Edward Bartolo <edb...@gmail.com> writes:
> Now I have several exercises from "The C programming language"
> (Kernighan & Ritchie). Some question look rather challenging...
>
> I learnt arrays are always passed by reference to function calls.
As was already pointed out, all arguments to C functions are passed via
value copies.
Take the following, contrived example:
--------
#include <stdio.h>
#define n_of(a) ((sizeof(a) / sizeof(*a)))
void double_them(int *a, unsigned n)
{
do {
--n;
a[n] += a[n];
} while (n);
}
void print_them(char *msg, int *a, signed n)
{
int i;
puts(msg);
i = 0;
do printf("\t%d\n", a[i]); while (++i < n);
putchar('\n');
}
int main(void)
{
int a[] = { 1, 2, -3, -4, 5, 6 };
print_them("before", a, n_of(a));
double_them(a, n_of(a));
print_them("after", a, n_of(a));
return 0;
}
-------
The reason this works is because an expression whose type is 'array of
/type/', eg, a, type 'array of int', is automatically converted to a
value of type 'pointer to /type/' ('pointer to int' here) pointing to
the first element of the array (exceptions to this rule are the sizeof
operator, as shown in the n_of definition above, the &-operator and
'string literal used to initialize an array').
The 'double_them' function modifies the elements of the original array
by accessing them indirectly through the pointer passed to the first of
them.
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