CODEWITHSUNDEEP
cpointersmemory-address
Mike M
Mike M

Reputation: 752

Displacement with C pointers

I am looking to implement a stack in a C program with which I was thinking about using an unsigned char * as a base. This would be the bottom of the stack and all other registers and information would be a displacement of this address. However, I cannot seem to understand how to do this properly. I was thinking of doing something like this...

//Info: Store 20 at address x0000007c

unsigned char * base = 0;
int address = x0000007c;
(base + address) = 20;

The C compiler does not like this so I was wondering how to fix this or do something similar without losing my pointer.

Upvotes: 0

Views: 789

Answers (2)

Jim Balter
Jim Balter

Reputation: 16406

The compiler doesn't like your code because it's conceptually wrong; base + address is not an lvalue, and even if it were it has the wrong type: you can't store an int into a char*. But this is logically correct:

 base[address] = 20;

or, equivalently,

 *(base + address) = 20;

although it isn't functionally correct because base doesn't point to valid memory. You need to allocate your stack, either as a static array or via malloc, and assign it to base, e.g.,

 unsigned char* base = malloc(STACKSIZE);
 if (!base) out_of_memory();

Upvotes: 3

luser droog
luser droog

Reputation: 19504

For a stack, the easiest is probably to use two pointers,

enum { STACKSZ = 100 };
char *base = malloc(STACKSZ);
char *top = base;

Then you can push with

*top++ = 'H';

and pop with

char x = *--top;

and get size with

int sz = top-base;

Note this code doesn't check for stack over-/under-flow.

Upvotes: 0

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