Passing array into a function

As we've seen throughout this chapter, it's straightforward to manipulate the elements of an array using pointers, so there's no particular insurmountable difficulty if getline receives a pointer. One question remains, though: we had been defining getline with its line parameter declared as an array:

Code:

[COLOR="#FF0000"]int getline(char line[], int max)
	{
	...
	}[/COLOR]

We mentioned that we didn't have to specify a size for the line parameter, with the explanation that getline really used the array in its caller, where the actual size was specified. But that declaration certainly does look like an array--how can it work when getline actually receives a pointer?

The answer is that the C compiler does a little something behind your back. It knows that whenever you mention an array name in an expression, it (the compiler) generates a pointer to the array's first element. Therefore, it knows that a function can never actually receive an array as a parameter. Therefore, whenever it sees you defining a function that seems to accept an array as a parameter, the compiler quietly pretends that you had declared it as accepting a pointer, instead. The definition of getline above is compiled exactly as if it had been written

Code:

[COLOR="#FF0000"]int getline(char *line, int max)
	{
	...
	}[/COLOR]

Click to expand...

unsigned char pat[8] = {0xC3,0xC3,0xC3,0xFF,0xFF,0xC3,0xC3,0xC3,}

void display(unsigned char fun_pat[]) {

}

alexan_e said:

He doesn't need to change his code, it is perfectly operational as it is , you are just doing the same thing using different coding.
*(pat+cnt)
and
pat[cnt] are exactly the same

and unsigned char pat is already declared as a pointer from the compiler so using unsigned char* pat doesn't make a difference

Alex

Click to expand...