我会尽量说清楚。
我在一个名为"esercizio.h"的文件中有这个结构体(由于我的老师,我不能改变它):
#ifndef ESERCIZIO_H
#define ESERCIZIO_H
struct ElemSCL{
int info;
struct ElemSCL* next;
};
typedef struct ElemSCL NodoSCL;
typedef NodoSCL* TipoSCL;
void accoda(TipoSCL* pscl, int i);
#endif
函数"accoda"必须在循环链表(由TipoSCL* pscl指向)的末尾添加一个元素(int i)。我尝试在一个名为"esercizio.c"的文件中编写函数体:
#include <stdio.h>
#include <stdlib.h>
#include "esercizio.h"
void accoda(TipoSCL* pscl, int i){
NodoSCL* temp = (NodoSCL*) malloc(sizeof(NodoSCL));
temp->info = i;
temp->next = temp;
if (pscl == NULL){
return;}
while ((*pscl)->next != *pscl){
*pscl = (*pscl)->next;}
(*temp)->next = (*pscl)->next; //Problems starts here
(*pscl)->next = *temp;
}
正如我让你注意到的,在我的代码中,一切都是好的,如果我不添加最后两行。如果函数中没有TypeSCL*而是NodeSCL*,我会使用:
temp->next = pscl->next;
pscl->next = temp;}
但是我的老师决定用TypeSCL* pscl代替NodeSCL* pscl。
我有一个"test.h"文件…
#include "esercizio.h"
#ifndef TEST_H
#define TEST_H
char* toString(TipoSCL scl);
#endif
…还有一个"test.c"文件,其中包含main()函数和所有让我检查代码是否正常工作的输入:
#include <stdlib.h>
#include <string.h>
#include "../libtest/libtest.h"
#include "test.h"
#include "esercizio.h"
const int NTEST=5;
TipoSCL input[5];
int add[5] = {1,2,3,4,5};
TipoSCL expected[5];
int main(int argc, char** argv){
input[0] = NULL;
input[1] = (TipoSCL) malloc(sizeof(NodoSCL));
input[1] -> info = 1;
input[1] -> next = input[1];
input[2] = (TipoSCL) malloc(sizeof(NodoSCL));
input[2] -> info = 1;
input[2] -> next = (TipoSCL) malloc(sizeof(NodoSCL));
input[2] -> next -> info = 2;
input[2] -> next -> next = input[2];
input[3] = (TipoSCL) malloc(sizeof(NodoSCL));
input[3] -> info = 1;
input[3] -> next = (TipoSCL) malloc(sizeof(NodoSCL));
input[3] -> next -> info = 2;
input[3] -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
input[3] -> next -> next -> info = 3;
input[3] -> next -> next -> next = input[3];
input[4] = (TipoSCL) malloc(sizeof(NodoSCL));
input[4] -> info = 1;
input[4] -> next = (TipoSCL) malloc(sizeof(NodoSCL));
input[4] -> next -> info = 2;
input[4] -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
input[4] -> next -> next -> info = 3;
input[4] -> next -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
input[4] -> next -> next -> next -> info = 4;
input[4] -> next -> next -> next -> next = input[4];
expected[0] = (TipoSCL) malloc(sizeof(NodoSCL));
expected[0] -> info = 1;
expected[0] -> next = expected[0];
expected[1] = (TipoSCL) malloc(sizeof(NodoSCL));
expected[1] -> info = 1;
expected[1] -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[1] -> next -> info = 2;
expected[1] -> next -> next = expected[1];
expected[2] = (TipoSCL) malloc(sizeof(NodoSCL));
expected[2] -> info = 1;
expected[2] -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[2] -> next -> info = 2;
expected[2] -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[2] -> next -> next -> info = 3;
expected[2] -> next -> next -> next = expected[2];
expected[3] = (TipoSCL) malloc(sizeof(NodoSCL));
expected[3] -> info = 1;
expected[3] -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[3] -> next -> info = 2;
expected[3] -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[3] -> next -> next -> info = 3;
expected[3] -> next -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[3] -> next -> next -> next -> info = 4;
expected[3] -> next -> next -> next -> next = expected[3];
expected[4] = (TipoSCL) malloc(sizeof(NodoSCL));
expected[4] -> info = 1;
expected[4] -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[4] -> next -> info = 2;
expected[4] -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[4] -> next -> next -> info = 3;
expected[4] -> next -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[4] -> next -> next -> next -> info = 4;
expected[4] -> next -> next -> next -> next = (TipoSCL) malloc(sizeof(NodoSCL));
expected[4] -> next -> next -> next -> next -> info = 5;
expected[4] -> next -> next -> next -> next -> next = expected[4];
test_reset();
for (int i = 0; i < NTEST; i++) {
print_test_start(i+1);
printf("Funzione: accodan");
printf("Input: %sn", toString(input[i]));
accoda(&input[i],add[i]);
test_compare_strings(toString(expected[i]),toString(input[i]));
print_test_end();
print_n_success("#Test superati: ");
}
print_test_result("Percentuale test corretti:");
}
char* toString(TipoSCL scl){
char* res = (char*) malloc(200*sizeof(char));
res[0] = '[';
res[1] = ' ';
TipoSCL aux = scl;
if (aux != NULL) {
char buf[10];
do{
sprintf(buf,"%d->",aux -> info);
strcat(res,buf);
aux = aux -> next;
}
while(aux != scl);
sprintf(buf,"|%d",aux -> info);
strcat(res,buf);
aux = aux -> next;
}
strcat(res,"]");
return res;
}
我的意思是"一切都好,如果我不添加最后两行"到我的代码?当我运行不带
的程序时(感谢终端和cd 和make )(*temp)->next = (*pscl)->next; //Problems starts here
(*pscl)->next = *temp;
测试运行没有问题(当然,它告诉我,我没有一个正确的结果。)但是,如果我将这两行添加到我的代码中,我得到了"分割错误:11"。
尝试将add函数更改为::
void add(TypeSCL* pscl, int i){
NodeSCL* temp = (NodeSCL*) malloc(sizeof(NodeSCL));
temp->info = i;
temp->next = temp;
if (*pscl == NULL){
*pscl = temp;
return;}
NodeSCL *tempCheck = *pscl;
while(tempCheck->next != *pscl) {
tempCheck = tempCheck->next;
}
tempCheck->next = temp;
temp->next = (*pscl);
}
你做错的事情::你需要意识到C中的所有东西都是按值传递的,所以如果你传递一个指针,那个指针也是按值传递的。所以,当你的老师告诉你使用TypeSCL* pscl
时,它意味着NodeSCL **pscl
,她是对的,因为你不能用NodeSCL *pscl
来做,这可能有助于你理解我为什么这么说::在链表中添加节点时使用双指针的原因是什么?
此外,在您的情况下,当pscl == NULL
首先应该是*pscl == NULL
时,您需要将*pscl
设置为新节点。
接下来,如果你想在末尾添加一个新节点,你应该使用一个while循环,正如@Neha Chauhan所提到的。
接下来,您使用的最后两个语句::
(*temp)->next = (*pscl)->next;
(*pscl)->next = *temp;
您将新添加节点的next
设置为链表的第二个元素,这没有任何意义。
我已经使用变量NodeSCL *tempCheck
移动到链表的最后一个节点!
最后两行应该是
tempCheck->next = temp;
temp->next = (*pscl);
最后,兄弟,你需要在指针的基础上工作。(不介意!)
编辑::
为什么你的测试没有最后两行::因为,用你写的代码,你的链表总是大小为1,没有元素添加到第一个元素之后!所以,测试运行了,但是失败了!
我认为你的PSCL指针必须走到链接列表的末尾,然后它应该插入temp元素。你能做的是
void add(TypeSCL* pscl, int i){
NodeSCL* temp = (NodeSCL*) malloc(sizeof(NodeSCL));
temp->info = i;
temp->next = temp;
if (pscl == NULL){
return;
while(pscl->next != pscl)
pscl = pscl->next;
temp->next = (*pscl)->next; //THINGS GO WRONG HERE...
(*pscl)->next = temp; //...AND HERE
}
"