pointer to an array

Why Doesnt This code work in Linux??

Code:

---

main()
{
        int a[3][4] = {1,2,3,4,
                        5,6,7,8,
                        0,9,1,6
                                  };
        show(a,3,4);
}

show(int (*q)[4],int row, int col)
{
        int i,j,*p;
        for(i=0;i<row;i++)
        {
        p = q+i;
        for(j=0;j<col;j++)
        printf("%d",*(p+j));
        printf("\n");
        }
        printf("\n");
}

---

Well that depends on how you define "doesn't work".

Perhaps it's the lack of a prototype?
Perhaps it's the broken array notation in the function?
Perhaps it's Penry, the mild mannered janitor.

If you enable warnings (with -Wall) you will get a heap of warnings for your code.

I managed to make it work with (on Windows, but using gcc which is the same compiler as you'd use on Linux, so the code generated should match) a few minor changes.

--
Mats

Following version of your program is working fine with gcc( if we ignore a few warnings that it through):

Code:

---

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

int main()
{
        int a[3][4] = {1,2,3,4,
                      5,6,7,8,
                                        0,9,1,6
                                  };
        show(a,3,4);
        return(0);
}

show(int (*q)[4],int row, int col)
{
        int i,j,*p;
        for(i=0;i<row;i++)
        {
        p = q+i;
        for(j=0;j<col;j++)
        printf("&#37;d",*(p+j));
        printf("\n");
        }
        printf("\n");
}

---

Folloing is the output that i get:

Code:

---

1234
5678
0916

---

What do you get when you try to compile it?

Quote:

---

Originally Posted by jas_atwal

Following version of your program is working fine with gcc( if we ignore a few warnings that it through):

Code:

---

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

void show(int (*q)[4],int row, int col);

int main()
{
        int a[3][4] = {1,2,3,4,
                                5,6,7,8,
                                0,9,1,6
                                };
        show(a,3,4);
        return(0);
}

void show(int (*q)[4],int row, int col)
{
        int i,j;
        int (*p)[4];
        for(i=0;i<row;i++)
        {
                p = &q[i];
                for(j=0;j<col;j++)
                        printf("&#37;d",p[j]);
                printf("\n");
        }
        printf("\n");
}

---

Folloing is the output that i get:

Code:

---

1234
5678
0916

---

What do you get when you try to compile it?

---

You realize it's much easier to read when indented properly, like above?
And x+y and *(x+y) is harder to read than &x[y] and x[y].

And jas_atwal, you just had to go pad with spaces when the OP used tabs, didn't you? Sheesh.
Other errors are show should have a return value, and if you return nothing, it should be void. Main should be int as pointed out.
p should be declared as int (*)[4]. Because int* is not the type of your pointer q!
I also modified the arrays to use [] instead of x + y and *(x + y). It's easier to read.

...And don't mix tabs and spaces! Use one, stick with one.

And I did not test the code, of course.