我有一个结构数组,我想写入二进制文件。我有一个write.c程序和一个read.c程序。 write.c程序似乎工作正常,但是当我运行read.c程序时,我遇到了分段错误。我是C的新手所以如果有人可以查看我的代码以查找任何明显的错误,那将会很棒。我保证不会太久:)
为write.c:
#include <stdlib.h>
#include <stdio.h>
struct Person
{
char f_name[256];
char l_name[256];
int age;
};
int main(int argc, char* argv[])
{
struct Person* people;
int people_count;
printf("How many people would you like to create: ");
scanf("%i", &people_count);
people = malloc(sizeof(struct Person) * people_count);
int n;
for (n = 0; n < people_count; n++)
{
printf("Person %i's First Name: ", n);
scanf("%s", people[n].f_name);
printf("Person %i's Last Name: ", n);
scanf("%s", people[n].l_name);
printf("Person %i's Age: ", n);
scanf("%i", &people[n].age);
}
FILE* data;
if ( (data = fopen("data.bin", "wb")) == NULL )
{
printf("Error opening file\n");
return 1;
}
fwrite(people, sizeof(struct Person) * people_count, 1, data);
fclose(data);
return 0;
}
read.c:
#include <stdlib.h>
#include <stdio.h>
struct Person
{
char f_name[256];
char l_name[256];
int age;
};
int main(int argc, char* argv[])
{
FILE* data;
if ((data = fopen("data.bin", "rb")) == NULL)
{
printf("Error opening file\n");
return 1;
}
struct Person* people;
fread(people, sizeof(struct Person) * 1/* Just read one person */, 1, data);
printf("%s\n", people[0].f_name);
fclose(data);
return 0;
}
感谢您的帮助!
答案 0 :(得分:5)
struct Person* people;
这只分配一个指向struct的指针,但是你没有为实际的struct内容分配任何空间。要么像编写程序一样使用malloc
,要么尝试类似:
struct Person people;
fread(&people, sizeof(people), 1, data);
答案 1 :(得分:3)
您需要先为此人分配内存。将struct Person* people;
更改为struct Person* people = malloc(sizeof(struct Person));
。不要忘记最后释放你的记忆:free(people);
。
答案 2 :(得分:1)
在执行malloc
之前,您需要people
内存到指针变量fread
,或者(更简单)直接读入本地变量:
struct Person people;
fread(&people, sizeof(struct Person) * 1/* Just read one person */, 1, data);
答案 3 :(得分:1)
您需要为正在阅读的数据分配空间:
people = malloc(sizeof(*people)*numPeople);