Thread: Question about array from a newbie

Hi all,
I'm learning C and running into this problem with array that I try to understand. When I try the following code to print an array of 5 interger:
for(i = 0; i<5; i++)
{
printf("%.1lf %.1lf %.1lf \n", source[i], target1[i], *target2++);
}
I get the error "lvalue required as increment operand" for using *target2++. However, if it use: *(target2+i), then i'm fine. Can someone explain to me why I can't use increment to print like that?

Thanks alot in advance.

target2 is an array?

Originally Posted by Babkockdood

target2 is an array?

Yeah. I declare it: int target2[5];

Here's my code:

Code:

#include <stdio.h>

void copy_arr(const double source[], double target[], int COLS);
void copy_ptr(const double source[], double target[], int COLS);
int main(void)
{
	int i;
	double source[5] = {1.1, 2.2, 3.3, 4.4, 5.5};
	double target1[5], target2[5];
	
	copy_arr(source, target1, 5);
	copy_ptr(source, target2, 5);
	printf("source     target 1      target 2\n");
	for(i = 0; i<5; i++)
	{
		printf("%.1lf        %.1lf              %.1lf \n", source[i], target1[i], *target2++); // *target2++ gives compile error. 
		
	}
	return 0; 
}

void copy_arr(const double source[], double target[], int COLS)
{
	int i; 
	for (i=0; i < COLS; i++)
		target[i] = source[i];
}

void copy_ptr(const double source[], double target[], int COLS)
{
	int i;
	for (i=0; i<COLS; i++)
	{	
		*target++ = *source++;
	}
}

Like quzah said, array names are constants, not variables. Can't touch them.

You've probably read that array name is the address of the first element of the array (e.g. arrayName == &arrayName[0]), which can get one thinking that it is equivalent to a pointer when it is not.

Originally Posted by msh

Like quzah said, array names are constants, not variables. Can't touch them.

You've probably read that array name is the address of the first element of the array (e.g. arrayName == &arrayName[0]), which can get one thinking that it is equivalent to a pointer when it is not.

Thank you. I think I kinda of understand. But how come this statement work: "*target++ = *source++;"

In function definitions, AND ONLY, type arrayName[] and type* arrayName are equivalent.

E.g., both of the below definitions are equivalent.

Code:

void copy_ptr(const double source[], double target[], int COLS)
{
    // Code
}

void copy_ptr(const double* source, double* target, int COLS)
{
    // Code
}

Look at the arguments you call your functions with and what they represent, and think about it.

You can't increment an array itself. You can increment a value in an array like so.

Code:

printf("%.1lf %.1lf %.1lf \n", source[i], target1[i], target2[i]++);

Originally Posted by msh

In function definitions, AND ONLY, type arrayName[] and type* arrayName are equivalent.

E.g., both of the below definitions are equivalent.

Code:

void copy_ptr(const double source[], double target[], int COLS)
{
    // Code
}

void copy_ptr(const double* source, double* target, int COLS)
{
    // Code
}

Look at the arguments you call your functions with and what they represent, and think about it.

I think I got it. An array name is like a constant pointer, so "target2++" would change the address, so it's not allowed. However, when I pass the address &target2[0] into a function, in the scope of the function, it becomes variable pointer thus i can use increment with it.

Originally Posted by Elysia

An array is not a pointer.
That is an important concept to grasp.
With a few exceptions, the standard says that when you put the array's name somewhere, it will decay to a pointer to its first element. Passing an array to a function is not one of those exceptions, so it decays. It's that simple.

Ok, so it decays to a pointer, but that did not answer the user's question and I am interested since I have a similar question.

Why is it ok in the following case? Due to the fact that arr is a copy that belongs to foo()? Output is 1 2 3 4

Code:

#include <stdio.h>

void foo(int * arr) {
	int i;
	for(i = 0; i < 4; ++i) {
		printf("%d ", *arr);
		arr++;
	}
}

int main(void) {

	int arr[] = {1, 2, 3, 4};
	foo(arr); // Decays to a pointer to the first element
			  // in arr[], which is 1

	return 0;
}

Ah I understand with the code example, thanks

I understand it now. Thank you very much guys.