我需要在C中创建一个“通用”堆。
我有一个包含比较功能的主文件。本质上,数组的基地址,元素数量,每个元素的大小以及比较函数都被传递到堆栈函数中。我遇到了两个问题,一个程序没有排序,所以我想知道是否有人可以看到代码有什么问题。两个我在1710个元素之后得到了一个分段错误。
#include <stdio.h>
#include <string.h>
#include "srt.h"
void srtheap(void *, size_t, size_t, int (*)(const void *, const void *));
void heapify(void *, size_t, size_t, size_t, int (*)(const void *, const void *));
void srtheap(void *base, size_t nelem, size_t size, int (*compar)(const void *, const void *)) {
char *p1, *p2;
char *qb=base;
void *last = qb + (size*(nelem-1));
for (size_t curpos = nelem-1; curpos>0; curpos=-2){
p1 = qb + ((curpos-1)/2)*size;
if(compar(last, (last-size)) >= 0){
if(compar(last, p1) > 0){
swap(last, p1, size);
heapify(qb, nelem, curpos, size, compar);
}
}
else { //LEFT>RIGHT
if(compar(last-size, p1) > 0){
swap(last-size, p1, size);
heapify(qb, nelem, curpos-1, size, compar);
}
//otherwise, parent is greater than LEFT & RIGHT,
//or parent has swapped with child, iteration done, repeat through loop
} //end else, children have been compared to parent
//end check for two children, only left child if this loop is skipped
last = last-(2*size);
}
/*
**Now heapify and sort array
*/
while(nelem > 0){
last = qb + (size*(nelem-1));
swap(qb, last, size);
nelem=nelem-1;
heapify(qb, nelem, 0, size, compar); //pass in array, #elements, starting pos, compare
}
}
void heapify(void *root, size_t numel, size_t pos, size_t sz, int (*compar)(const void *, const void *)){
void *rc, *lc, *p1;
while(pos < numel){
rc = root+((pos+1)*2)*sz; //right child
lc = root+(((pos+1)*2)-1)*sz; //left child
p1 = root+(pos*sz); //parent
if((pos+1)*2 < numel){ //check if current element has RIGHT
if (compar(rc, lc)>=0){
if(compar(rc, p1)>0) {
swap(rc, p1, sz);
pos=(pos+1)*2; //move to RIGHT, heapify
}
else {
pos = numel; //PARENT>LEFT&RIGHT, array is heapified for now
}
} //end RIGHT>LEFT
else { //LEFT>RIGHT
if(compar(lc, p1) >0 ) {
swap(lc, rc, sz);
pos=((pos+1)*2)-1; // move to LEFT, heapify
}
else {
pos = numel; //PARENT>LEFT&RIGHT, array is heapified for now
} //end inner if, else
}//end LEFT,RIGHT comparison
}//end check for RIGHT
else if (((pos+1)*2)-1 < numel){ //else, check if element has LEFT
if(compar(lc, p1)>0){
swap(lc, p1, sz);
pos=((pos+1)*2)-1; //move to LEFT, continue heapify
}
else {
pos = numel; //PARENT>LEFT, array is heapified for now
}
}//end check for LEFT
else { //current element has no children, array is heapified for now
pos = numel;
}
}
}
答案 0 :(得分:2)
您的代码和问题与2010年的this question非常相似,所以我怀疑这是一项家庭作业。
关于崩溃,问题出现在srtheap
:
for (size_t curpos = nelem-1; curpos>0; curpos=-2){
循环的更新表达式curpos=-2
错误。你可能想要curpos-=2
。但是,如果是这种情况,仍然存在问题。 size_t
是无符号类型,因此如果原始数组中包含偶数个元素,那么最终程序将达到curpos
等于1
的点。当它试图减去2
时,结果将是一个非常大的正数,而不是您可能期望的-1
。因此,循环条件curpos>0
将评估为true,并且循环将尝试访问远远超出数组末尾的数组元素,从而触发崩溃。
我不愿意弄清楚为什么你的代码没有正确排序,因为它看起来不必要复杂。这是一个基于this implementation的工作通用heapsort。
void srtheap(void *base, size_t nelem, size_t size, int (*compar)(const void *, const void *))
{
char *cb = (char *)base;
size_t i = (nelem / 2);
while (1) {
heapify(cb, size, i, nelem-1, compar);
if (i == 0) break;
i--;
}
for (size_t i = nelem-1; i >= 1; i--) {
swap(cb, cb+(size * i), size);
heapify(cb, size, 0, i-1, compar);
}
}
void heapify(char *base, const size_t size, size_t root, const size_t bottom, int (*compar)(const void *, const void *))
{
size_t maxChild = root * 2 + 1;
if (maxChild < bottom) {
size_t otherChild = maxChild + 1;
maxChild = (compar(base + (otherChild * size), base + (maxChild * size)) > 0) ? otherChild : maxChild;
} else {
if (maxChild > bottom) return;
}
if (compar(base + (root * size), base + (maxChild * size)) >= 0) return;
swap(base + (root * size), base + (maxChild * size), size);
heapify(base, size, maxChild, bottom, compar);
}
此版本使用递归,但将其转换为循环是一项简单的练习。我会把它留给读者。