Passing Arrays into function. How do i fix the code?

Question

#include <stdio.h>
void display(int *);
void show(int *);
int main()
{
int num[]={1,2,3,4,5}, s;
for(s=0; s<5; s++){
display(&num[s]);
}

return 0;
}
void display(int *x)
{
show(&x);
}
void show(int *a)
{
printf("%d\n", *a); //this is not printing the value.
}

I've tried using **a but the compiler doesn't execute, it throws a warning of incompatible pointer type.

Answer 1

Write in line 3
void show(int **);

and in line 17
void show( int **a)

The problem is as I mentioned before that x holds an address and show expects an address not the address to an address which is what you wanted to do by &x

Answer 2

write show(x) instead of write(&x)

Answer 3

did it not work?

Answer 4

@AKC actually the book that I'm learning from requires me to write only the show function. It had show(&x) and then said to write a show function which prints all the values.

Answer 5

ok that's a little bit confusing for me because x should be the the address not the value.

Answer 6

I know!

Answer 7

So I tried it without the "&" Then it works... But I try to figure out what the problem is when I dont modify at this position

Answer 8

And that worked! Thank you!