我有一个名为islands.txt的文件,内容是:
islandone
islandtwo
islandthree
这是我的代码:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct island{
char *name;
struct island *previous;
} island;
void printIsland(island is){
printf("%s", is.name);
if(is.previous && is.previous->name[0] != ' '){
printf("%s", is.previous->name);
}
}
int main(){
// the file to be read.
FILE *islandsFile = fopen("islands.txt","r");
// temporary location to store the name read from the file.
char name[40];
// temporary pointer to an island which has been already read for linking.
island *previousIsland;
while(fscanf(islandsFile,"%s",name) != EOF){
// allocate space for a new island and point to it with (*newIsland) pointer
island *newIsland =malloc(sizeof(island));
// assign name
newIsland->name = name;
// if previousIsland pointer is not null
// it means there is an island that was read before newIsland in the file
if(previousIsland){
// newIsland.previous should hold the address of this previously read island..
newIsland->previous = previousIsland;
}
// now previousIsland is the newIsland..
previousIsland = newIsland;
printIsland(*newIsland);
puts("");
}
fclose(islandsFile);
}
我对输出的期望是:
islandone
islandtwoislandone
islandthreeislandtwo
相反,我得到的只是分割错误。我什么都试过了,但我被卡住了。这里的分段错误在哪里?我对C还很陌生,不知道如何调试。
是的,您还需要为名称分配内存。您只为结构分配
typedef struct island{
char *name;
struct island *previous;
} island;
所以这个
// assign name
newIsland->name = name;
将指针设置到堆栈上的数组,但每次循环迭代时,指针都是相同的地址。
而是做一些类似的事情
newIsland->name = strdup(name);
或者如果你更喜欢
newIsland->name = malloc( strlen( name ) + 1 );
strcpy( newIsland->name, name );
这里有几个问题。除了CyberSpock提到的代码外,您还有以下代码:
island *previousIsland;
while(fscanf(islandsFile,"%s",name) != EOF){
/* some code omitted */
if(previousIsland){
newIsland->previous = previousIsland;
}
previousIsland变量未初始化,第一次if可能为true,因此前一个指针指向无效内存。然后,当您在printIsland中到达末尾时,它将继续跟随未初始化的指针,进入无效内存。我也看到你没有释放()任何内存,但这可能是因为你不喜欢这样一个小例子。
要调试C程序,调试器就是你的朋友。现在您不知道使用哪种操作系统和编译器,但如果使用gcc,gdb就是匹配的调试器。