Thread: Have problems with copying my array!

You should not use strcpy() on anything that isn't STRINGS. If you want to use a library function to copy generic data, then you should use memcpy().

Edit: And of course, with code like this:

Code:

   int array[ SIZE ]; /* declare the array of ints to be sorted (test with bublesort)*/
   int copy[SIZE];/*copy of original array to test with selectionsort*/
   strcpy(array ,copy );

even if we replace strcpy with something that copies the array properly, it will just copy garbage, since array and copy contain nothing useful until you have filled them in with something meaningful. So you are just copying random gunk, really [and you're just lucky that the program doesn't crash at that point - obviously there is a zero somewhere in the data before you've done too much damage by using strcpy to copy something that isn't a string].

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Mats

Originally Posted by AvaGodess

Yes sorry love, i just finished reading that part. So memcpy would solve the problem?

Please read my edit as well.

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Mats

Originally Posted by AvaGodess

I know at that point it has basically nothing. But arent I feeding it some data in this part

Code:

for ( i = 0; i < SIZE; i++ )
      array[ i ] = rand() % 90 + 10; /* give each element a value */
       
   printf( "Unsorted array:\n" );
    
   for ( i = 0; i < SIZE; i++ ) /* print the array */
      printf( "%d  ", array[ i ] );

I'm just asking for some help thats all, how could this little program be settled. Without doing what you say. Basically i wanted to make copies of the elements given to the array, and work with them in the function.

Yes, but that is AFTER you called strcpy(). My point was that if we assume that you replace strcpy() with something that actually copies array into copy, it would still contain garbage, since array has NOT YET been filled with something - it contains whatever happens to be in that location of memory [1], and it's very unlikely that this matches your random number sequence.

[1] Some compilers will actually fill unused memory with a recognisable pattern so that we can "see" that the memory has not been filled in, but the content is still not what you would like to have, so it's pretty irrelevant really.

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Mats

Originally Posted by AvaGodess

I know at that point it has basically nothing. But arent I feeding it some data in this part

Code:

for ( i = 0; i < SIZE; i++ )
      array[ i ] = rand() % 90 + 10; /* give each element a value */
       
   printf( "Unsorted array:\n" );
    
   for ( i = 0; i < SIZE; i++ ) /* print the array */
      printf( "%d  ", array[ i ] );

I'm just asking for some help thats all, how could this little program be settled. Without doing what you say. Basically i wanted to make copies of the elements given to the array, and work with them in the function.

The two arrays are not "magically" linked together.
They are separate arrays, with separate allocated space.
What you do with one, does not affect the other.
Therefore, if you want to duplicate the array, you should do so AFTER you've input the data.

Originally Posted by AvaGodess

Oo im begning to love C, but its a bit tricky i think im gona enjoy this language!

If you like C, then you're going to love C++.
Just a friendly advice

Originally Posted by AvaGodess

Yes sorry love, i just finished reading that part. So memcpy would solve the problem?

Think simple.

Code:

copy[i] = array[ i ] = rand() % 90 + 10;

da da!

Is your nickname inspired from Sin City (curious) ?

EDIT: Yes, memcpy() would work. But, in any case, you can always just use a loop and copy each individual element of the array.