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?
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.
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.
argv[0] is the execution path of the program, argv[1] is the first parameter to the program
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?
Short answer is yes, the array contains all the options passed into the program.
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
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.
Related
I want to better understand what's going on under the hood with the command line arguments when a C or C++ program is launched. I know, of course, that argc and argv, when passed to main(), represent the argument count and argument vector, respectively.
What I'm trying to figure out is how the compiler knows to interpret int argc as the number of arguments passed from the command line. If I write a simple function that attempts to mimic main() (e.g. int testfunc(int argc, char* argv[])), and pass in a string, the compiler complains, "Expected 'int' but argument is of type char*" as I would expect. How is this interpreted differently when command line arguments are passed to main()?
In common C implementations, main is not the first routine called when your process starts. Usually, it is some special entry point like _start that is provided by the C library built into your program when you link it. The code at this special entry point examines the command line information that is passed to it (in some way outside of C, particular to the operating system) and constructs an argument list for main. After that and other work, it calls main.
You don't pass argc value on your own (from the command line, for example), it is supplied by your environment (runtime), just like the exact content for argc.[Note below]
To elaborate, C11, chapter ยง5.1.2.2.1, (indicators mine)
The value of argc shall be nonnegative.
argv[argc] shall be a null pointer.
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. The
intent is to supply to the program information determined prior to program startup
from elsewhere in the hosted environment. [Note start]If the host environment is not capable of
supplying strings with letters in both uppercase and lowercase, the implementation
shall ensure that the strings are received in lowercase.[Note end]
Everytime I create a project (standard command line utility) with Xcode, my
main function starts out looking like this:
int main(int argc, const char * argv[])
What's all this in the parenthesis? Why use this rather than just
int main()?
main receives the number of arguments and the arguments passed to it when you start the program, so you can access it.
argc contains the number of arguments, argv contains pointers to the arguments.
argv[argc] is always a NULL pointer. The arguments usually include the program name itself.
Typically if you run your program like ./myprogram
argc is 1;
argv[0] is the string "./myprogram"
argv[1] is a NULL pointer
If you run your program like ./myprogram /tmp/somefile
argc is 2;
argv[0] is the string "./myprogram"
argv[1] is the string "/tmp/somefile"
argv[2] is a NULL pointer
Although not covered by standards, on Windows and most flavours of Unix and Linux, main can have up to three arguments:
int main(int argc, char *argv[], char *envp[])
The last one is similar to argv (which is an array of strings, as described in other answers, specifying arguments to the program passed on the command line.)
But it contains the environment variables, e.g. PATH or anything else you set in your OS shell. It is null terminated so there is no need to provide a count argument.
These are for using the arguments from the command line -
argc contains the number of arguments on
the command line (including the program name), and argv is the list of
actual arguments (represented as character strings).
These are used to pass command line paramters.
For ex: if you want to pass a file name to your process from outside then
myExe.exe "filename.txt"
the command line "filename.txt" will be stored in argv[], and the number of command line parameter ( the count) will be stored in argc.
main() is a function which actually can take maximum three parameters or no parameters.
The following are the parameters that main() can take is as follows:-
1) int argc : It holds the number of arguments passed (from the command prompt) during the execution of program or you can say it is used ot keep a track of the number of variables passed during the execution of program. It cannot hold the negative value. Eg. If you pass your executable file "./a.out" then that will be considered as a parameter and hence argc value will be 0 i.e 1 value is passed.
2) char *argv[] : it is an array of character pointers which holds the address of the strings(array of characters) that are passed from the command prompt during execution of program. Eg. in above example if you wrote argv[argc] i.e argv[0] in cout then it will give ./a.out as output.
3) char*envp[] : it is also an array of character pointer which is used to hold the address of the environments variables being used in the program. Environment variables are the set of dynamic named value that can affect the way the running process will behave on the computer. For example running process want to store temporary files then it will invoke TEMP environment variables to get a suitable location.
I know basic c++ coding and command line syntax. Often, the main() function looks like this:
int main(int argc, char *argc[])
When I execute my program via command line, I just do this:
cd c:\path\to\foo 1
Where 1 is the argument. And of course, arc would be equal to '2' and the '1' would be at element '1' of the array argc (as opposed to 0).
I've seen the dash used in a lot of places. Specifically:
gcc -v
And if I type just 'gcc', it says there are no arguments. And if I type 'gcc v' I get "error: no such file or directory". But when I take a look at the minGW bin folder where gcc is, there is no folder 'v'.
This is just a style of setting options in programs. It comes from the getopt library provided in Unix/Linux etc. The -- form that kc7zax mentions in their answer is the long-form of these options (allowing a long identifier instead of a single character).
There's nothing magic about these. They are simply parsed out of the argv array. You can implement similar or identical functionality yourself if you want. But it's a pain. That's why libraries exist.
It is simply a standard for identifying the name of a command line parameter from the parameter value that may exist. e.g. gcc -o myfile.o
The windows world usually sees a - or a / as the marker. I've often encountered a -- in the unix/bsd world as well.
I am having trouble using getopt_long() function with custom argc and argv.
I receive my arguments in a string instead of the real command line args. Then a new_argc and new_argv was built from this string to be used with getopt_long(). But getopt_long() fails on the first call itself. returns EOF and optarg = NULL.
string is "-c 10.30.99.41"
new_argc = 3
new_argv[0]=>./prog_name
new_argv[1]=>-c
new_argv[2]=>10.30.99.41
getopt_long works OK for me if I pass command line args. So my short and long options are correct. But if I pass the new_argc and new_argv it fails.
I am sure my short and long options are right and the argv is NULL terminated. I apologize I cant post more code here.
I doubt if getopt_long can be used with a custom argc and argv. I suspect it works only with a real argc and argv because it must be referencing some other code in libc related to argc,argv. Any comments?
option = getopt_long( new_argc, new_argv, short_options, long_options, NULL );
EDIT:
"The variable optind is the index of the next element to be processed in argv. The system initializes this value to 1. The caller can reset it to 1 to restart scanning of the same argv, or when scanning a new argument vector."
So, yes. You can use getopt_long to scan the arguments or another argument list again. However, if someone has called getopt_long previously, you have to set the global optind variable to back to 1.
Remember that the argv in main() is NULL terminated and argc long, that is; argv[argc] == NULL. So you likely have to make sure the last element in your own new_argv is a NULL pointer.
(Note, please show all the relevant code when posting, it's hard to guess what the error is, e.g. showing what short_options, long_option is, how you actually build your new_argv, variable declarations etc.)
This question already has answers here:
Closed 11 years ago.
Duplicate:
What is the proper declaration of main?
What do we mean by the arguments in this main function? What are they trying to tell us?
int main(int argc, char** argv)
UPDATE: And, is the preceding line of code similar to this int main(int argc, char* argv[])? If so, how can we say that char** argv is similar to char* argv[] as they don't look similar from an array point of view?
How is it compared with int main() which does not have any arguments?
Thanks.
The argc parameter is the number of command line options specified, including the executable name, when the executable was invoked. The individual command line options are found in the argv array, which is NULL terminated (the name and path used to invoke the executable is in argv[0]).
The difference between the two versions is simply if you want to parse command line arguments or not - if you are not interested in them then you can ignore them using the second form.
Wikipedia offers a good explanation. The first parameter gives you the number of command line arguments, and the second gives you the actual arguments.
They represent the command line parameters.
argc is the number of command line parameters, including the name of the executable.
argv is an array of null-terminated strings, where argv[0] is the command line parameter, and argv[i] is the ith parameter after that, argv[argc-1] being the last one and argv[argc] is actually well defined and a NULL pointer.
Thus:
foo bar baz
on the command line will have argc=3, argv[0]="foo" argv[1]="bar" argv[2]="baz" argv[3] = NULL
Note that there is no special attachment placed for "flag" arguments.
grep -i foo bar.cpp bar.h
would have 4 arguments (argc=5 including grep itself), -i being one of them and this would apply even if the next parameter was a "value" attached to the flag.
Note if you did a wildcard
grep -i foo *
in UNIX at least, the * would be expanded before the call into grep and thus each file that matched would be an argument.
Incidentally
char** argv and char* argv[]
do the same thing.
Also while the standard says you must use one of these signatures (you shouldn't even add in any consts) there is no law you have to use those two variable names, but it is so conventional now that they are pretty much universal. (i.e. you could use argCount and argValues if you want).
argc gives you the number of arguments and argv gives you those arguments. The first one is the path to the .exe used to run your program, the following ones are arguments the caller of your .exe provided on the command line like this:
my.exe arg1 arg2
Whereas
int main() {}
just ignores the arguments.
argv is an array holding the commandline parameters passed to the application. argc tells you the number of elements contained in that array.