How to pass array to function by reference

I have one array with value I want to pass the content of array to function by reference (pointer). I do not understand how to do that

Code:

---

#include <stdio.h>

#define  Y    '0'
#define  N    '1'


#define  A    '0'


char MyFunction(char text, int i, char *buffer);  /* Prototype */


char MyFunction( char text )  /* Called function definition */
{
  if (text == A)
  {
    //store content here//
    printf("success \n"); 
    return Y;
  }
  else
  {
      printf("Failure \n"); 
    return N;
  }
}


int main( void )  /* Calling function */
{
  int size = 10;
 
  char name[size] = {'L','e','s','s','o','n','s','\0'};
 
  char function =    MyFunction(A, size, name);
 
  printf("%c\n", function); 
 
  return 0;
}

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Code:

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char MyFunction(char text, int i, char *buffer);  /* Prototype */
 
 
char MyFunction( char text )  /* Called function definition */

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Lying to the compiler ends badly!

You said MyFunction has 3 parameters and then then you used only one in the function body.

Tim S.

Quote:

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Originally Posted by stahta01

Code:

---

char MyFunction(char text, int i, char *buffer);  /* Prototype */
 
 
char MyFunction( char text )  /* Called function definition */

---

Lying to the compiler ends badly!

You said MyFunction has 3 parameters and then then you used only one in the function body.

Tim S.

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my typing mistake

Code:

---

#include <stdio.h>
#define  Y    '0'
#define  N    '1'
 
 
#define  A    '0'
 
 
char MyFunction(char text, int i, char *buffer);  /* Prototype */
 
 
char MyFunction(char text, int i, char *buffer);
{
  if (text == A)
  {
    //store content here//
    printf("success \n"); 
    return Y;
  }
  else
  {
      printf("Failure \n"); 
    return N;
  }
}
 
 
int main( void )  /* Calling function */
{
  int size = 10;
 
  char name[size] = {'L','e','s','s','o','n','s','\0'};
 
  char function =    MyFunction(A, size, name);
 
  printf("%c\n", function); 
 
  return 0;
}

---

I suggest you try to compile your code.

Tim S.