c - sqlite3 sqlite3_prepare_v2 段错误

Question

在测试程序中:

#include <sqlite3.h>
#include <stdio.h>
#include <stdlib.h>

int     main(void) 
{
   sqlite3        *conn;
   sqlite3_stmt   *res;
   int            rc          =   0;
   int            rec_count   =   0;
   const char     *errMSG;
   const char     *tail;
   const char     *sql;

   rc = sqlite3_open("ljdata.sl3", &conn);
   if (rc) 
   {   puts("Can not open database");
      exit(0);
   }
   printf ("database open\n");

   sql   =   "create table if not exists people (id int, firstname varchar(20), lastname varchar(20), phonenumber char(10))";
   rc   =   sqlite3_exec(conn, sql, 0, 0, 0);
   if   (rc != SQLITE_OK)
   {   printf   ("Cannot access table, rc = %d,\n%s\n", rc, sqlite3_errmsg (conn));
      printf   ("sql = %s", sql);
      return   -2;
   }
   printf("db table (?)created\n");

   rc = sqlite3_exec(conn,
               "update people set phonenumber=\'5055559999\' where id=3",
               0, 0, 0);
   //printf ("1st update done with rc = %d\n", rc);

   rc = sqlite3_prepare_v2(conn,
               "select lastname,firstname,phonenumber,id from people order by id",
               1000, &res, &tail);
   printf ("prepare statement executed\n");

   if (rc != SQLITE_OK) 
   {   puts("We did not get any data!");
      exit(0);
   }

   puts("==========================");

   while (sqlite3_step(res) == SQLITE_ROW) 
   {   printf("%s|", sqlite3_column_text(res, 0));
      printf("%s|", sqlite3_column_text(res, 1));
      printf("%s|", sqlite3_column_text(res, 2));
      printf("%u\n", sqlite3_column_int(res, 3));

      rec_count++;
   }

   puts("==========================");
   printf("We received %d records.\n", rec_count);

   sqlite3_finalize(res);

   sqlite3_close(conn);

   return 0;
}

如果我注释掉:
printf ("1st update done with rc = %d\n", rc); // <<== THIS STATEMENT
我遇到段错误。否则我似乎会正常完成。gcc -v -->
使用内置规范。
COLLECT_GCC=gcc
COLLECT_LTO_WRAPPER=/usr/lib/gcc/x86_64-linux-gnu/4.7/lto-wrapper
目标:x86_64-linux-gnu
配置为:../src/configure -v --with-pkgversion='Debian 4.7.2-5' --with-bugurl=file:///usr/share/doc/gcc-4.7/README.Bugs - -enable-languages=c,c++,go,fortran,objc,obj-c++ --prefix=/usr --program-suffix=-4.7 --enable-shared --enable-linker-build-id --with- system-zlib --libexecdir=/usr/lib --without-included-gettext --enable-threads=posix --with-gxx-include-dir=/usr/include/c++/4.7 --libdir=/usr/lib --enable-nls --with-sysroot=/--enable-clocale=gnu --enable-libstdcxx-debug --enable-libstdcxx-time=yes --enable-gnu-unique-object --enable-plugin --enable-objc-gc --with-arch-32=i586 --with-tune=generic --enable-checking=release --build=x86_64-linux-gnu --host=x86_64-linux-gnu --目标=x86_64-linux-gnu
线程模型:posix
gcc 版本 4.7.2 (Debian 4.7.2-5)

但是，我在 Debian 测试中也遇到了一个(分析较少的)段错误。
有什么建议么？ (我注意到“您需要为 sql 字符串分配空间，但我使用的是文字，因此不应该适用。”回答了类似的错误。)

Answer 1

我没想到，sqlite3_prepare_v2 调用失败了，因为你在字符串长度上撒了谎:

rc = sqlite3_prepare_v2(conn,
           "select lastname,firstname,phonenumber,id from people order by id",
           1000, &res, &tail);

如果您包含 printf 调用，它似乎可以工作，因为字符串文字将与您说谎的字符串以及任何 sqlite3_prepare_v2 放在同一部分正在做的只是溢出了另一个字符串。

这里实际上是缓冲区溢出。