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In shell scripts one specifies language interpreter on shebang(#!) line. As far as I know, it is recommended to use #!/usr/bin/env bash because env is always located in /usr/bin directory while location of bash may vary from system to system. However, are there any technical differences if bash is started directly with /bin/bash or through env utility? In addition, am I correct that if I do not specify any variables for env, the bash is started in unmodified environment?

marked as duplicate by cuonglm, Anthon, chaos, G-Man, casey May 29 '15 at 18:18

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In one sense, using env could be considered "portable" in that the path to bash is not relevant (/bin/bash, /usr/bin/bash, /usr/local/bin/bash, ~/bin/bash, or whatever path) because it is specified in the environment. In this way, a script author could make his script easier to run on many different systems.

In another sense, using env to find bash or any other shell or command interpreter is considered a security risk because an unknown binary (malware) might be used to execute the script. In these environments, and sometimes by managerial policy, the path is specified explicitly with a full path: #!/bin/bash.

In general, use env unless you know you are writing in one of these environments that scrutinize the minute details of risk.

When Ubuntu first started using dash, some time in 2011, many scripts were broken by that action. There was discussion about it on askubuntu.com. Most scripts were written #!/bin/sh which was a link to /bin/bash. The consensus was this: the script writer is responsible for specifying the interpreter. Therefore, if your script should always be invoked with BASH, specify it from the environment. This saves you having to guess the path, which is different on various Unix/Linux systems. In addition, it will work if tomorrow /bin/sh becomes a link to some other shell like /bin/newsh.

Another difference is that the env method won't allow the passing of arguments to the interpreter.

  • Regarding env as a security risk to find bash. Do you mean that if I use #!/usr/bin/env bash shebang, then bash is searched from my $PATH and for example if my $PATH is "/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin" and I have a malware named bash in /usr/local/sbin directory(/usr/local/sbin/bash is executed instead of /usr/bin/bash because /usr/local/sbin is before /usr/bin in $PATH), then this is a security risk? – Martin Jun 4 '15 at 8:17
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    Yes, but of more concern, the user's PATH could place the malware first in the PATH. Not every system is so rigorously scrutinized, but, for example, consider a system that has mission-critical services running in an environment where threats could be launched from anywhere. In some environments, like any organization under the DoD, an Information Assurance Officer may flag it while scrutinizing a system that is being tested for its Authority to Operate. Not all vulnerabilities are known, and specifying a full path mitigates the FUD. – Christopher Jun 4 '15 at 11:22
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Apart from that using /usr/bin/env is somewhat slower there is no difference, if you start a program which such a shebang. (Unless /bin/bash doesn't exists but bash is somewhere (else) in the PATH)

env has the possibility to modify the environment of the invoked command, but that can only be used when env is started from the commandline, with in shebang line (depending on the OS) the options cannot be specified.

The source of env is rather small so you can verify what it is doing if you are familiar with C. Since execvp is used to call the program to be executed. There will not even be an env parent process if you inspect the process tree.

  • So in a nutshell /usr/bin/env bash, is exactly the same as to search for bash from $PATH? – Martin Jun 4 '15 at 8:31

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