The use of array as a function parameter and return value in C language simple introduction _c language

function is used as an argument by an array

If you want to use a one-dimensional array as the parameter of a function, you must declare a function-form argument, which produces a similar result with one of all three declarative methods in the following three ways, because each way tells the compiler that an integer pointer will be received. In a similar way, you can use multidimensional array form parameters.

Way-1
The form parameter is the pointer below. The next chapter will learn what pointers are.

void MyFunction (int *param)
{
.
.
.
}
Way-2
The size of the array parameter as follows:

void MyFunction (int param[10])
{
.
.
.
}
Way-3
Array parameters as variable sizes as follows:

void MyFunction (int param[])
{
.
.
.
}
Example
Now, consider the following function, which will require an array as another parameter, and, depending on the parameters passed, it returns the average of the number of values in the group, as follows:

Double getaverage (int arr[], int size)
{
 int  i;
 Double avg;
 Double sum;

 for (i = 0; i < size; ++i)
 {
  sum + = Arr[i]
 ;

 } avg = sum/size;

 return avg.
}

Now, let's call the above function as follows:

 #include <stdio.h>
 
/* Function declaration *
/double getaverage (int arr[], int size);

int main ()
{/
  * is int array with 5 elements */
  int balance[5] = {1000, 2, 3, m};
  Double avg;

  /* Pass Yiibaier
  to the array as a argument * * avg = getaverage (balance, 5);
 
  /* Output The returned value *
  /printf ("Average value is:%f", avg);
  
  return 0;
}

When the above code is compiled and executed, it produces the following results:

Average value is:214.400000

As you can see, the length of the array is not important, as long as the C language function does not check the bounds of formal arguments.

function returns an array
C language Programming does not allow the return of the entire array as a parameter to the function. However, you can return an array of pointers by specifying an array name without an index. If you want to return a one-dimensional array from a function, you must declare that you return a pointer, as in the following example:

int * MyFunction ()
{
.
.
.
}
The 2nd to remember is that the C language does not advocate that local variables are returned outside the function, so the local variable must be defined as a static variable.

Now, consider the following function, which produces 10 random digits and returns them using an array, and calls this function as follows:

#include <stdio.h>

/* function to generate and return random numbers
*/int * Getrandom ()
{
 static int r[10];
 int i;

 /* Set the seed */
 Srand ((unsigned) time (NULL));
 for (i = 0; i < ++i)
 {
   R[i] = rand ();
   printf ("r[%d] =%d
", I, R[i]);

 return r;
}

/* Main function to call above defined function
/int main ()
{/
  * a yiibaier to a int */
  int *p;
   int i;

  p = getrandom ();
  for (i = 0; i < i++)
  {
    printf ("* (P +%d):%d
", I, * (P + i));
  }

  return 0;
}

When the above code is compiled and executed, it produces some of the following:

R[0] = 313959809
r[1] = 1759055877
r[2] = 1113101911
r[3] = 2133832223 r[4
] = 2073354073 r[5
] = 167288 147
R[6] = 1827471542
r[7] = 834791014 r[8
] = 1901409888 r[9
] = 1990469526
* (p + 0): 313959809
* (P + 1): 1759055877 * (P +
2): 1113101911
* (P + 3): 2133832223
* (P + 4): 2073354073
* (P + 5): 167 288147
* (P + 6): 1827471542 * (P +
7): 834791014
* (P + 8): 1901409888
* (P + 9): 1990469526