C 最小堆结构中的哨兵控制循环

我对堆的定义如下：

文件：heap.h

#ifndef __HEAP__
#define __HEAP__

typedef struct heap heap;

heap* heap_new();
void  heap_delete(heap* h);
void  heap_insert(heap* h, int x);
void  heap_delete_last(heap* h);

#endif

文件：heap.c

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

#ifndef MIN_SIZE
#define MIN_SIZE 4
#endif

struct heap {
    int* v;
    int n;
    int size;
};


// ---------------------------------------------------------------------
// expand
// ---------------------------------------------------------------------
// expand heap size by 50%
static void expand(heap* h) {
    h->size += h->size/2;
    h->v = realloc(h->v, h->size*sizeof(int));
}


// ---------------------------------------------------------------------
// heap_new
// ---------------------------------------------------------------------
// create new empty heap object
heap* heap_new() {
    heap* h = malloc(sizeof(heap));
    h->size = MIN_SIZE;
    h->v = malloc(h->size*sizeof(int));
    h->n = 0;
    return h;
}


// ---------------------------------------------------------------------
// heap_delete
// ---------------------------------------------------------------------
// delete heap object
void heap_delete(heap* h) {
    free(h->v);
    free(h);
}


// ---------------------------------------------------------------------
// heap_dump
// ---------------------------------------------------------------------
// dump heap content to stdio
void heap_dump(heap* h) {
    int i;
    for (i=0; i<=h->n; i++)
        printf("[%d] %d\n", i, h->v[i]);
}


// ---------------------------------------------------------------------
// heap_check
// ---------------------------------------------------------------------
// test heap correctness
int heap_check(heap* h) {
    int i;
    for (i=1; i<=h->n; i++)
        if (  i   >  1    && h->v[i] < h->v[i/2]  ) return 0;
        if (2*i   <= h->n && h->v[i] > h->v[2*i]  ) return 0;
        if (2*i+1 <= h->n && h->v[i] > h->v[2*i+1]) return 0;
    return 1;
}


// ---------------------------------------------------------------------
// heap_insert
// ---------------------------------------------------------------------
// insert key into heap
void heap_insert(heap* h, int x){
    if (++h->n == h->size) expand(h);
    int *v = h->v;
    unsigned long c = h->n, p = c >> 1;
    while (p > 0 && v[p] > x) {
        v[c] = v[p];
        c = p;
        p >>= 1;
    }
    v[c] = x;
}


// ---------------------------------------------------------------------
// heap_delete_last
// ---------------------------------------------------------------------
// remove last heap element
void  heap_delete_last(heap* h) {
    if (h->n > 0) h->n--;
}

find_value

具有sentinel功能（避免一次测试）：

int find_值（int*v，int n，int x）{
int i=0；
如果（v[n-1]==x）返回1；//x在最后一个位置
int last=v[n-1]；//保存最后一个元素
v[n-1]=x；//在v中至少有一个x元素（test[ii如果您使代码更具可读性（停止在一行上放置多个逻辑操作），您可能会得到更多帮助。“如果”获取它自己的行，每行一个变量，不要在其他逻辑操作中使用+/--in等。代码行的减少并不是以复杂性为代价的改进。您可以忽略第一个if条件。剩余的代码在我看来不是不可读的。什么意思，“通过哨兵控制它”？您的意思是“如果其他变量为true，但
v[p]
不大于
x
，仍执行循环体”？请参见我编辑的问题中的示例。
    /*Returns 0 if x doesn't exist in v, 1 otherwise */
int find_value(int *v, int n, int x) {
    int i;
            //loop with 2 tests
    for (i=0; i<n; i++)
        if (v[i]==x) return 1;
    return 0;
}

 int find_value(int *v, int n, int x) {
    int i=0;
    if(v[n-1] == x) return 1; //x is in the last spot
    int last = v[n-1]; //save last element
    v[n-1] = x; //one x element in v at least (test [i<n] id unnecessary)
    while(v[i++] != x); 
    v[n-1] = last; 

    return i < n-1; 
}