Thread: what type of parameters can be pass to function

What type of parameters can be pass to function

I wrote following function. I think it can pass the given parameters to function.

Code:

void function (char name[10], int age, char grade, int *intPointer, char *charPointer, struct record){
	
}

Does it a valid function ? if not then what's wrong in the function ?
can we pass function to function as parameter ?

Originally Posted by Salem

> Does it a valid function ? if not then what's wrong in the function ?
The 'struct record' is incomplete.
Is record the name of the struct, and the variable name unknown?
Or is record the name of the variable of an unknown struct type.
.

record is name of structure and id and height is structure variable and name is structure member

Code:

struct record { int id; float height; } name;

does it valid function that pass the structure ?

Code:

void function ( struct record){
    
}

Originally Posted by Salem

> does it valid function that pass the structure ?
No, because it's basically the same as your first attempt.

I understand now what types of parameter can be pass to function. I can pass integer to function

I am struggling to pass array

Code:

#include <stdio.h>

int main()
{
   int array [5];  // array declration 
   
   function (value) // call function 
	
    return 0;
}


void function( int value[5])
{
	int i;
    for( i = 0; i < 5; i++)
        value[i] = value[i] + 10;
}

Originally Posted by Salem

> function (value)
Try
function (array) ;

Code execute without any error.

I don't think its useful example. I want to understand how to pass array with useful example

Code:

#include <stdio.h>

void function( int value[5])
{
	int i;
    for( i = 0; i < 5; i++)
	{
        value[i] = value[i] + 10;
	    printf("\n value %d ", value[i]);
	}
} 


int main()
{
   int array [5];  // array declration 
   
   function (array); // call function 
	
    return 0;
}

value 4200965
value 4200874
value 10
value 3133450
value 4194442

Originally Posted by Salem

So initialise the array properly in main then

Thank you salem , How argument pass by value or reference?

Passing a pointer to function

Code:

#include <stdio.h>   
void fun(int *pointer) 
{ 
   int i; 
   for ( i = 0; i < 8; i++) 
     printf("%d  ", pointer[i]); 
} 
   
int main() 
{ 
   int array[] = {1, 2, 3, 4, 5, 6, 7, 8}; 


   fun(array); 
   
   return 0; 
}

I do not understand what happen when we call the function ?

C++ Pointers and Arrays

Read more about Arrays and Pointers and you'll easily understand what's happening.

When you declare "array", you create a pointer called 'array' which know it's size is 8 because you provide it with 8 elements in initialisation. So, you're indexes from 0-7 are initialised from 1-8. When you call fun(array), you need to know two things.
First, what does 'fun' expect? It expects a pointer to an integer.
Second, 'array' is a pointer to the first element of your array (which is an integer), i.e., to your element at index [0] which hold the number 1. This array has a size of 8 integers as mentioned above.

So, you basically passed a pointer to a function expecting a pointer.

> How argument pass by value or reference?

Google.
Passing by Value vs. by Reference Visual Explanation with animations and diagrams
Passing by pointer Vs Passing by Reference in C++ - GeeksforGeeks

I need your help

Code:

#include <stdio.h>
void fun(int *pointer)
{
   int i;
   for ( i = 0; i < 5; i++)
     printf("%d  ", pointer[i]);
}
 
int main()
{
   int array[] = {1, 2, 3, 4, 5}; 

   fun(array);  // I don't understand this line. there is no & operator 
 
   return 0; }

What happen when we call function fun(array) ?

Why we can't write fun(&array) instead of fun(array) ?