[SOLVED] argv in a C program.

stf92 · 03-11-2015, 06:23 PM

In C, what are the strings in argv[][] in main(int argc, *argv[]). Are they null terminated strings? Are they terminated by a line terminator?

Of course argv could be any other name.

maciuszek · 03-11-2015, 06:32 PM

Edit: deleted
Didn't read it through

maciuszek · 03-11-2015, 06:41 PM

Yeah they're terminated with a \0 all "strings" in the second dimension.

The first dimension I'm not sure of I haven't ever not based it on the count I don't think.. Although 0 will return the program call .. Hrm look how execl works for a better understanding . I would assume the final index to be a null pointer in the first dimension of the vector.. But not sure.

maciuszek · 03-11-2015, 06:45 PM

I was right:

Quote:

The list of arguments must be terminated by a NULL pointer, and, since these are variadic functions, this pointer must be cast (char *) NULL.

Edit: also I advocate the use of stdin over such arguments.. Where it I sent regressive or incredibly impraticle to do so

rknichols · 03-11-2015, 08:22 PM

Quote:

Originally Posted by maciuszek

I was right:

Quote:

The list of arguments must be terminated by a NULL pointer, and, since these are variadic functions, this pointer must be cast (char *) NULL.

[pedantic-mode]That just says how those members of the exec() family are called. It says nothing about what is actually passed to the invoked program.[/pedantic-mode]

The C Language Standard does state in section 5.1.2.2.1, "argv[argc] shall be a null pointer".

perbh · 03-11-2015, 09:19 PM

argv is kind-of doubly defined, both by argc and the final NULL-pointer.
You can loop through the arguments in two ways:

Code:

int i = 1;
char *s;
while (1) {
  s = argv[i++];
  if (s == NULL) break;
  /* ... */
}

Code:

int i;
char *s;
for (i=1; i<argc; i++) {
  s = argv[i];
  /* ... */
}

Note that a 'string' in c is always zero-terminated. You can easily have a byte-string with zero's in it, but then you also have to carry along a length-entity, and the usual c-library string-functions are no longer valid (well, not invalid but inappropriate)

SoftSprocket · 03-11-2015, 10:35 PM

Quote:

Originally Posted by perbh

Code:

int i = 1;
char *s;
while (1) {
  s = argv[i++];
  if (s == NULL) break;
  /* ... */
}

Or more succinctly:

Code:

while (*(++argv)) {
    printf ("%s\n", *argv);
}

NevemTeve · 03-12-2015, 12:19 AM

Quote:

In C, what are the strings in argv[][] in main(int argc, *argv[]). Are they null terminated strings? Are they terminated by a line terminator?

Yes. No (if you mean \n).

AnanthaP · 03-12-2015, 04:30 AM

Why do you worry?

Inside your programme, you are dealing with \0 terminated strings like any other string.

I mean when you type it in at the command line, it is a series of strings (quoted if there are intermediate spaces) with each argument terminated by a space.

OK

jpollard · 03-12-2015, 06:38 AM

Quote:

Originally Posted by NevemTeve

Yes. No (if you mean \n).

Unless the string was something like "a line terminator\n"...

for the OP:

But it does have to be explicitly stated. By default command parameters do not have newlines.

However, consider something like the awk parameters:

Code:

$  awk '
> BEGIN {count = 25;}
> {count++;}
> END {print count;}
> '
1
2
ctrl-d
27
$

The first parameter to awk does have a "\n" at the end...

The point is that argv[] are null terminated strings - which may include any valid character (even UTF-8 characters).

rtmistler · 03-16-2015, 12:33 PM

Quote:

Originally Posted by stf92

In C, what are the strings in argv[][] in main(int argc, *argv[]). Are they null terminated strings? Are they terminated by a line terminator?

Of course argv could be any other name.

Do all the varieties of more complex answers solve your question?

Why did you ask? Are you looking to write a program which uses command line arguments and wish to ensure that you get things correct? There are a few ways. Firstly argc tells you how lengthy the argv array is, or should be. And further you can use library functions like the getopt(3) to assist you. I've done both and depending on how I want the invocation line to look, I either parse the arguments manually or choose to use getopt(), and honestly as good as getopt() is, I always have to look it up, thus wasting a bit of time. Well ... at least I've managed to save an example or two over the years so I know what works and what doesn't.

stf92 · 03-17-2015, 07:47 AM

I did printf("%s", argv[1]) and it printed argv[1]. I saw then that argv[n] is a null terminated string for any n. Thanks also for the replies, which were very helpful.