loksh | A Linux port of OpenBSD 's ksh

by dimkr C Version: 7.3 License: No License

X-Ray Key Features Code Snippets Community Discussions(1)Vulnerabilities Install Support

kandi X-RAY | loksh Summary

loksh is a C library. loksh has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.

loksh is a Linux port of OpenBSD's ksh. Unlike other ports of ksh, loksh targets only one platform, follows upstream closely and keeps changes to a minimum. loksh does not add any extra features; this reduces the risk of introducing security vulnerabilities and makes loksh a good fit for resource-constrained systems.

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Quality

Security

License

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Support

loksh has a low active ecosystem.

It has 114 star(s) with 6 fork(s). There are 9 watchers for this library.

It had no major release in the last 12 months.

There are 0 open issues and 15 have been closed. On average issues are closed in 49 days. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of loksh is 7.3

Quality

loksh has no bugs reported.

Security

loksh has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

License

loksh does not have a standard license declared.

Check the repository for any license declaration and review the terms closely.

Without a license, all rights are reserved, and you cannot use the library in your applications.

Reuse

loksh releases are available to install and integrate.

Installation instructions, examples and code snippets are available.

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loksh Key Features

No Key Features are available at this moment for loksh.

loksh Examples and Code Snippets

No Code Snippets are available at this moment for loksh.

Community Discussions

Trending Discussions on loksh

How should a backslash resulting from variable expansion be treated?

QUESTION

How should a backslash resulting from variable expansion be treated?

Asked 2020-Feb-04 at 07:10

I ran the following command (*sh being the name of a sh implementation) with all the shells I could find; although I was expecting all to print match, I got inconsistent results. I don't know which behavior is correct and reliable.

...

ANSWER

Answered 2019-Dec-16 at 13:43

Perhaps this bit of reference can clear things up for your question. From the POSIX's Shell Command Language specification - under Pattern Matching Notation the clause states,

2.13.1 Patterns Matching a Single Character

A character shall escape the following character. The escaping shall be discarded. If a pattern ends with an unescaped , it is unspecified whether the pattern does not match anything or the pattern is treated as invalid

The above rule applies to patterns while using the case statements and when using the the == glob match operator in bash using test construct.

So since your $2 remains unquoted while undergoing pattern match, the literal value of f\\oo is lost forever and gets interpreted as f\oo.

Just to clear things up, the shell has preserved your literal value during quote removal and just when applying the passed argument in this pattern matching rule, this behavior is exhibited.

Source https://stackoverflow.com/questions/59348035

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install loksh

loksh is included in many distributions:. For example, under Alpine Linux, loksh can be installed using apk add loksh.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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