Suppose I have a function, int function(int N, int c[N]){...}, taking as parameters an integer and an array. Suppose now I have a double array **c of size 'N times 2' and suppose, I want to apply the function function to one column of the double arrow, c[i][0], i varying from to N-1. How am I supposed to use this function. Does it looks like something like function(N,*c[0]) ?
Does anyone can help ?
An array in C always decays to a pointer to its first element when passed into a function. This is good to know in some situations.
As an example, you could write
int list[10];
func (int *x) {
int i;
for (i = 0; i < 10; i++) {
printf("%d", x[i]);
}
}
x[i] is really just syntactic sugar. In C, when you use bracket notation to access an element of an array, it gets converted to *(x + i), where x is the name of the array.
This works because of pointer arithmetic. If x is the name for an array of 10 integers, then the value of x in an expression is the address of the first integer of the array.
x + i will always point to the i-th element after x (C takes into account the size of the element type stored in the array, and increments the pointer accordingly).
Thus, when passing 2d arrays, the array decays to a pointer to its first element - which is an array.
A function signature taking a 2d array can be written as
func(int x[][columns] {
...
}
// but could also be written as
func(int (*x)[columns]) {
...
}
which indicates that x is a pointer to an array of integers.
Sometimes you need to write a function to accept a 2 dimensional array where the width is not known until run time. In this case, you can pass a pointer to the [0][0] element and also pass the two dimensions, and then use pointer arithmetic to get the right element. For example,
print_2d_array (int *x, height, width) {
int i, j;
for (i = 0; i < height; i++) {
for (j = 0; j < width; j++) {
printf("%d", x[i * width + j]);
}
}
}
int list[10][10];
print_2d_array (&list[0][0], 10, 10);
would work for a dynamically allocated 2d array.
In c langauge, for single dimentional array you no need to mention the size of the array in the function arguments while passing an array to that function.
If it is a single dimentional array
...
int a[10];
func(10, a);
...
void func(int size, int x[]) //no need to mention like int x[10]
{
//here x is not an array. Its equivalent to int *
printf("%d", sizeof(x)); // this will print 4 or 8 not 40
}
If it is 2D array
...
int a[10][5];
func (10, a);
...
void (int rows, int x[][5]) //here int x[][] is invalid
{
}
N in the first dimension, but it can be considered being good practice to show the intention of the code. And also for fixed sizes there is the notation int x[static 10] that requires that the array that is passed through the pointer has at least 10 elements.
**c) completely different beasts and not compatible.