strange error! with pointer

#include <stdio.h>

Code:

int main(){
	int a[10];
	int *pa = a; //this is OK but...
	*pa = a; //this is EXC_BAD_ACCESS 
	/* (EXC_BAD_ACCESS is telling you is that you are 
	 attempting to reference a pointer that is no longer valid.
	*/
	//WHY?
}

as you can see above almost I have the same definitions but one doesn't give me an error and other one it does... WHY?

This:

Code:

int *pa = a; //this is OK but...

Is the same as this:

Code:

int *pa;
pa = a;

This:

Code:

	*pa = a; //this is EXC_BAD_ACCESS

Isn't assigning a pointer, it's dereferencing a variable and assigning a value to where you're pointing.


Quzah.

*pa = a; //this is EXC_BAD_ACCESS

Ritly said by quzah, depending on your compiler, that value can be 0 (or some other garbage value out of program memory), which means NULL. So attempting pointer which is not valid.

But could you please explain why I get this warnings here?

Code:

#include <stdlib.h>
#include <stdio.h>

main(){
	//define two strings: a and b
	char a[]="Cool";
	char b[]="things";
	//define a pointer 
	char *ptrA;
	ptrA=a;
	char d;
	d=*(ptrA+2);
	int *pp;
	int zz=4;
	pp=&zz;
	*pp=33;
	
	
	printf("%s", ptrA);
	printf("\n %p", ptrA);
	printf("\n %p", ++*ptrA); // format '%p' expects type 'void *', but argument 2 has type 'int'
        printf("\n %p", *(ptrA+2)); // format '%p' expects type 'void *', but argument 2 has type 'int'
	printf("\n %p", d); // format '%p' expects type 'void *', but argument 2 has type 'int'
	printf("\n %c", d);
	printf("\n %d", zz);
}

Try to read the help. According to warnings, printf() takes void*, where as you are providing it int.
I hope you have some knowledge of pointers. ptrA is a pointer, and *ptrA is value pointer is pointing to.

with format specifier %p, you need to provide a void* pointer to printf(), not any value.
How to fix your issues.
Use printf() with format specifier %d for integer, %c for char, and many more.