Thread: How To pass 2 dimensional array of strings to a function

You might write the function as:

Code:

void print_strings(char arr[][30], size_t size)
{
    size_t i;
    for (i = 0; i < size; ++i)
    {
        printf("&#37;s\n", arr[i]);
    }
}

You would then call it as:

Code:

print_strings(temp, 10);

Incidentally, please read our wiki FAQ concerning void main and [url=http://cpwiki.sourceforge.net/Gets]gets[/man] to find out what is wrong with them. You should also indent your code better.

probably becasu temp is declared as char temp[10][30]
print_strings should be declared as

Code:

void print_strings(char arr[][30], size_t size);

and not char**

PS. And i should be declared as size_t not int

print_strings should be declared as

And i should be declared as size_t not int

Indeed, I overlooked both, so I shall make the fixes.

Ok now can anybody tell me how to do the same job with integer array??i mean please can anybody give me code for that? i have done it by 1 method but anybody knows a sweet and short method for it then please tell me?

Code:

#include <stdio.h>
void print(int *a,int m,int n)
{
 int i,j;
 for(i=0;i<m;i++)
 {
   for(j=0;j<n;j++)
   {
     printf("&#37;d ",*(a+i*n+j));
   }
     printf("\n");
 }

}

int main(void)
{
 int a[3][3],i,k=1,j;
 for(i=0;i<3;i++)
 {
   for(j=0;j<3;j++)
   {
     a[i][j]=k++;
   }
 }
 print(&a[0][0],3,3);
 return 0;
}

Also explain me the funtion of size_t i haven't come across that function

If you base your new code on my corrected example, you should have concluded that:

Code:

void print(int *a,int m,int n)

should be:

Code:

void print(int a[][3],int m,int n)

and that:

Code:

print(&a[0][0],3,3);

should be:

Code:

print(a, 3, 3);

Also, you can use a normal array subscript with a[i][j] instead of trying to compute the array index yourself.

By the way, I think that an indentation of just one space may be too little. Typically about four spaces would be optimal.

one dimention parameter is enough - second is hardcoded

Code:

void print(int a[][3],int m)

Originally Posted by manav

do this:

Code:

void func(char** names, n)
{
for (int i=0; i<n; i++) puts(names[i]);
}

// then call like this:
func(temp, n);

A double pointer won't work as a replacement for a 2D array - it only works the other way around, that you can make a 2D array by using a double pointer. If you are going to use a double pointer, you will have to allocate "rows" number of pointers, then assign each pointer to the first element of the rows.

--
Mats

> but anybody knows a sweet and short method for it then please tell me?
Yeah, it's called copy and paste.

If this is your array,
int arr[2][3][4];

Then this is your function prototype
void func ( int arr[2][3][4] );


Yes, you can then fiddle with the prototype to change it into either of these forms, but you don't buy a lot. Some compilers will warn about the left-most dimension not being empty, but apart from that, it's good.
Alternatives:
void func ( int arr[][3][4] );
void func ( int (*arr)[3][4] );

Irrespective of the form you choose to write the function prototype / declaration, it would be called with
func ( arr );

And the usual array access subscripts you would use if the array were in scope will still do what you want inside the function.