C 结构指针的分段错误
我是C语言的新手,我编写了以下代码C 结构指针的分段错误,c,segmentation-fault,C,Segmentation Fault,我是C语言的新手,我编写了以下代码 #include <stdlib.h> #include<stdio.h> typedef struct { int name1; }check1; typedef struct { int name2; }check2; int main() { check1 *test1; check2 *test2; test1->name1=1; test2->name2=2;
#include <stdlib.h>
#include<stdio.h>
typedef struct
{
int name1;
}check1;
typedef struct
{
int name2;
}check2;
int main()
{
check1 *test1;
check2 *test2;
test1->name1=1;
test2->name2=2;
return 0;
}
在gdb中:-
Program received signal SIGSEGV, Segmentation fault.
0x000000000040045e in main ()
原因可能是什么
谢谢。您声明了两个指针,但尚未为它们分配任何内存。指针指向无效内存 试试这个:
check1 *test1 = malloc(sizeof(*test1));
if (test1 == NULL)
// report failure
check2 *test2 = malloc(sizeof(*test2));
if (test2 == NULL)
// report failure
您已经声明了两个指针,但尚未为它们分配任何内存。指针指向无效内存 试试这个:
check1 *test1 = malloc(sizeof(*test1));
if (test1 == NULL)
// report failure
check2 *test2 = malloc(sizeof(*test2));
if (test2 == NULL)
// report failure
您还可以在堆栈上声明变量,并将其地址分配给指针
check checka;
check* pcheck = &checka;
printf("%i",pcheck->name1);
您还可以在堆栈上声明变量,并将其地址分配给指针
check checka;
check* pcheck = &checka;
printf("%i",pcheck->name1);