Allocation of Memory in Heap Error

Not sure what is the error in this code. This is from the book called "Understanding and Using C Pointers". This is from page number 36. It is supposed to demonstrate that we allocate 6 bytes in memory bt we use 8 bytes.

Code:

---

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

int main() {
    char *pc = (char*) malloc(6);

    for(int i=0; i<8; i++) {
        *pc[i] = 0;
    }
}

---

However, when I run it, I get the following error.

Quote:

---

error: invalid type argument of unary '*' (have 'int')

---

Change *pc[i] to pc[i] as pc is a pointer to char, so pc[i] is a char, and you cannot dereference a char.

Quote:

---

Originally Posted by laserlight

Change *pc[i] to pc[i] as pc is a pointer to char, so pc[i] is a char, and you cannot dereference a char.

---

Thanks for the response.
When you say we can't dereference a char, why not?
I am kind of new to pointers.
But the in the following code snippet I can dereference a pointer, or am I missing something.

Code:

---

int main() {
    char c = 'a';
    char *pc = &c;
    printf("%c\n", *pc);
}

---

Quote:

---

Originally Posted by josnow

When you say we can't dereference a char, why not?

---

A char isn't a pointer.

Quote:

---

Originally Posted by josnow

But the in the following code snippet I can dereference a pointer, or am I missing something.

---

That just proves that you can dereference a pointer to char. It doesn't prove that you can dereference a char. Try to compile this:

Code:

---

int main() {
    char c = 'a';
    printf("%c\n", *c);
}

---

OK, I got it.
Thanks again.

Quote:

---

Originally Posted by josnow

When you say we can't dereference a char, why not?
I am kind of new to pointers.

---

Think of pointers as any normal variable, but instead of containing a "value" it contains a memory address.
The declaration:

Code:

---

char *p;

---

Declares a variable called 'p' (not '*p') of 'char *' type. So 'p' will store a memory address.
"Dereference" is the operation to use the "reference" (the pointer variable) as a memory address and get data 'pointed' by it.
In a crude way:

Code:

---

char *p; // declare p as a pointer to a char.
p = 0x6001a3;  // Puts the address 0x6001a3 in p.
*p = 'a';  // puts the character 'a' in the address contained in p.

---

But, of course, instead of initializing 'p' with a literal address you'll get the "address of" another symbols, for example:

Code:

---

char s[] = "fred";
char *p;

p = &s[1]; // p is assigned with the 'address of' s[1].
putchar( *p ); // will print 'r'

---