Is “argv[0] = name-of-executable” an accepted standard or just a common convention?

问题

When passing argument to main() in a C or C++ application, will argv[0] always be the name of the executable? Or is this just a common convention and not guaranteed to be true 100% of the time?

回答1:

Guesswork (even educated guesswork) is fun but you really need to go to the standards documents to be sure. For example, ISO C11 states (my emphasis):

If the value of argc is greater than zero, the string pointed to by argv[0] represents the program name; argv[0][0] shall be the null character if the program name is not available from the host environment.

So no, it's only the program name if that name is available. And it "represents" the program name, not necessarily is the program name. The section before that states:

If the value of argc is greater than zero, the array members argv[0] through argv[argc-1] inclusive shall contain pointers to strings, which are given implementation-defined values by the host environment prior to program startup.

This is unchanged from C99, the previous standard, and means that even the values are not dictated by the standard - it's up to the implementation entirely.

This means that the program name can be empty if the host environment doesn't provide it, and anything else if the host environment does provide it, provided that "anything else" somehow represents the program name. In my more sadistic moments, I would consider translating it into Swahili, running it through a substitution cipher then storing it in reverse byte order :-).

However, implementation-defined does have a specific meaning in the ISO standards - the implementation must document how it works. So even UNIX, which can put anything it likes into argv[0] with the exec family of calls, has to (and does) document it.

回答2:

Under *nix type systems with exec*() calls, argv[0] will be whatever the caller puts into the argv0 spot in the exec*() call.

The shell uses the convention that this is the program name, and most other programs follow the same convention, so argv[0] usually the program name.

But a rogue Unix program can call exec() and make argv[0] anything it likes, so no matter what the C standard says, you can't count on this 100% of the time.

回答3:

According to the C++ Standard, section 3.6.1:

argv[0] shall be the pointer to the initial character of a NTMBS that represents the name used to invoke the program or ""

So no, it is not guaranteed, at least by the Standard.

回答4:

This page states:

The element argv[0] normally contains the name of the program, but this shouldn't be relied upon - anyway it is unusual for a program not to know its own name!

However, other pages seem to back up the fact that it is always the name of the executable. This one states:

You’ll notice that argv[0] is the path and name of the program itself. This allows the program to discover information about itself. It also adds one more to the array of program arguments, so a common error when fetching command-line arguments is to grab argv[0] when you want argv[1].

回答5:

ISO-IEC 9899 states:

5.1.2.2.1 Program startup

If the value of argc is greater than zero, the string pointed to by argv[0] represents the programname; argv[0][0] shall be the null character if the program name is not available from the host environment. If the value of argc is greater than one, the strings pointed to by argv[1] through argv[argc-1] represent the program parameters.

I've also used:

#if defined(_WIN32)
  static size_t getExecutablePathName(char* pathName, size_t pathNameCapacity)
  {
    return GetModuleFileNameA(NULL, pathName, (DWORD)pathNameCapacity);
  }
#elif defined(__linux__) /* elif of: #if defined(_WIN32) */
  #include <unistd.h>
  static size_t getExecutablePathName(char* pathName, size_t pathNameCapacity)
  {
    size_t pathNameSize = readlink("/proc/self/exe", pathName, pathNameCapacity - 1);
    pathName[pathNameSize] = '\0';
    return pathNameSize;
  }
#elif defined(__APPLE__) /* elif of: #elif defined(__linux__) */
  #include <mach-o/dyld.h>
  static size_t getExecutablePathName(char* pathName, size_t pathNameCapacity)
  {
    uint32_t pathNameSize = 0;

    _NSGetExecutablePath(NULL, &pathNameSize);

    if (pathNameSize > pathNameCapacity)
      pathNameSize = pathNameCapacity;

    if (!_NSGetExecutablePath(pathName, &pathNameSize))
    {
      char real[PATH_MAX];

      if (realpath(pathName, real) != NULL)
      {
        pathNameSize = strlen(real);
        strncpy(pathName, real, pathNameSize);
      }

      return pathNameSize;
    }

    return 0;
  }
#else /* else of: #elif defined(__APPLE__) */
  #error provide your own implementation
#endif /* end of: #if defined(_WIN32) */

And then you just have to parse the string to extract the executable name from the path.

回答6:

Applications of having argv[0] != executable name

many shells determine if they are a login shell by checking argv[0][0] == '-'. Login shells have different properties, notably that they source some default files such as /etc/profile.

It is typically the init itself or getty that adds the leading -, see also: https://unix.stackexchange.com/questions/299408/how-to-login-automatically-without-typing-the-root-username-or-password-in-build/300152#300152
multi-call binaries, perhaps most notably Busybox. These symlink multiple names e.g. /bin/sh and /bin/ls to a single exebutable /bin/busybox, which recognizes which tool to use from argv[0].

This makes it possible to have a single small statically linked executable that represents multiple tools, and will work on basically on any Linux environment.

See also: https://unix.stackexchange.com/questions/315812/why-does-argv-include-the-program-name/315817

Runnable POSIX execve example where argv[0] != executable name

Others mentioned exec, but here is a runnable example.

a.c

#define _XOPEN_SOURCE 700
#include <unistd.h>

int main(void) {
    char *argv[] = {"yada yada", NULL};
    char *envp[] = {NULL};
    execve("b.out", argv, envp);
}

b.c

#include <stdio.h>

int main(int argc, char **argv) {
    puts(argv[0]);
}

Then:

gcc a.c -o a.out
gcc b.c -o b.out
./a.out

Gives:

yada yada

Yes, argv[0] could also be:

NULL: When can argv[0] have null?
empty: Can argv[0] contain an empty string?

Tested on Ubuntu 16.10.

回答7:

I'm not sure whether it is a nearly universal convention or a standard, but either way you should abide by it. I've never seen it exploited outside of Unix and Unix-like systems, though. In Unix environments - and maybe particularly in the old days - programs might have significantly different behaviors depending on the name under which they are invoked.

EDITED: I see from other posts at the same time as mine that someone has identified it as coming from a particular standard, but I'm sure the convention long predates the standard.

回答8:

If you start an Amiga program by Workbench argv[0] will not be set, only by CLI.

来源：https://stackoverflow.com/questions/2050961/is-argv0-name-of-executable-an-accepted-standard-or-just-a-common-conventi

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