DaniWeb Forum Index > Software Development > C

Arrays and Pointers

Please support our C advertiser: Programming Forums - DaniWeb Sister Site

•

Join Date: Jul 2008

Posts: 6

Reputation: java_girl is an unknown quantity at this point

Solved Threads: 0

java_girl

Offline

Newbie Poster

Arrays and Pointers

Dec 1st, 2008

Hi.

I recently read that using the name of an array without brackets was one method for accessing the <address> of the array's first element. E.g:

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
#include <stdio.h>
 
int arr[5] = { 2, 4, 6, 8, 10 };
 
int main( void ) {
 
printf( "The address of the '2' is %d\n", arr );
 
return 0;
}
 
char* message1 = "C";
char* message2 = "is the";
char* message3 = "best";
char* message4 = "programming";
char* message5 = "language!";
 
 puts( message1 );
 puts( message2 );
 puts( message3 );
 puts( message4 );
 puts( message5 );
 
/* message1 points to the 'C' and using puts (message1) will */
/* display the letter as opposed to the address. Why is this? */
/* I know the puts() function receives a char pointer as an */
/* argument. Is this why one need not dereference it by */
/* using *message1 unless one is using */
/* printf( "%c\n", *message1 ); */
/* Essentially, I'm slightly confused by when one pointer is */
/* pointing to the address of the first element in an array and when */
/* it is pointing to the first value */
 
return 0;
#include <stdio.h>

int arr[5] = { 2, 4, 6, 8, 10 };

int main( void ) {

printf( "The address of the '2' is %d\n", arr );

return 0;
}

char* message1 = "C";
char* message2 = "is the";
char* message3 = "best";
char* message4 = "programming";
char* message5 = "language!";

 puts( message1 );
 puts( message2 );
 puts( message3 );
 puts( message4 );
 puts( message5 );

/* message1 points to the 'C' and using puts (message1) will */
/* display the letter as opposed to the address. Why is this? */
/* I know the puts() function receives a char pointer as an */
/* argument. Is this why one need not dereference it by */
/* using *message1 unless one is using */
/* printf( "%c\n", *message1 ); */
/* Essentially, I'm slightly confused by when one pointer is */
/* pointing to the address of the first element in an array and when */
/* it is pointing to the first value */

return 0;

Thanks for the help,

java_girl

Last edited by Narue; Dec 1st, 2008 at 3:41 pm. Reason: fixed code tags

•

Join Date: Aug 2008

Posts: 77

Reputation: mahlerfive is an unknown quantity at this point

Solved Threads: 16

mahlerfive mahlerfive is offline

Offline

Junior Poster in Training

Re: Arrays and Pointers

Dec 1st, 2008

You are right that the array name is the address of the first element of that array. This works exactly the same for arrays of char as well.

In C, there is no "string" type, so we are forced to use arrays of char. There is one special difference between a string (array of char) and other arrays, and that is that the last element of a string MUST be the null terminating character '\0'. This signifies the end of the string.

The puts() function will take a char*, assume it is a string ( an array of char), and output all the chars in the array up to the null terminating character.

By using printf in the way you are using it, you are telling it to output one character (%c) and giving it the first character of the string (since you dereference the array, giving you the first element). If you want to print the whole string, use %s and don't dereference the array. You could also print any single character within a string by just giving the index of the character. For example:

 Help with Code Tags 
 C Syntax (Toggle Plain Text) 
printf( "%c\n", message2[3] ); // this will print 't'
printf( "%c\n", message2[3] ); // this will print 't'

•

Join Date: Jul 2008

Posts: 2,001

Reputation: ArkM has much to be proud of

Solved Threads: 343

ArkM

Offline

Postaholic

Re: Arrays and Pointers

Dec 1st, 2008

Some additions:
1. To print pointer value use %p (not %d) format specifier. As usually, it's a senseless operation

.
2. True type of a string literal is const char* - a constant pointer to the 1st element of a char array contained a string with terminated null character. You can't change string literal via pointer to it. Constructs like

Help with Code Tags

c Syntax (Toggle Plain Text)

char* message1 = "C";

are valid (for backward compatibility with older C programs) but modern compilers place literals in read-only memory.
3. An array name is not converted to a pointer to the 1st element of an array (&array[0]) in sizeof array_name (operator sizeof argument position).

Last edited by ArkM; Dec 1st, 2008 at 6:04 am.

•

Join Date: Aug 2006

Posts: 319

Reputation: Luckychap is on a distinguished road

Solved Threads: 42

Luckychap

Offline

Posting Whiz

Re: Arrays and Pointers

Dec 1st, 2008

if u want to print the address of first character in the string it is same as you did with other array. For Ex.

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
int intArray[3] = {1,2,4};
char *str = "C for Cat";
 
//This will print the address of 1 in inArray array
printf("The address of intArray[1] is %d", intArray);
 
//And this will print the address of 'C' in str
printf("The address of 'C' is %d", str);
int intArray[3] = {1,2,4};
char *str = "C for Cat";

//This will print the address of 1 in inArray array
printf("The address of intArray[1] is %d", intArray);

//And this will print the address of 'C' in str
printf("The address of 'C' is %d", str);