你的問題是指針。如果你使用n1，那麼void *總是包含最後的n1值，因爲指向n1。然後，當你釋放列表（queue_remove_double），總是顯示N1（最近）的值...
一些解決方案：
（1）陣列 - 下面顯示 - 僅限....
（2）用的變量（malloc的兩倍）的副本新雙重價值 - 「ilimited」（你的內存的限制...）
在新的雙重情況：
List* insert_queue_end (List* end, void* v) {
List* p = (List*) malloc(sizeof(List));
double *data=(double*)malloc(sizeof(double)); /*New double new address*/
*data=*((double*)v); /*Copy the VALUE and not the address to new double in new and fixed address*/
/*void* points to a new double with a new address in memory */
p->info=data;
p->prox=NULL;
if(end!=NULL) {
end->prox=p;
}
return p;
}
不要忘了從刪除功能中的新雙內存中釋放內存！
一些解決方案：
SOLUTION 1 - 雙打的陣列：
#include "stdio.h"
#include "stdlib.h"
typedef struct list {
void *info;
struct list *prox;
} List;
typedef struct queue {
List* begin;
List* end;
} Queue;
Queue* create_queue (void) {
Queue* f = (Queue*) malloc(sizeof(Queue));
f->begin = f->end = NULL;
return f;
}
/*Yes the type of return is List* and not Queue* - check your code */
List* insert_queue_end (List* end, void* v) {
List* p = (List*) malloc(sizeof(List));
///If you want create a new double this is the local - don't use the same address in v !
p->info=v;
p->prox=NULL;
if(end!=NULL) {
end->prox=p;
}
return p;
}
double queue_empty_double(Queue* f) {
return (f->begin==NULL);
}
double queue_remove_double(Queue* f) {
List* t;
double v;
double isRemoved=0;
if(queue_empty_double(f)) {
exit(1);
}
t=f->begin;
v=*((double*)(t->info));
f->begin=t->prox;
if (f->begin==NULL) {
f->end = NULL;
}
free(t);
printf("%.3lf\n",v);
}
void insert_queue(Queue* f, void* v) {
f->end = insert_queue_end(f->end,v);
if(f->begin==NULL) {
f->begin=f->end;
}
}
void print_queue_double(Queue* f) {
List* i;
for(i=f->begin;i!=NULL;i=i->prox)
printf("%.3lf\n",*((double*)i->info));
}
int main() {
Queue* f;
f = create_queue();
char ent1[100];
///One Double Allocated - only one address in memory !!!
///double n1;
double listOfDoubles[20]; ///All doubles allocated and one per one have a address in memory !!!
int countDoubles=0;
scanf("%s",ent1);
while(ent1[0]!='X' && countDoubles < 20)
{
if(ent1[0]=='E') {
///IN YOUR CODE:
///Fill the same one double...
///scanf("%lf",&n1);
///Testing new code
scanf("%lf",&listOfDoubles[countDoubles]); ///a new double with your address
///IN YOUR CODE:
///The same address of one variable. The same result.
///Remember void * is a pointer and use the address of n1, and then, print the most recent &n1 content...
///insert_queue(f,&n1);
///A new double in a new address, the content is not equal
insert_queue(f,&listOfDoubles[countDoubles]);
countDoubles++;
}
else if (ent1[0]=='D') {
///free the address and decrement the index of array to reuse a double
queue_remove_double(f);
countDoubles--;
}
else
{
///Print the list - and see the result "debug"
print_queue_double(f);
}
scanf("%s",ent1);
}
free(f);
}
一些解決方案：
SOLUTION 2 - 一種新的雙用的值
#include "stdio.h"
#include "stdlib.h"
typedef struct list {
void *info;
struct list *prox;
} List;
typedef struct queue {
List* begin;
List* end;
} Queue;
Queue* create_queue (void) {
Queue* f = (Queue*) malloc(sizeof(Queue));
f->begin = f->end = NULL;
return f;
}
/*Yes the type of return is List* and not Queue* check your code */
List* insert_queue_end (List* end, void* v) {
List* p = (List*) malloc(sizeof(List));
double *data=(double*)malloc(sizeof(double)); /*New double new address*/
*data=*((double*)v); /*Copy the VALUE and not the address to new double in new and fixed address*/
/*void* points to a new double with a new address in memory */
p->info=data;
p->prox=NULL;
if(end!=NULL) {
end->prox=p;
}
return p;
}
double queue_empty_double(Queue* f) {
return (f->begin==NULL);
}
double queue_remove_double(Queue* f) {
List* t;
double v;
double isRemoved=0;
if(queue_empty_double(f)) {
exit(1);
}
t=f->begin;
v=*((double*)(t->info));
f->begin=t->prox;
if (f->begin==NULL) {
f->end = NULL;
}
free(t->info); ///free the double
free(t);
printf("%.3lf\n",v);
}
void insert_queue(Queue* f, void* v) {
f->end = insert_queue_end(f->end,v);
if(f->begin==NULL) {
f->begin=f->end;
}
}
void print_queue_double(Queue* f) {
List* i;
for(i=f->begin;i!=NULL;i=i->prox)
printf("%.3lf\n",*((double*)i->info));
}
int main() {
Queue* f;
f = create_queue();
char ent1[100];
double n1;
int countDoubles=0;
scanf("%s",ent1);
while(ent1[0]!='X' && countDoubles < 20)
{
if(ent1[0]=='E') {
scanf("%lf",&n1);
insert_queue(f,&n1);
}
else if (ent1[0]=='D') {
queue_remove_double(f);
}
else
{
///Print the list - and see the result "debug"
print_queue_double(f);
}
scanf("%s",ent1);
}
free(f);
}
副本通過'prox'你的意思是「下一個」？你是在傳遞一個double *還是'double'？這會有所作爲。你怎麼知道它不適用於雙值？編譯器是否會引發錯誤？ –
你怎麼知道什麼時候返回值是一個'int'（或指向'int'）與一個'double'（或指向'double'）？如果你要返回一個值，爲什麼你需要'next'（'prox'？）指針？如果你要返回一個值列表，就這樣吧，但也許你應該這樣說。 「info」成員指向的空間分配情況如何？一個節點（或整個節點列表）如何被釋放？ –
這裏似乎有一個[XY問題]（http://mywiki.wooledge.org/XyProblem）潛伏在這裏。雖然你想要做的事情可以完成，但目前還不清楚它是否可以在你嘗試完成時完成，而不清楚你爲什麼試圖做到這一點。所以，也許你需要拿出一個例子來說明如何/爲什麼這會有好處。 –