C Arguments character Array

Question

I would like to pass in arguments and output them as an array of characters or string. What is the easiest way to do this? Once I have them as an array of characters I want to be able to manipulate the array. I was able to pass them in and use a character pointer to traverse through the string but I'd like to use a char array. Is there any easy way to do this? Im pretty new to C, here is what I have so far.

    if(argc != 3){
            printf("incorrect number of arguments, program needs 3\n");
            exit(1);
    }

    char s1[20];
    s1 = (char*)argv[1];
    char s2[20];
    s2 = (char*)argv[2];

    printf("String1 is %s\n", s1);
    printf("String2 is %s\n", s2);

    exit(0);

strcpy (argv[1],&s1); or, better: answer by Kerrek SB

Johan Lundberg
– Johan Lundberg

2012-02-06 20:41:58 +00:00
Commented Feb 6, 2012 at 20:41 — Johan Lundberg
– Johan Lundberg, Commented Feb 6, 2012 at 20:41

ouah · Accepted Answer · 2012-02-06 20:38:20Z

1

The strings pointed to by the argv array are modifiable, you can just modify the strings pointed to by the argv array, there is no need to copy them.

From the C standard:

(C99, 5.1.2.2.1p2) "The parameters argc and argv and the strings pointed to by the argv array shall be modifiable by the program, and retain their last-stored values between program startup and program termination."

answered Feb 6, 2012 at 20:38

ouah

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6 Comments

Kerrek SB Over a year ago

Do you have a standard reference for that?

Greg Hewgill Over a year ago

That's not generally a good idea, as changing the length of the argv strings can definitely be a problem.

ouah Over a year ago

@KornelKisielewicz yes, same quote in C90 in 5.1.2.2.1

ouah Over a year ago

@KerrekSB the quote says the elements of the argv array are modifiable AND the strings are modifiable.

Nicolas Fiorini Over a year ago

Thank you for the help. How do I access each argument? Can I have example of how I would print character 2 of argument 1?

|

Richard J. Ross III · Accepted Answer · 2012-02-06 20:43:38Z

1

You can simply print the values of argv directly:

printf("String1 is %s\n", argv[1]);
printf("String2 is %s\n", argv[2]);

If you need to modify the strings, here is how you would do that

char *newString1 = malloc(strlen(argv[1]) + 1 +  charactersToAdd); // add one for null termination.
strcpy(newString1, argv[1]);

// manipulate newString1 however you choose
printf("String1 is %s\n", newString1);
free(newString1);

edited Feb 6, 2012 at 20:43

answered Feb 6, 2012 at 20:37

Richard J. Ross III

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2 Comments

Johan Lundberg Over a year ago

He is asking for how to get modifiable variables.

Richard J. Ross III Over a year ago

@JohanLundberg ah, I must have misread the OP. I updated my answer now.

Kerrek SB · Accepted Answer · 2012-02-06 20:48:25Z

1

argv[1] and argv[2] are already pointers to character arrays, so you can use those directly. Moreover (see @ouah's post), the strings are already mutable, so you might not need anything beyond this at all.

If you really want to have a scope-local array, you need to use the C99 feature called "variable-length arrays", since the string lengths are only known at runtime:

#include <string.h>

int main(int argc, char * argv[])
{
    char s1[strlen(argv[1]) + 1];
    strncpy(s1, argv[1], sizeof s1);

    // ...
}

Alternatively you could malloc enough space for a copy of the string (say char * s1 = malloc(strlen(argv[1]) + 1);), but then you wouldn't have a local array after all (and you might as well use the original strings, unless you really wanted to make a copy).

edited Feb 6, 2012 at 20:48

answered Feb 6, 2012 at 20:40

Kerrek SB

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Comments

Kornel Kisielewicz · Accepted Answer · 2012-02-06 20:41:48Z

0

argv is already char* (actually an array of an array), however, they are noted const as in "do not modify" ;). What you want is actually making a copy of them so you can manipulate them freely:

char str[40];
strcpy (str,argv[1]);

Now you can manipulate str as you wish.

answered Feb 6, 2012 at 20:41

Kornel Kisielewicz

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3 Comments

Richard J. Ross III Over a year ago

But what happens if argv[1]'s length is greater than or equal to 40 characters?

Kornel Kisielewicz Over a year ago

@Richard - sure we can worry about that, and we probably always do. But did you worry when you were just starting out to learn C? It's good to keep things simple in the beginning :).

Richard J. Ross III Over a year ago

'Keeping things simple' is all fine and dandy in theory, but perhaps later on this code gets copied into production code and produces a bug that is hard to track down? Personally, I would do strncpy, and pass 39 bytes, and then set the null terminator personally.

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