2012-04-20 52 views
2

Valgrind的报告一个SQLite错误:Valgrind报告了一个SQLite错误,为什么?

 
==11614== Jump to the invalid address stated on the next line 
==11614== at 0x0: ??? 
==11614== by 0x6E8CF: sqlite3MallocSize (sqlite3.c:19034) 
==11614== by 0x6E472: mallocWithAlarm (sqlite3.c:18870) 
==11614== by 0x6E520: sqlite3Malloc (sqlite3.c:18895) 
==11614== by 0x6ED56: sqlite3MallocZero (sqlite3.c:19159) 
==11614== by 0x6DA46: pthreadMutexAlloc (sqlite3.c:18039) 
==11614== by 0x6D779: sqlite3MutexAlloc (sqlite3.c:17353) 
==11614== by 0xFA4B7: sqlite3_initialize (sqlite3.c:112588) 
==11614== by 0xFD895: openDatabase (sqlite3.c:114531) 
==11614== by 0xFDF13: sqlite3_open (sqlite3.c:114780) 
==11614== by 0x579C1: SqlLiteConnection_Connect(char*) (SqlLite.cpp:14) 
==11614== Address 0x0 is not stack'd, malloc'd or (recently) free'd 

我传递一个有效的价值sqlite3_open。任何人有任何想法,为什么发生这种情况?

P.S.我正在使用SQLite Amalgamation 3.7.11版。我正在运行Mac OSX 10.7.3。 SQLite编译为gcc -c -g -m32 sqlite3.c

回答

0

我遇到了与valgrind 3.7.0完全相同的问题。升级到3.8.1后,问题消失了。

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