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Copy argv to string in C(newbie question)

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Copy argv to string in C(newbie question)

Oct 24th, 2006

hello. i'm new here and new to C programming so would appreciate any help with a problem i have. how can i copy the contents of a command line argument e.g. argv[1] to a string declared within main() e.g. filename[30]

cheers.....
PS this isn'y my homework:lol:

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Re: Copy argv to string in C(newbie question)

Oct 24th, 2006

Try memcpy or strcpy.

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Originally Posted by http://www.acm.uiuc.edu/webmonkeys/book/c_guide/2.14.html

2.14.4 memcpy

Declaration:

void *memcpy(void *str1, const void *str2, size_t n);

Copies n characters from str2 to str1. If str1 and str2 overlap the behavior is undefined.

Returns the argument str1.

2.14.13 strcpy

Declaration:

char *strcpy(char *str1, const char *str2);

Copies the string pointed to by str2 to str1. Copies up to and including the null character of str2. If str1 and str2 overlap the behavior is undefined.

Returns the argument str1.

Last edited by andor; Oct 24th, 2006 at 8:33 am.

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Re: Copy argv to string in C(newbie question)

Oct 24th, 2006

i've tried strcpy() but when i compile and run it the prog hangs then crashes. i'm probably doing something basically wrong in the fubction maybe.

 Help with Code Tags 
 C Syntax (Toggle Plain Text) 
int main(int argc, char *argv[]){
 
    char filename[30];
    strcpy(filename, argv[1]);
 
}
int main(int argc, char *argv[]){
    
    char filename[30];
    strcpy(filename, argv[1]);
    
}

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Re: Copy argv to string in C(newbie question)

Oct 24th, 2006

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Originally Posted by toxic

i've tried strcpy() but when i compile and run it the prog hangs then crashes. i'm probably doing something basically wrong in the fubction maybe.
Help with Code Tags

C Syntax (Toggle Plain Text)
int main(int argc, char *argv[]){
 
    char filename[30];
    strcpy(filename, argv[1]);
 
}
int main(int argc, char *argv[]){ char filename[30]; strcpy(filename, argv[1]); }

Yes probably U R doing something wrong. What R U passing as arg[1]? Probably nothing and thats why it crashes.

 Help with Code Tags 
 C Syntax (Toggle Plain Text) 
#include <stdio.h>
#include <string.h>
 
int main(int argc, char *argv[]){
 
   char filename[30];
   if (argc != 2)
   {
      printf("Dude are you passing something at all");
      return 1;   
   }
   strcpy(filename, argv[1]);
   printf("filename: %s\n", filename);
   return 0;
}
#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]){
    
   char filename[30];
   if (argc != 2)
   {
      printf("Dude are you passing something at all");
      return 1;   
   }
   strcpy(filename, argv[1]);
   printf("filename: %s\n", filename);
   return 0;
}

This works for me.
PS: memcpy is safer

Last edited by andor; Oct 24th, 2006 at 8:59 am.

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Re: Copy argv to string in C(newbie question)

Oct 24th, 2006

D'oh! just going to crawl into a hole and die to save the embarassment!!:eek:
cheers for your help....

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Re: Copy argv to string in C(newbie question)

Oct 24th, 2006

Its not a big deal, everyone mistakes and we learn from it.

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Re: Copy argv to string in C(newbie question)

Oct 31st, 2007

I also want to read filename, but in addition I need to read some interger parameter. I used this, now want to change. How can I handle: ./progname infilename integer ?

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
int main(){    
         ifstream input;                                  
		 input.open("filename");                           
		 if (!input)
		 {
                cout<<"File not open";
                exit(1);
                }
     getline(input, str);
     return 0;
}
int main(){    
         ifstream input;                                  
		 input.open("filename");                           
		 if (!input)
		 {
                cout<<"File not open";
                exit(1);
                }
     getline(input, str);
     return 0;
}

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Re: Copy argv to string in C(newbie question)

Nov 1st, 2007

> PS: memcpy is safer
No it's not. It will try to copy n characters whether it hits an architecture-dependent read boundary or not. Use strncpy(). It is fully standard and stops in the right place.

To convert a string to an integer #include <cstdlib> and use the atoi() function:
int i = atoi( "42" );

Also, this is the C forum. If you plan to use C++ please post in the C++ forum. While the same people frequent both, the advice you get will vary depending on the language. (C and C++ are significantly different languages.)

Hope this helps.

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Re: Copy argv to string in C(newbie question)

Nov 1st, 2007

>i'm probably doing something basically wrong in the fubction maybe.
Are you passing any arguments to your program? If not, argv[1] isn't guaranteed to be there, and if it is, it's pretty much sure to be NULL. If you're passing something, make sure that it's less than 30 characters or you'll overflow the buffer. strncpy is the logical choice, but it's really poorly named and doesn't do quite what you're expecting. I'd say use strncat instead:

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
#include <string.h>
 
int main ( int argc, char *argv[] )
{
  char buffer[30];
 
  if ( argc > 1 ) {
    /* strncat won't work if buffer isn't already a string */
    buffer[0] = '\0';
    strncat ( buffer, argv[1], 29 );
  }
 
  return 0;
}
#include <string.h>

int main ( int argc, char *argv[] )
{
  char buffer[30];

  if ( argc > 1 ) {
    /* strncat won't work if buffer isn't already a string */
    buffer[0] = '\0';
    strncat ( buffer, argv[1], 29 );
  }

  return 0;
}

>It will try to copy n characters whether it hits an
>architecture-dependent read boundary or not.
An architecture-dependent read boundary? Can you describe what you mean by this?

>Use strncpy(). It is fully standard and stops in the right place.
memcpy is fully standard. Also, strncpy always copies n characters. If the string isn't shorter than the buffer, strncpy will pad the buffer all the way to the count with null characters. If the string is longer than the buffer, strncpy will fail to null terminate the buffer. Despite the name, strncpy shouldn't be used as a "count restricted strcpy".

>To convert a string to an integer #include <cstdlib> and use the atoi() function
I wouldn't recommend atoi unless you've completely validated the input string first. It's basically impossible to tell if atoi failed without prior validation, and if the string isn't representable as an integer then calling atoi produces undefined behavior. It's too risky, so you should use strtol instead:

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
char *end;
long temp;
int x;
 
errno = 0;
temp = strtol ( buffer, &end, 0 );
 
if ( end == buffer )
  puts ( "No conversion made" );
else if ( *end != '\0' )
  puts ( "Partially invalid string" );
else if ( temp < INT_MIN || temp > INT_MAX )
  puts ( "Out of range value" );
else
  x = (int)temp;
char *end;
long temp;
int x;

errno = 0;
temp = strtol ( buffer, &end, 0 );

if ( end == buffer )
  puts ( "No conversion made" );
else if ( *end != '\0' )
  puts ( "Partially invalid string" );
else if ( temp < INT_MIN || temp > INT_MAX )
  puts ( "Out of range value" );
else
  x = (int)temp;

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Re: Copy argv to string in C(newbie question)

#10

Nov 1st, 2007

I never said memcpy() wasn't standard. Only that strncpy() was also.
The difference in choice is this: do you own/did you allocate the memory you are accessing? If not, you may very well be playing with fire to try reading (or writing) past the end of memory you have not allocated.

The strncpy() function always pads the target out to n bytes. However, it stops reading the source when it hits the end of the string, which the C specification guarantees to be there in argv[].

You are correct about watching the end of string, as strncpy() does not guarantee that the target is null-terminated.

I will agree that the use of strncat() is the best choice so far, since it addresses both issues nicely.

The strtol() has the same validation problems as stroi(). Just because it returns something different... But you are right in that it is a little more convenient for catching incomplete conversions...

Cheers.