How can resort a string array in C using pointers only

Question

char * players[3]={"Player 1","Player 2","Player 3";
char *p = *players;
char *temp;
for(int t = 0; t < 2; t++)
     {
         temp = *(p+t);
         *(p+t) = *(p+t+1);
         *(p+t+1) = temp;
     }

How can I resort the array using pointers only ? I wan't it to be like this : "Player 2","Player 3", "Player 1"

Resort based on what ordering ? Or do you mean reorder by rotating everything left, and the initial left-most pointer placed on the sequence tail ? — WhozCraig
– WhozCraig, Commented Dec 11, 2018 at 1:34
qsort() only uses pointers. Oh, and the code is missing a } on line 1. — Kingsley
– Kingsley, Commented Dec 11, 2018 at 1:35
I wanna reorder by replacing every element with the right element next to it. @WhozCraig — Yamen Massalha
– Yamen Massalha, Commented Dec 11, 2018 at 1:44

Kingsley · Accepted Answer · 2018-12-11 03:27:14Z

1

It looks like you want to do a left-shift (rotation).
This is a pretty simple operation

Keep a pointer to the head
For every element, make that spot point to the next element
put the head on the tail

The Code:

#include <stdio.h>

void rotateLeft( char **arr, int elements )
{
    // keep a pointer to the head
    char *head = arr[0];

    // shift every element left
    for (int i=0; i<(elements-1); i++)
    {
        arr[i] = arr[i+1];
    }

    // put the head on the tail
    arr[elements-1] = head;
}

// EDIT - rotate without [] notation.
void rotateLeftNoArrayNotation( char **arr, int elements )
{
    // keep a pointer to the head
    char *head = *arr; //arr[0];

    // shift every element left
    for (int i=0; i<(elements-1); i++)
    {
        *(arr+i) = *(arr+i+1); //arr[i] = arr[i+1];
    }

    // put the head on the tail
    *(arr+elements-1) = head; //arr[elements-1] = head;
}

int main( void )
{
    char *players[3] = { "Player 1", "Player 2", "Player 3" };

    for (int i=0; i<3; i++)
        printf( "players[%u] = \"%s\"\n", i, players[i] );

    rotateLeftNoArrayNotation( players, 3 );

    for (int i=0; i<3; i++)
        printf( "players[%u] = \"%s\"\n", i, players[i] );

    return 0;
}

Which gives:

$ ./rotLeft
players[0] = "Player 1"
players[1] = "Player 2"
players[2] = "Player 3"
players[0] = "Player 2"
players[1] = "Player 3"
players[2] = "Player 1"

edited Dec 11, 2018 at 3:27

answered Dec 11, 2018 at 1:44

Kingsley

15k5 gold badges38 silver badges60 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Yamen Massalha Over a year ago

Yes, exactly, But I wanna do it using pointers, without using any of [ ].

WhozCraig Over a year ago

@YamenMassalha subscripting via [ ] is just as applicable against a pointer as it is against an array (which converts to pointer-to-first-element when use in an expression context, i.e. almost everywhere except decl). The expression *(p+i) is synonymous with p[i] whether p is an array or a pointer. So the statement "I wanna do it using pointers" is irrelevant. what you're really saying is you want to do it without using array subscript notation (apparently), applied anywhere (pointer or otherwise). Is that correct ?

Kingsley Over a year ago

@YamenMassalha - I updated the answer, see rotateLeftNoArrayNotation() - basically it's the same function, slightly different syntax. Syntax that would not pass a code review ;)

Collectives™ on Stack Overflow

How can resort a string array in C using pointers only

1 Answer 1

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related