Array shifting help

[viclaster]

Can someone help me with array shifting. I am trying to shift an array a number of times that someone enters.

Example: The array is abcdef and I enter 1 to shift the array once. The output is bcdefa. If I enter 2 then the output is cdefab.

Here is my code so far. Can someone help me find out what I'm doing wrong. Thank you.

#include <stdio.h> 
char shift( int yy, char loop[7]);
int main() 
{
char loop[7] = "abcdef";
char arrloop[7];
int y;
printf("How many times do you want to shift this string?\n");
scanf("%d", y);

shift( y, char arrloop[7]);

printf("%s\n", loop);

}

{char shift(int yy, char loop[7]);
while( yy != 0)
{char x;
x = loop[5];
loop[5] = loop[4];
loop[4] = loop[3];
loop[3] = loop[2];
loop[2] = loop[1];
loop[1] = loop[0];
loop[0] = x;
yy--;}

printf("%s", loop);
getchar();
getchar();
return 0;
}

[laserlight]

I suggest you indent your code, and use the code bbcode tags.
I believe that your code blocks are somewhat misaligned.

btw, I think your idea is to keep shifting the places by one until the number of shifts is satisfied.
There should be a faster (and cleaner) way of doing it, by calculating the final place of each element, given the number of shifts.

[Salem]

> scanf("%d", y);
> shift( y, char arrloop[7]);

Should be
scanf("%d", &y);
shift( y, loop );

> {char shift(int yy, char loop[7]);
It's a bit mis-placed
Try
char shift(int yy, char loop[7]) {

[viclaster]

Thanks guys, you are a big help to me. I've changed a few things that helped, but now it prints out the array as many times as the scanf input. I just want it to print one time after the shifting takes place. How do I go about doing that. I even pulled the printf out of the function into the main and it still prints as many times as the scanf input. Here is what I have:

#include <stdio.h> 
char shift( int y, char loop[7]);
int main() 
{
    char loop[7] = "abcdef";
    int y;
    printf("How many times do you want to shift this string?\n");
    scanf("%d", &y);

       shift(y,  loop);

    printf("%s\n", loop);
    
getchar();
getchar();
return 0;
}

char shift(int y, char loop[7]) {

while( y != 0)
{
    char x;
    x = loop[5];
    loop[5] = loop[4];
    loop[4] = loop[3];
    loop[3] = loop[2];
    loop[2] = loop[1];
    loop[1] = loop[0];
    loop[0] = x;
    y--;
}

Thanks again.

[Sebastiani]

Store the length of the string. That way, the function will work with any size of array. Here's an example:

char *
 shift(char str[], int times)
{
 int end = strlen(str)-1;

 int i, loops;

 char value, temp;

    for(loops = 0; loops < times; ++loops)
   {
    value = str[0];

       for(i = end; i >= 0; --i)
      {
       temp = str[i];
       str[i] = value;
       value = temp;
      }
   }

 return str;
}

[Sebastiani]

>> but now it prints out the array as many times as the scanf input.

Recompile. The code you posted should only print once.

[viclaster]

Okay, I thik I'm getting closer. I have narrowed it down to one error I think. It says line 33 parse error at end of input. Strange that it has only 32 lines of code.

Sebastiani I tried fooling around with what you posted but I got more errors and crashes than I can shake a stick-of-ram at. I'm still new at this and I think you were using pointers. Lord knows how bad I can mess pointers up. If my current code method does not work for me then I will try your method again.

#include <stdio.h> 
char shift( int y, char loop[7]);
int main() 
{
    char loop[7] = "abcdef";
    int y;
    printf("How many times do you want to shift this string?\n");
    scanf("%d", &y);

       shift(y,  loop);

    printf("%s\n", loop);
    
getchar();
getchar();
return 0;
}

char shift(int y, char loop[7])
 {
    while( y != 0)
    {
    char x;
    x = loop[5];
    loop[5] = loop[4];
    loop[4] = loop[3];
    loop[3] = loop[2];
    loop[2] = loop[1];
    loop[1] = loop[0];
    loop[0] = x;
    y--;
}
  **// I get a "parse error at end of input" here for some reason.//**

[viclaster]

I also get a " [Warning] In function `char shift(int, char*)':"
Is this what you were talking about with using the pointers? I'm a bit confused.

[quzah]

char shift(int y, char loop[7])
 {
    while( y != 0)
     {
    char x;
    x = loop[5];
    loop[5] = loop[4];
    loop[4] = loop[3];
    loop[3] = loop[2];
    loop[2] = loop[1];
    loop[1] = loop[0];
    loop[0] = x;
    y--;
 }

See the problem?

Quzah.

[viclaster]

hehe, what can I say? Thank you very much. But I still get the error even after I recompile this:

#include <stdio.h> 
char shift( int y, char loop[7]);
int main() 
{
    char loop[7] = "abcdef";
    int y;
    printf("How many times do you want to shift this string?\n");
    scanf("%d", &y);

       shift(y,  loop);

    printf("%s\n", loop);
    
getchar();
getchar();
return 0;
}

char shift(int y, char loop[7])
 {
    while( y != 0)
    {
    char x;
    x = loop[5];
    loop[5] = loop[4];
    loop[4] = loop[3];
    loop[3] = loop[2];
    loop[2] = loop[1];
    loop[1] = loop[0];
    loop[0] = x;
    y--;
    }
}**// I get a "parse error at end of input" below this line that does not exist for some reason.//**

[thefroggy]

That code compiles fine for me. The only change I made was to make the shift function return void instead of char (since you're not returning anything from it)

[viclaster]

Thanks for the help so far. Okay here is what I've got. Does anyone else have any problems compiling and running this program?

#include <stdio.h> 
void shift( int y, char loop[7]);
int main() 
{
    char loop[7] = "abcdef";
    int y;
    printf("How many times do you want to shift this string?\n");
    scanf("%d", y);

       shift(y,  loop);

    printf("%s\n", loop);
    
getchar();
getchar();
return 0;
}

void shift(int y, char loop[7])
 {
    while( y != 0)
    {
    char x;
    x = loop[5];
    loop[5] = loop[4];
    loop[4] = loop[3];
    loop[3] = loop[2];
    loop[2] = loop[1];
    loop[1] = loop[0];
    loop[0] = x;
    y--;
    }
}

[Salem]

> scanf("%d", y);
This started off wrong,
Sometime later, you got it right
and now it's wrong again

[viclaster]

Okay, I've got it now. Thank you so much.