C changing variable and array value [closed] - c

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At first code a[0] becomes 31 but in the second one a still 55. Can someone explain me the difference? What is the difference between changing a variable's and array's value in a function?
#include <stdio.h>
int h(int x[]);
int main()
{
printf("Hello, World!\n");
int a[0] = 55;
h(a);
printf("%d", a[0]);
return 0;
}
int h(int x[]) {
x[0] = 31;
}
#include <stdio.h>
int h(int x);
int main()
{
printf("Hello, World!\n");
int a= 55;
h(a);
printf("%d", a);
return 0;
}
int h(int x) {
x = 31;
}
And What about this code? This array is changed. I am really confused.
#include <stdio.h>
int h(int x[], int length);
int main()
{
int i;
int a[]= {5, 5, 5, 5};
h(a, 4);
printf("Array After Function Call\n");
for(i = 0; i < 4; i++) {
printf("%d ", a[i]);
}
return 0;
}
int h(int x[], int length) {
int i;
for(i = 0; i < length; i++) {
x[i] += x[i];
}
printf("Array in Function\n");
for(i = 0; i < length; i++) {
printf("%d ", x[i]);
}
printf("\n");
return 0;
}

Answer 1

And any decent book should have mentioned that arguments to functions are passed by value, meaning the value in the call is copied into the local function argument variable.
Modifying this local variable will only modify the local copy, not the value used in the call to the function.
One can overcome this problem by emulating pass by reference, which is done by passing a pointer to the variable:
void f(int *x)
{
*x = 123; // Dereference the pointer, to set the value of where it's pointing
}
int main(void)
{
int a = 0;
f(&a); // Pass a pointer to the variable a
}
This is really what's happening in the first program you show, you pass a pointer, and modify the value of where the pointer is pointing (the zero-sized array notwithstanding).
It seems that one of your major confusions is about your "array" argument.
When you declare a function as
void f(int a[]);
you don't actually declare an array argument: The compiler will treat it as a pointer and parse it as
void f(int *a);
Another point is that when you use array indexing, like
x[i] += x[i];
you are actually dereferencing the pointer, and write the values to the memory where the pointer is pointing. In fact, for any pointer or array x and index i, the expression x[i] is exactly equal to *(x + i).