我试图让一个C程序工作,但我很生气。这是我简化的代码查找错误:
#include <stdio.h>
#include <unistd.h>
#include <sqlite3.h>
int main(){
sqlite3 *conn;
sqlite3_stmt *res;
const char *tail, *sqlresult;
sqlite3_open("cubecat", &conn);
char buffer,query;
int id;
id= 1;
buffer = 'a';
if(buffer == 'a') snprintf(&query,100,"SELECT start FROM payloads WHERE id=%d", id);
printf("%s",&query);
int error = sqlite3_prepare_v2(conn, &query, 100, &res, &tail);
printf("%d",error);
}
错误正是发生在"sqlite_prepare_v2"函数上,因为如果我对该行进行注释,则不存在Segmentation Fault。
提前谢谢!
char query;
snprintf(&query,100,"SELECT start FROM payloads WHERE id=%d", id);
这就是问题所在。query
只为一个字符保留内存。snprintf()
的第二个参数指定大小是有原因的。这个代码应该这样修改:
char query[100];
snprintf(query, sizeof(query), "SELECT start FROM payloads WHERE id=%d", id);