pointer to an array in a function

[powerb23]

Hi,
I wanted to find out how i would point to the array arr in function get(char *p) so that i ould print the contents of this array.
Your help will be appreciated.



int main(){
char * ptr = NULL:

get(ptr);
printf("contents: %s:, ptr);
}

void get(char *p){
char [5]mm = NULL;
int i;

for(ii = o;i <5;i++){
mm = i;
}

p = mm; /*I'm not sure about this part */
}

 

[manvid]

iam afraid i didnt get ur question/code rite.. 'but i have a hunch that u need sumthing like dis

main()
{
  int a[]={1,2,3,4};

   get(a);
}
void get (int *p)
{
 int i = 0;
 for (i=0; i < 3; ++i)
   printf(&quot;%d\n&quot;, p[i]);

this is jus d basic stuff. u may also pass the array max subscript in the function so that u know when the loop must terminate.
here my function get already &quot;knows&quot; that index
the same thing can be done for a character array..
here there is 1 advantage
the loop conditon can be replaced by

for (i = 0 ; p[i]!='\0' ; ++i)
or
for (i = 0 ; p[i] ; ++i)

hope this helps

 

[raghu345]

I thought you wanted to pass a charcter pointer to a sub-function, assign values to it & print it back in the main function. Please find below a sample function which achieves the same.

#include <stdio.h>
#include <malloc.h>

void get(char **p)
{ 
	unsigned int i; 
	char *q;
	q=*p;
	
	for(i=0;i<5;i++)
	{ 
		*q = 'a';
		q++;
	} 
	*q = '\0';
}

int main()
{ 
	char *ptr;
	unsigned int i=0;

	ptr = (char*)malloc(sizeof(char) * 6);

	if(ptr != NULL)
	{
		get(&ptr); 
		printf(&quot;contents: %s\n&quot;, ptr); 
		free(ptr);
	}
	return 1;
} 
[\code]

 

[powerb23]

Thanks alot for the help. i used the for loop and got the program up and running.
All your help has been greatly appreciated