welcome!)
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              unsign has a low active ecosystem.
              It has 369 star(s) with 53 fork(s). There are 17 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              There are 4 open issues and 2 have been closed. On average issues are closed in 79 days. There are 2 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of unsign is current.

            kandi-Quality Quality

              unsign has 0 bugs and 0 code smells.

            kandi-Security Security

              unsign has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              unsign code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              unsign does not have a standard license declared.
              Check the repository for any license declaration and review the terms closely.
              OutlinedDot
              Without a license, all rights are reserved, and you cannot use the library in your applications.

            kandi-Reuse Reuse

              unsign releases are not available. You will need to build from source code and install.
              Installation instructions are not available. Examples and code snippets are available.

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            Community Discussions

            QUESTION

            Efficient overflow-immune arithmetic mean in C/C++
            Asked 2022-Mar-10 at 14:02

            The arithmetic mean of two unsigned integers is defined as:

            ...

            ANSWER

            Answered 2022-Mar-08 at 10:54

            The following method avoids overflow and should result in fairly efficient assembly (example) without depending on non-standard features:

            Source https://stackoverflow.com/questions/71019078

            QUESTION

            Why is the XOR swap optimized into a normal swap using the MOV instruction?
            Asked 2022-Mar-08 at 10:00

            While testing things around Compiler Explorer, I tried out the following overflow-free function for calculating average of 2 unsigned 32-bit integer:

            ...

            ANSWER

            Answered 2022-Mar-08 at 10:00

            Clang does the same thing. Probably for compiler-construction and CPU architecture reasons:

            • Disentangling that logic into just a swap may allow better optimization in some cases; definitely something it makes sense for a compiler to do early so it can follow values through the swap.

            • Xor-swap is total garbage for swapping registers, the only advantage being that it doesn't need a temporary. But xchg reg,reg already does that better.

            I'm not surprised that GCC's optimizer recognizes the xor-swap pattern and disentangles it to follow the original values. In general, this makes constant-propagation and value-range optimizations possible through swaps, especially for cases where the swap wasn't conditional on the values of the vars being swapped. This pattern-recognition probably happens soon after transforming the program logic to GIMPLE (SSA) representation, so at that point it will forget that the original source ever used an xor swap, and not think about emitting asm that way.

            Hopefully sometimes that lets it then optimize down to only a single mov, or two movs, depending on register allocation for the surrounding code (e.g. if one of the vars can move to a new register, instead of having to end up back in the original locations). And whether both variables are actually used later, or only one. Or if it can fully disentangle an unconditional swap, maybe no mov instructions.

            But worst case, three mov instructions needing a temporary register is still better, unless it's running out of registers. I'd guess GCC is not smart enough to use xchg reg,reg instead of spilling something else or saving/restoring another tmp reg, so there might be corner cases where this optimization actually hurts.

            (Apparently GCC -Os does have a peephole optimization to use xchg reg,reg instead of 3x mov: PR 92549 was fixed for GCC10. It looks for that quite late, during RTL -> assembly. And yes, it works here: turning your xor-swap into an xchg: https://godbolt.org/z/zs969xh47)

            xor-swap has worse latency and defeats mov-elimination

            with no memory reads, and the same number of instructions, I don't see any bad impacts and feels odd that it be changed. Clearly there is something I did not think through though, but what is it?

            Instruction count is only a rough proxy for one of three things that are relevant for perf analysis: front-end uops, latency, and back-end execution ports. (And machine-code size in bytes: x86 machine-code instructions are variable-length.)

            It's the same size in machine-code bytes, and same number of front-end uops, but the critical-path latency is worse: 3 cycles from input a to output a for xor-swap, and 2 from input b to output a, for example.

            MOV-swap has at worst 1-cycle and 2-cycle latencies from inputs to outputs, or less with mov-elimination. (Which can also avoid using back-end execution ports, especially relevant for CPUs like IvyBridge and Tiger Lake with a front-end wider than the number of integer ALU ports. And Ice Lake, except Intel disabled mov-elimination on it as an erratum workaround; not sure if it's re-enabled for Tiger Lake or not.)

            Also related:

            If you're going to branch, just duplicate the averaging code

            GCC's real missed optimization here (even with -O3) is that tail-duplication results in about the same static code size, just a couple extra bytes since these are mostly 2-byte instructions. The big win is that the a path then becomes the same length as the other, instead of twice as long to first do a swap and then run the same 3 uops for averaging.

            update: GCC will do this for you with -ftracer (https://godbolt.org/z/es7a3bEPv), optimizing away the swap. (That's only enabled manually or as part of -fprofile-use, not at -O3, so it's probably not a good idea to use all the time without PGO, potentially bloating machine code in cold functions / code-paths.)

            Doing it manually in the source (Godbolt):

            Source https://stackoverflow.com/questions/71382441

            QUESTION

            The fastest way to swap the two lowest bits in an unsigned int in C++
            Asked 2022-Feb-19 at 11:39

            Assume that I have:

            ...

            ANSWER

            Answered 2021-Oct-28 at 10:51

            QUESTION

            Is if(A | B) always faster than if(A || B)?
            Asked 2022-Feb-11 at 05:03

            I am reading this book by Fedor Pikus and he has some very very interesting examples which for me were a surprise.
            Particularly this benchmark caught me, where the only difference is that in one of them we use || in if and in another we use |.

            ...

            ANSWER

            Answered 2022-Feb-08 at 19:57

            Code readability, short-circuiting and it is not guaranteed that Ord will always outperform a || operand. Computer systems are more complicated than expected, even though they are man-made.

            There was a case where a for loop with a much more complicated condition ran faster on an IBM. The CPU didn't cool and thus instructions were executed faster, that was a possible reason. What I am trying to say, focus on other areas to improve code than fighting small-cases which will differ depending on the CPU and the boolean evaluation (compiler optimizations).

            Source https://stackoverflow.com/questions/71039947

            QUESTION

            Is it safe to bind an unsigned int to a signed int reference?
            Asked 2022-Feb-09 at 07:17

            After coming across something similar in a co-worker's code, I'm having trouble understanding why/how this code executes without compiler warnings or errors.

            ...

            ANSWER

            Answered 2022-Feb-09 at 07:17

            References can't bind to objects with different type directly. Given const int& s = u;, u is implicitly converted to int firstly, which is a temporary, a brand-new object and then s binds to the temporary int. (Lvalue-references to const (and rvalue-references) could bind to temporaries.) The lifetime of the temporary is prolonged to the lifetime of s, i.e. it'll be destroyed when get out of main.

            Source https://stackoverflow.com/questions/70712797

            QUESTION

            Why does iteration over an inclusive range generate longer assembly in Rust?
            Asked 2022-Jan-15 at 11:19

            These two loops are equivalent in C++ and Rust:

            ...

            ANSWER

            Answered 2022-Jan-12 at 10:20

            Overflow in the iterator state.

            The C++ version will loop forever when given a large enough input:

            Source https://stackoverflow.com/questions/70672533

            QUESTION

            Clang generates strange output when dividing two integers
            Asked 2021-Dec-07 at 09:57

            I have written the following very simple code which I am experimenting with in godbolt's compiler explorer:

            ...

            ANSWER

            Answered 2021-Dec-07 at 09:52

            The assembly seems to be checking if either num or den is larger than 2**32 by shifting right by 32 bits and then checking whether the resulting number is 0. Depending on the decision, a 64-bit division (div rsi) or 32-bit division (div esi) is performed.

            Presumably this code is generated because the compiler writer thinks the additional checks and potential branch outweigh the costs of doing an unnecessary 64-bit division.

            Source https://stackoverflow.com/questions/70257914

            QUESTION

            Compiler changes the type variable type from uin16_t to int when it's marked as constexpr
            Asked 2021-Oct-15 at 12:32

            I've encountered a weird problem trying to flip all bits of my number.

            ...

            ANSWER

            Answered 2021-Oct-15 at 12:32

            This is caused by IMHO a quite unfortunate C++ rule about integer promotion. It basically states that all types smaller than int are always promoted to int if int can represent all values of the original type. Only if not, unsigned int is chosen. std::uint16_t on standard 32/64-bit architectures falls into the first category.

            int is guaranteed to be at least 16-bit wide, if that would happen to be the case, unsigned int would have been chosen, so the behaviour of the code is implementation-defined.

            I do not know precisely why the compiler issues the warning only for constexpr values, most likely because it could easily propagate that constant through the ~. In other cases, someone might change DefaultValueForPortStatus to some "safe" value that won't overflow when negated and converted from int back to std::uint16_t. But the issue is there regardless of constness, you can test it with this code:

            Source https://stackoverflow.com/questions/69464759

            QUESTION

            Why is movl preferred to movb when translating a C downcast from unsigned int to unsigned char?
            Asked 2021-Jul-18 at 10:01

            Considering a pared-down example of down-casting unsigned to unsigned char,

            ...

            ANSWER

            Answered 2021-Jul-17 at 13:15

            Writing to an 8 bit x86 register possibly incurs an extra merge µop when the new low byte is merged with the old high bytes of the corresponding 32/64 bit register. This can also cause an unexpected data dependency on the previous value of the register.

            For this reason, it is generally a good idea to only write to 32/64 bit variants of general purpose registers on x86.

            Source https://stackoverflow.com/questions/68420792

            QUESTION

            Why does unsigned char have different default initialization behaviour than other data types?
            Asked 2021-Jul-02 at 14:48

            I am reading through the cppreference page on default initialization and I noticed a section that states something along these lines:

            ...

            ANSWER

            Answered 2021-Jul-02 at 07:10

            The difference is not in initialisation behaviour. The value of uninitialised int is indeterminate and default initialisation leaves it indeterminate. The value of uninitialised unsigned char is indeterminate and default initialisation leaves it indeterminate. There is no difference there.

            The difference is that behaviour of producing an indeterminate value of type int - or any other type besides the exceptional unsigned char or std::byte - is undefined (unless the value is discarded).

            The exception for unsigned char (and later std::byte) was added to the language in C++14 when indeterminate value was properly defined (although since the change was a defect resolution, to my understanding it applies to the official standard at the time, C++11).

            I could not find a documented rationale for that design choice. Here is a timeline of the definitions (all standard quotes are from drafts):

            C89 - 1.6 DEFINITIONS OF TERMS

            Undefined behavior --- behavior, upon use of ... indeterminately-valued objects

            C89 - 3.5.7 Initialization - Semantics

            ... If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.

            There are no exceptions for any type. You'll see why C standard is relevant when reading C++98 standard.

            C++98 - [dcl.init]

            ... Otherwise, if no initializer is specified for an object, the object and its subobjects, if any, have an indeterminate initial value

            There was no definition for what indeterminate value means or what happens when you use it. The intended meaning may presumably have been same as C89, but it is underspecified.

            C99 - 3. Terms, definitions, and symbols - 3.17.2

            3.17.2 indeterminate value

            either an unspecified value or a trap representation

            3.17.3 unspecified value

            valid value of the relevant type where this International Standard imposes no requirements on which value is chosen in any instance

            NOTE An unspecified value cannot be a trap representation.

            C99 - 6.2.6 Representations of types - 6.2.6.1 General

            Certain object representations need not represent a value of the object type. If the stored value of an object has such a representation and is read by an lvalue expression that does not have character type, the behavior is undefined. If such a representation is produced by a side effect that modifies all or any part of the object by an lvalue expression that does not have character type, the behavior is undefined. 41) Such a representation is called a trap representation.

            C99 - J.2 Undefined behavior

            The behavior is undefined in the following circumstances:

            • ...
            • The value of an object with automatic storage duration is used while it is indeterminate
            • A trap representation is read by an lvalue expression that does not have character type
            • A trap representation is produced by a side effect that modifies any part of the object using an lvalue expression that does not have character type
            • ...

            C99 introduced the term trap representation, and which also have UB when used, just like indeterminate values. Character types (which are char, unsigned char and signed char) don't have trap representations, and may be used to operate on trap representations of other types without UB.

            C++ core language issue - 616. Definition of “indeterminate value”

            The C++ Standard uses the phrase “indeterminate value” without defining it. C99 defines it as “either an unspecified value or a trap representation.” Should C++ follow suit?

            Proposed resolution (October, 2012):

            [dcl.init] paragraph 12 as follows:

            If no initializer is specified for an object, the object is default-initialized. When storage for an object with automatic or dynamic storage duration is obtained, the object has an indeterminate value, and if no initialization is performed for the object, that object retains an indeterminate value until that value is replaced (5.17 [expr.ass]). [Note: Objects with static or thread storage duration are zero-initialized, see 3.6.2 [basic.start.init]. —end note] If an indeterminate value is produced by an evaluation, the behavior is undefined except in the following cases:

            • If an indeterminate value of unsigned narrow character type (3.9.1 [basic.fundamental]) is produced by the evaluation of:
            • the second or third operand of a conditional expression (5.16 [expr.cond]),
            • the right operand of a comma (5.18 [expr.comma]),
            • the operand of a cast or conversion to an unsigned narrow character type (4.7 [conv.integral], 5.2.3 [expr.type.conv], 5.2.9 [expr.static.cast], 5.4 [expr.cast]), or
            • a discarded-value expression (Clause 5 [expr]),

            then the result of the operation is an indeterminate value.

            If an indeterminate value of unsigned narrow character type (3.9.1 [basic.fundamental]) is produced by the evaluation of the right operand of a simple assignment operator (5.17 [expr.ass]) whose first operand is an lvalue of unsigned narrow character type, an indeterminate value replaces the value of the object referred to by the left operand.

            If an indeterminate value of unsigned narrow character type (3.9.1 [basic.fundamental]) is produced by the evaluation of the initialization expression when initializing an object of unsigned narrow character type, that object is initialized to an indeterminate value.

            The proposed change was accepted as a defect resolution with some further changes (issue 1213) but has remained mostly the same (similar enough for purposes of this question). This is where the exception for unsigned char seems to have been introduced into C++. The core language issue has no public comments or notes about the rationale for the exception as far as I could find.

            Source https://stackoverflow.com/questions/68219526

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