CODEWITHSUNDEEP
c++arraysprogram-entry-pointargv
Simplicity
Simplicity

Reputation: 48916

int main(int argc, char *argv[])

If I have this:

int main(int argc, char *argv[])

In the body, you can sometimes find programs using argv[1].

When do we use argv[1] over argv[0]? Is it only when we just want to read the second argument in the command line?

Upvotes: 4

Views: 10472

Answers (7)

Alnitak
Alnitak

Reputation: 339816

Yes, that's mostly it, argv[1] is the second command line parameter. The first command line parameter is the name of the program itself.

Alternatively, to avoid the semantic mess that this answer originally had, and the comments from others, it might make sense to call argv[0] the zeroth parameter, so that argv[1] would now be the "first" of the user supplied values.

In any event, this comes from the exec() family of functions, e.g. execl which has usage:

 int execl(const char *path, const char *arg0, ... /*, (char *)0 */);

In the (Unix) shell when you type in a command, if necessary the shell first resolves the command name (using $PATH) to find the real absolute path. The (absolute or relative) path is supplied for path, and the command as originally typed-in is supplied as arg0, eventually becoming argv[0] in your program.

The remaining command line parameters then end up as argv[1], etc.

Upvotes: 0

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361472

argv is an array of pointers, and each pointer in this array stores one argument from command line. So argv[0] is the first argument (that is the executable/program itself), argv[1] is the second argument, and so on!

The total number of arguments is determined by argc.

Upvotes: 6

Oscar Mederos
Oscar Mederos

Reputation: 29823

Let's suppose your C++ executable file is:

/home/user/program (or C:\program.exe in Windows)

if you execute:

./home/user/program 1 2 (or C:\program.exe 1 2 in Windows)

argv[0] = /home/user/program (C:\program.exe)
argv[1] = 1
argv[2] = 2

That is because:

  • argv[0] is the path of the executable file
  • argv[1] is the 1st argument

Edit:

Now I see that argv[0] isn't necessarily the path of the executable file.
Read the following SO question: Is args[0] guaranteed to be the path of execution?

Upvotes: 2

DhruvPathak
DhruvPathak

Reputation: 43235

as argv[0] is filepath of the program itself. Extra command line parameters are in further indexes, argv[1],argv[2].. You can read more here : http://www.site.uottawa.ca/~lucia/courses/2131-05/labs/Lab3/CommandLineArguments.html

Upvotes: 0

user541686
user541686

Reputation: 210455

By convention, argv[0] is the current program's name (or path), and argv[1] through argv[argc - 1] are the command-line arguments that the user provides.

However, this doesn't have to be true -- programs can OS-specific functions to bypass this requirement, and this happens often enough that you should be aware of it. (I'm not sure if there's much you can do even if you're aware of it, though...)

Example:

gcc -O3 -o temp.o "My file.c"

would (should) produce the following arguments:

argc: 5
argv: ["gcc", "-O3", "-o", "temp.o", "My file.c"]

So saying argv[0] would refer to gcc, not to -O3.

Upvotes: 12

Bogatyr
Bogatyr

Reputation: 19323

argv[0] is the execution path of the program, argv[1] is the first parameter to the program

Upvotes: 2

Nim
Nim

Reputation: 33655

Short answer is yes, the array contains all the options passed into the program.

Upvotes: 0

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