K&R section 8.7 for loop in free() - c

Question

In K&R(2nd) section 8.7, I think there is an unexpected infinite for loop in free() and its test part seems wrong.
I inserted four // comments. malloc() has one, morecore() has another and free() has the others.
typedef long Align;
union header {
struct {
union header* ptr;
unsigned size;
}s;
Align x;
};
typedef union header Header;
static Header base;
static Header *freep = NULL;
/* malloc: general-purpose storage allocator */
void *malloc(unsigned nbytes)
{
Header *p, *prevp;
Header *morecore(unsigned);
unsigned nunits;
nunits = (nbytes + sizeof(Header) - 1) / sizeof(Header) + 1;
if ((prevp = freep) == NULL) { /* no free list yet */
base.s.ptr = freep = prevp = &base;
base.s.size = 0;
}
for (p = prevp->s.ptr; ; prevp = p, p = p->s.ptr) {
if (p->s.size >= nunits) { /* big enough */
if (p->s.size == nunits) /* exactly */
prevp->s.ptr = p->s.ptr;
else { /* allocate tail end */
p->s.size -= nunits;
p += p->s.size;
p->s.size = nunits;
}
freep = prevp;
return (void *)(p + 1);
}
if (p == freep) /* wrapped around free list */
// base.s.ptr = &base, freep == &base
if ((p = morecore(nunits)) == NULL)
return NULL; /* none left */
}
}
#define NALLOC 1024 /* minimum #units to requst */
/* morecore: ask system for more memory */
static Header *morecore(unsigned nu)
{
char *cp, *sbrk(int);
Header *up;
if (nu < NALLOC)
nu = NALLOC;
cp = sbrk(nu * sizeof(Header));
if (cp == (char *) -1) /* no space at all */
return NULL;
up = (Header *) cp;
up->s.size = nu;
// base.s.ptr = &base, freep == &base
free((void *)(up+1));
return freep;
}
/* free: put block ap in free list */
void free(void *ap)
{
Header *bp, *p;
bp = (Header *)ap - 1; /* point to block header */
// base.s.ptr = &base, freep == &base
for (p = freep; !(bp > p && bp < p->s.ptr); p = p->s.ptr)
if (p >= p->s.ptr && (bp > p || bp < p->s.ptr))
break; /* freed block at start or end of arena */
// for (p = freep; !(0); )\
if (p >= s.ptr && (0))\
break;
if (bp + bp->s.size == p->s.ptr) { /* join to upper nbr*/
bp->s.size += p->s.ptr->s.size;
bp->s.ptr = p->s.ptr->s.ptr;
} else
bp->s.ptr = p->s.ptr;
if (p + p->s.size == bp) { /* join to lower nbr */
p->s.size += bp->s.size;
p->s.ptr = bp->s.ptr;
} else
p->s.ptr = bp;
freep = p;
}
What I want to say is
void free(void *ap) {
/* ... */
for (p = freep; !(bp > p && bp < p->s.ptr); p = p->s.ptr)
if (p >= p->s.ptr && (bp > p || bp < p->s.ptr))
break; /* freed block at start or end of arena */
/* ... */
}
this for loop is infinite. Because bp > p && bp < p->s.ptr is the same as 0 because each value of p and p->s.ptr is &base. And p = p->s.ptr does nothing. p points to base, base.s.ptr points to itself. So the value of bp > p && bp < p->s.ptr won't be changed.
Moreover, the pointer bp points to the header of memory received by sbrk(). Is it possible to compare bp with p?
I think the function free() only makes sense when we already got a valid memory block using malloc() and call it to free it. The function call free((void *)(up+1)); in morecore() and the body of free() doesn't match(I think).
Q1. Why is the for loop infinite?
Q2. Is it valid to compare bp with p in that for loop? If so, how and where is bp(which points to the memory allocated by sbrk()) placed?

Answer 1

Q1. Why is the for loop infinite?
No, it's not infinite.
void free(void *ap) {
/* ... */
for (p = freep; !(bp > p && bp < p->s.ptr); p = p->s.ptr)
if (p >= p->s.ptr && (bp > p || bp < p->s.ptr))
break; /* freed block at start or end of arena */
/* ... */
}
When base.s.ptr = &base and freep == &base, above code is the same as
void free(void *ap) {
/* ... */
for (p = freep; !(0); p = p->s.ptr)
if (1 && (1))
break; /* freed block at start or end of arena */
/* ... */
}
(I was confused with the meaning of || operator. I thought bp > p || bp < p->s.ptr should be the same as 0 when I asked the question.)
Q2. Is it valid to compare bp with p in that for loop? If so, how and where is bp(which points to the memory allocated by sbrk()) placed?
Yes, comparing those two is valid. sbrk() returns a previous program break value. So bp and p are comparable.