pointers and referencing

**nata** · 10-21-2002

The question on my problem set asks to compute the dot product with one of the arguments being a pointer to a pointer
the two vectors are stored as rows of an array 2xn
My code looks like this but its giving me fatal errors:

Code:

#include <stdio.h>
#include <stdlib.h>

double dotprod1(int n, double **a) /*n is the vector length*/
{
    int i, j;
    double dtprd1;

 a = calloc(n, sizeof(double *));
  
 printf("Please enter vectors:\n");
 
 for (i=0; i<2; ++i)
    for (j=0; j<n; ++j)
       scanf("%lf", a[i][j]);

for (j=0; j<n; ++j)
  dtprd1 += a[1][j] * a[2][j];

return dtprd1;
}

any clues where i went wrong i must admit i dont understand pointers and arrays well

Code Tags added by kermi3

outoftown · 10-21-2002

for starters u dont have a main function

**nata** · 10-21-2002

this is true! (i admit i dont know the basics yet) but the function that is there has something wrong with its referencing

***kermi3*** · 10-21-2002

I am posting this because you did not use code tags on this thread. In the furture please use Code Tags. They make your code MUCH easier to read and people will be much more likely to help you if you do. And they'll be happy about helping you

For example:

Without code tags:

for(int i=0;i<5;i++)
{
cout << "No code tags are bad";
}

With Code Tags:

Code:

for(int i=0;i<5;i++)
{
     cout << "This code is easy to read";
}

This is of course a basic example...more complicated code is even easier to read with code tags than without.

I've added code tags for you this time. They can be added by putting [code] at the beginning of your code and [/code] at the end. More information on code tags may be found at the link in my signature. I also suggest you take a look at the board guildlines if you have not done so already. Any further questions or ways I can help please feel free to PM me.

Good Luck,
Kermi3

**~~master5001~~** · 10-21-2002

Well the main() thing is really basic. But the reason you are having problems is because of this:

Code:

int **x = (int **)malloc(2 * sizeof(int *));
/*this allocates an array of pointers (2 pointers to be exact*/

x[0] = (int *)malloc(sizeof(int));
x[1] = (int *)malloc(sizeof(int));

/*this allocates the actual ints that you are going to work with*/

The point is that you are not allocating the memory required to an operation like
x[0][1] = 65;
You are must do the second malloc() or calloc().

Code:

int **x = (int **)calloc(2, sizeof(int *));

x[0] = (int *)calloc(1, sizeof(int));
x[1] = (int *)calloc(1, sizeof(int));

**Hammer** · 10-22-2002

Some reading on pointers and arrays for you.

Thread: pointers and referencing

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