这只是一个样本测试,我想把list Lb和La联合起来,不需要repeat元素。它不起作用,返回了-1073741510,我想可能是数组溢出,但我找不到问题所在
这是我的代码:
#include <stdio.h>
void Union(char *La, char *Lb);
int ListLength(char *L);
void GetElem(char *L, int i, char *e);
int LocateElem(char *L, char *e, int (*comp)(char a, char b));
int compare(char a, char b);
void ListInsert(char *, int, char *);
int main(){
char *La;
char *Lb;
int i;
for(i = 0; i <= 10; ++i){
La[i] = i;
Lb[i] = i + 5;
}
La[i] = '\0';
Lb[i] = '\0';
Union(La, Lb);
for(i = 0; La[i] != '\0'; ++i){
printf("%c\n", La[i]);
}
return 0;
}
//unite La and Lb without repeat elements
void Union(char *La, char *Lb){
int La_length = ListLength(La);
int Lb_length = ListLength(Lb);
int i = 0;
char *e;
for(i; i<= Lb_length; ++i){
GetElem(Lb, i, e);
if(!LocateElem(La, e, compare))
ListInsert(La, ++La_length, e);
}
}
//caculate the length of L
int ListLength(char *L){
int i;
for(i = 0; *(L + i) != '\0'; ++i);
return i;
}
void GetElem(char *L, int i, char *e){
*e = *(L + i);
}
//search the element e in L, if exist return the location, else return 0
int LocateElem(char *L, char *e, int (*comp)(char a, char b)){
int i;
for(i = 0; *(L + i) != '\0'; ++i){
if(comp(*(L + i), *e)) return i + 1;
}
return 0;
}
//compare the element a and b
int compare(char a, char b){
if(a == b) return 1;
return 0;
}
//if e doesn't exit in L, insert the e in L
void ListInsert(char *L, int i, char *e){
int j;
for(j = ListLength(L) - 1; j >= i; --j){
*(L + j + 1) = *(L + j);
}
L[ListLength(L)] = '\0';
*(L + i - 2) = *e;
}
最佳答案
首先,这是错误的:
char *La;
char *Lb;
int i;
for(i = 0; i <= 10; ++i){
La[i] = i;
Lb[i] = i + 5;
}
您需要为
La
和Lb
保留内存,例如,将它们声明为:char La[12];
char Lb[12];
然后这个:
char *e;
for(i; i<= Lb_length; ++i){
GetElem(Lb, i, e);
应改为:
char e;
for(; i<= Lb_length; ++i){
GetElem(Lb, i, &e); /* better yet: e=Lb[i] */
最后,你最有可能通过使用
<=
而不是<
作为for
退出条件来循环一次太多。关于c - 为什么此C编程无法正确运行?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/15708616/