我有一个像这样的列表
#include <stdio.h>
#include <stdlib.h>
struct ListItem
{
int x;
int y;
struct ListItem *next;
};
int main()
{
int x1 =0;
int y1 = 0;
printf("Please enter the x coordinate: ");
scanf("%d", &x1);
printf("Please enter the y coordinate: ");
scanf("%d", &y1);
struct ListItem root;
if( root.next == NULL )
{
root.x = x1;
root.y = y1;
//I dont know what should I assign here but I want to have about 30 locations
//root.next = struct ListItem next;
}
//WHAT SHOULD I DO HERE?
{
printf("Your location is : (%d,%d)\n", root.x, root.y);
}
}
现在我想编写一个循环来遍历它，以便我可以打印列表中的每个元素:) 基本上我要做的是，我想从用户那里获取位置，然后我将打印它们。 请帮忙。
答案 0 :(得分：1)
这是一个有效的玩具程序，用于演示存储在数组中的列表。 实际编译没有警告 - gcc -Wall -Werror -std = c99 l.c
使用calloc动态分配的版本，如果你想要动态内存，可以使用，所以你可以稍后添加到列表中。
ladm@ash:~/src/scratch> ./a.out|head | sed 's/^/ /'
array[ 0] : (1,1)
array[ 1] : (2,2)
array[ 2] : (3,3)
Your location is : (3,3)
Your location is : (2,2)
Your location is : (1,1)
array[ 2] : (3,3)
array[ 1] : (2,2)
array[ 0] : (1,1)
ladm@ash:~/src/scratch> cat l.c | sed 's/^/ /'
#include <stdio.h>
#include <assert.h>
typedef struct ListItem
{
int x;
int y;
struct ListItem *next;
} ListItem;
void getCoord(int *x, int *y) {
static int num = 1;
*x = *y = num++;
}
int main( int argc, char **argv) {
const int N = 3;
ListItem listN[ N];
/* ListItem *listN = calloc( N, sizeof( ListItem)); */ /* Dynamic allocation method */
/* First Coordinate */
listN[ 0].next = NULL; /* Add item at front of list */
getCoord( &listN[ 0].x, &listN[ 0].y); /* Does the scanf stuff */
/* Add new coords to the list */
for (int i=1; i < N; i++) {
getCoord( &listN[ i].x, &listN[ i].y); /* Does the scanf stuff */
listN[ i].next = &listN[ i-1]; /* Add item at front of list */
}
/* List built */
{
ListItem *first = &listN[ N-1];
/* Dump all the coords in backing store */
for (int i = 0; i < N; i++) {
printf("array[ %d] : (%d,%d)\n", i, listN[ i].x, listN[ i].y);
}
/* Print list following pointers - should be reversed */
for (ListItem *l = first; l != NULL; l = l->next) {
printf("Your location is : (%d,%d)\n", l->x, l->y);
}
/* Dump all the coords in backing store reversed */
for (int i = N-1; i >= 0; --i) {
printf("array[ %d] : (%d,%d)\n", i, listN[ i].x, listN[ i].y);
}
}
}
答案 1 :(得分：1)
链接列表。输入坐标，直到输入零。
#include <stdio.h>
#include <stdlib.h>
struct ListItem
{
int x;
int y;
struct ListItem *next;
};
int main()
{
int x1 =0;
int y1 = 0;
int iCount = 0; // keep count of the structures allocated
int iEach = 0;
struct ListItem root = { 0, 0, NULL};// declare and initialize the first structure
struct ListItem* pFreeListItem = NULL;// declare a pointer and initialize it null. use for freeing memory later
struct ListItem* pListItem = &root;// declare a pointer and initialize it to point to the first structure
while ( 1) { // the main loop
printf("Please enter the x coordinate: ");
scanf(" %d", &x1);
printf("Please enter the y coordinate: ");
scanf(" %d", &y1);
pListItem->x = x1; // use the pointer to assign the coordinate
pListItem->y = y1;
iCount++; // keep track of the number of structures
printf("Input complete for location number %d\n", iCount);
printf("Enter 0 to exit or any other number to continue: ");
scanf(" %d", &y1);
if ( y1 == 0) { // exit the loop if zero is entered
break;
}
else { // if zero was not entered
pListItem->next = malloc ( sizeof ( struct ListItem));// allocate memory for the next structure
if ( pListItem->next == NULL) {
//allocation failed
exit (1);
}
pListItem = pListItem->next; // set the pointer to point to the new 'next' structure
pListItem->next = NULL; // set this to null as no memory has yet been allocated
}
}
pListItem = &root; // set the pointer to the original structure root
for ( iEach = 0; iEach < iCount; iEach++) // loop through each structure. icount holds the number of structures
{
printf("Location number %d is : (%d,%d)\n", iEach + 1, pListItem->x, pListItem->y);
pListItem = pListItem->next; // set the pointer to the next structure
}
pListItem = root.next; // set the pointer to the first allocated structure
for ( iEach = 1; iEach < iCount; iEach++) // loop through each structure
//start with 1 as the first structure was not allocate and does not need to be freed. icount holds the number of structures
{
pFreeListItem = pListItem->next; // set the free pointer to the next structure
free ( pListItem); // free the memory for the structure
pListItem = pFreeListItem; // point to the free pointer
}
}
编辑：此代码将显示指针的地址，这可能有助于澄清发生了什么
else { // if zero was not entered
pListItem->next = malloc ( sizeof ( struct ListItem));// allocate memory for the next structure
if ( pListItem->next == NULL) {
//allocation failed
exit (1);
}
printf ( "pListItem points to %p and pListItem->next points to %p\n", pListItem, pListItem->next);
pListItem = pListItem->next; // set the pointer to point to the new 'next' structure
pListItem->next = NULL; // set this to null as no memory has yet been allocated
printf ( "NOW pListItem points to %p and pListItem->next points to %p\n", pListItem, pListItem->next);
}
答案 2 :(得分：0)
如果您知道将拥有的位置数量，最好使用数组或矩阵，例如：
int x [30]; int y [30];
或
int location [60];
或
int location [2] [30];
并使用for循环迭代元素。
如果您不知道位置的数量，那么您希望在C中搜索LinkedList（或ArrayList）实现。互联网上有大量材料教授如何使用这些数据结构。
答案 3 :(得分：0)
迭代列表的好方法是创建give_successor和is_after_last等函数，然后在for循环中使用它，就像这样：
struct ListItem *l;
struct ListItem *head;
for (l = head; !(is_after_last(l)); l = give_successor(l)) {
do_stuff(l);
}
is_after_last的实施可能会因列表的实施而异：您可能需要检查l == NULL或l是否指向列表末尾的人工元素。
函数give_successor看起来像这样：
struct ListItem *give_successor(l) {
if (l == NULL)
return NULL;
return l->next
}
但是，我建议使用those等表格和库。