Passing array into function in c - c

Question

I'm getting gcc errors when I compile my code. The errors are about "passing argument 1 of ‘print_path’ makes pointer from integer without a cast".
Here is my function prototype:
void print_path(int previous[], int desired_node_index);
Here is my function:
void print_path(int previous[], int desired_node_index)
{
if( previous[desired_node_index] != -1 )
print_path( previous[desired_node_index] );
printf("-> %d ", previous[desired_node_index]);
}
and here is where I call my function:
print_path(previous, dest_index);
I'm obviously passing it in wrong, or else I'm doing something incorrectly about how to pass an array into a function. Any help?
Thanks guys!

Answer 1

This is obviously a recursive function. Note that print_path() takes 2 parameters: the first is an int array, and the second is an index to a position inside that array.
Calling it:
print_path( previous[desired_node_index] );
is absolutely wrong (unless you have overloaded this function), because it expects 2 parameters and you are only passing it one. What you should be doing is:
print_path( previous, desired_node_index );
What you seem to be missing in this function is an operation to increase/decrease the index variable, else you will always be printing the same position in the array.
Without knowing what is exactly that you are trying to do, there's the possibility that you wanted to do this:
print_path( previous, previous[desired_node_index] );

Answer 2

An obvious error is:
print_path( previous[desired_node_index] );
I'm not sure what you're trying to do, but I guess you want something like:
#include <stdio.h>
void print_path(int *previous, int desired_node_index);
int main(void) {
int dest_index = 2;
int previous[5] = { -1, 0, 1, 2, 3};
print_path(previous, dest_index);
return 0;
}
void print_path(int *previous, int desired_node_index) {
if( previous[desired_node_index] != -1 )
print_path( previous, previous[desired_node_index]);
printf("-> %d ", previous[desired_node_index]);
}

Answer 3

void receive_array(int *temp_arr)
{
int i=0;
do
{
temp_arr[i]=temp_arr[i]+1;
i++;
}
while((char)temp_arr[i]!='\0');
}

Answer 4

here I have made some modifications. The array temp_arr2[] is a buffer array. In my actual program, I printed the array from the main(). Here, for doing the same thing, one need to store back the end result of some computation into temp_arr[].MAX can be a macro or a global variable. In the previous one, I just forgot to edit the lines: temp_arr[i]=temp_arr[i]+1; (my demo sample code) :)
void receive_array(int *temp_arr)
{
int i=0;
int temp_arr2[MAX];
do
{
temp_arr2[i]=temp_arr[i];
i++;
}
while((char)temp_arr[i]!='\0');
}

Answer 5

If you want to pass array to function and return after changing the elements from the function you can see the following example:
You may find the solution at: https://github.com/krishnabhat81/Send-and-return-array-from-function-in-C
#include <stdio.h>
/*
If you want to return a single-dimension array from a function, you would have to
declare a function returning a pointer as in the following example:
*/
int *getRandom(int arr[])
{
static int r[10];
/*Second point to remember is that C does not advocate to return the address of a
local variable to outside of the function so you would have to define the
local variable as static variable.*/
int i;
for ( i = 0; i < 10; ++i)
{
r[i] = arr[i]+1;//rand();
printf( "r[%d] = %d\n", i, r[i]);
}
return r;
}
/* main function to call above defined function */
int main ()
{
/* a pointer to an int */
int *p;
int i;
int arris[10] = {110,22,33,44,5,6,7,8,9,20};
p = getRandom(arris);
for ( i = 0; i < 10; i++ )
{
printf( "*(p + %d) : %d\n", i, *(p + i));
}
return 0;
}
return 0;
}