我正在尝试在链表上实现合并排序。遇到导致以下错误的问题：
student@wheezyupsec:~/CS305/HW4$ make
gcc -c sort.c
sort.c: In function ‘mergeSortHelper’:
sort.c:70:2: warning: passing argument 1 of ‘mergeSort’ from incompatible pointer type [enabled
by default]
sort.c:34:6: note: expected ‘struct listNode **’ but argument is of type ‘struct listNode *’
sort.c:70:2: warning: passing argument 1 of ‘mergeSort’ from incompatible pointer type [enabled
by default]
sort.c:34:6: note: expected ‘struct listNode **’ but argument is of type ‘struct listNode *’
sort.c:70:2: error: invalid use of void expression
sort.c:70:2: error: invalid use of void expression
make: *** [sort.o] Error 1
任何人都可以看到我的错误。它与mergeSortHelper有关，我在这里做了一个递归语句。它需要一个双指针而不是给定的单个指针。但是不知道如何做到这一点。我在下面放了我的sort.c代码。谢谢你的帮助。
/*
* sort.c
*
* Author: CS305 STUDENT ADD YOUR NAME HERE.
*
* Description: Contains sorting functions that operate on the linked
* list implementation for the company entry node found in
* list.[c,h].
*
*/
#include <stdio.h>
#include <stddef.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include "list.h"
#include "sort.h"
// Function prototypes for helper functions used only within this
// file. They are declared as static since they should be global
// function declarations available only to this file.
static listNode * mergeSortHelper(listNode * head, listNodeCompareFcn compare);
static listNode * merge(listNode * l1, listNode * l2, listNodeCompareFcn compare);
/* mergeSort()
*
* CS305 students should *not* alter this function for HW4. Nope
* Don't do it.
*/
void mergeSort(listNode ** listPtr, listNodeCompareFcn compare)
{
// mergeSort() calls mergeSortHelper which performs the actual
// merge sort algorithm. This function simply points the head
// of the list at the result of the sorting.
*listPtr = mergeSortHelper(*listPtr, compare);
}
// CS305 Students must implement the function stubs below. To maximize points earned
// on the homework, students should also supply function comment headers as well as
// document the function bodies with useful comments.
/* mergeSortHelper()
*
* CS305 students must implement this function for HW4.
*/
listNode * mergeSortHelper(listNode * head, listNodeCompareFcn compare)
{
listNode * chop = head;
listNode * other = head->next;
int n = count(head);
int i = 0;
while (i<((n/2) - 1))
{
//advance chop and other
chop = chop->next;
other = other->next;
i++;
}
chop->next = NULL;
//return head;
//return merge(mergesort(head, compare), mergesort(other, compare), compare);
return merge(mergeSort(head, compare), mergeSort(other, compare), compare);
}
/* merge()
* Parameters: 1. l1: the first linked list to be merged.
* 2. l2: the second linked list to be merged.
*
* Description: Merge two sorted linked lists and return the merged
* list.
*
* CS305 students must implement this function for HW4.
*
*/
listNode * merge(listNode * l1, listNode * l2, listNodeCompareFcn compare)
{
listNode * head;
//Base Case
if(l1 == NULL)
{
return l2;
}
if(l2 == NULL)
{
return l1;
}
//recursive case
if(compare(l1, l2))
{
head = l1;
head->next=merge(l1->next, l2, compare);
}
else
{
head = l2;
head->next = merge(l1, l2->next, compare);
}
return head;
}
/* alphabetCompare()
*
* Given two pointers to listNode, return 1 if the first one's
* companyName is lexicographically less than the second one.
* Otherwise, return 0.
*
* For example, if l1->companyName is 'aaaa' and l2->companyName is
* 'aaab' then l1->companyName is less than l2->companyName.
*
* CS305 students must implement this function for HW4.
*
*/
int alphabetCompare(listNode * l1, listNode * l2)
{
if(strcmp(l1->entryPtr->companyName,l2->entryPtr->companyName)<0)
{
return 1;
}
else
{
return 0;
}
}
/* distanceCompare()
*
* Given two pointers to listNode, return 1 if the first one's
* latitude + longitude place it closer to the University of Portland
* Bell Tower than the second one. Otherwise, return 0.
*
* CS305 students must implement this function for HW4.
*
* For full points, the comparison should be made based on the
* distance between two points on a sphere. For 80% credit
* a simple comparison can be made between two points on a plane.
*
*/
int distanceCompare(listNode * l1, listNode * l2)
{
//convert longitude and latitude of bell tower
//coordinates to radians
long upLonRadians = BELL_TOWER_LON/(180/PI);
long upLatRadians = BELL_TOWER_LAT/(180/PI);
//convert radians of bell tower coordinates
//to spherical coordinates (x,y,z)
long upX = cos(upLatRadians) * cos(upLonRadians);
long upY = cos(upLatRadians) * sin(upLonRadians);
long upZ = sin(upLatRadians);
//convert longitude and latitude of company
//coordinates in l1 and l2 to radians
long comp1LonRadians = (l1->entryPtr->latitude)/(180/PI);
long comp1LatRadians = (l1->entryPtr->longitude)/(180/PI);
long comp2LonRadians = (l2->entryPtr->latitude)/(180/PI);
long comp2LatRadians = (l2->entryPtr->longitude)/(180/PI);
//convert radians of company coordinates
//to spherical coordinates (x,y,z) for
//both companies
long comp1X = cos(comp1LatRadians) * cos(comp1LonRadians);
long comp1Y = cos(comp1LatRadians) * sin(comp1LonRadians);
long comp1Z = sin(comp1LatRadians);
long comp2X = cos(comp2LatRadians) * cos(comp2LonRadians);
long comp2Y = cos(comp2LatRadians) * sin(comp2LonRadians);
long comp2Z = sin(comp2LatRadians);
//find distance between bell tower and
//company 1 and distance between bell
//tower and company 2.
long dist1 = arccos((upX*comp1X)+(upY*comp1Y)+(upZ*comp1Z))*EARTH_RADIUS;
long dist2 = arccos((upX*comp2X)+(upY*comp2Y)+(upZ*comp2Z))*EARTH_RADIUS;
//compare distances between two companies.
//if company 1 distance is closer than
//company 2, return 1, otherwise return 0.
if(dist1<dist2)
{
return 1;
}
else
{
return 0;
}
}
答案 0 :(得分：1)
mergeSort()
void mergeSort(listNode ** listPtr, listNodeCompareFcn compare)
merge()
listNode * merge(listNode * l1, listNode * l2, listNodeCompareFcn compare)
这意味着，merge()的第一个参数应为listNode *类型，而在您的情况下，
return merge(mergeSort(head, compare), mergeSort(other, compare), compare);
其void。
第二个参数相同，不兼容。
提示：尝试显示与您的代码对应的行号并检查错误/警告消息。
答案 1 :(得分：1)
您可以将&header and &other传递给mergesort函数调用。它只会避免编译器警告，但您需要查看排序功能。
例如，您可以像mergesort(&head,compare)或mergesort(&other,compare)一样调用您的函数。
答案 2 :(得分：1)
在行中：
return merge(mergeSort(head, compare), mergeSort(other, compare), compare);
您正在调用mergeSort()的外部接口，返回void。您收到错误的部分原因是mergeSort()需要listNode **并且您传递了listNode *，部分原因是因为它没有返回值，但您尝试传递非另一个函数的存在价值。
您可以使用以下方法解决间接问题：
mergeSort(&head, compare);
mergeSort(&other, compare);
return merge(head, other, compare);
但使用内部接口mergeSortHelper()：
return merge(mergeSortHelper(head, compare), mergeSortHelper(other, compare), compare);