对'功能'的未定义引用 [英] undefined reference to 'Function'
本文介绍了对'功能'的未定义引用的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我附上了一个代码。对'SCardEstablishContext @ 16'和'SCardConnectA @ 24'的未定义引用不断出现。该怎么办?
SCARDCONTEXT hSC;
LONG lReturn;
// 建立上下文。
lReturn = SCardEstablishContext(SCARD_SCOPE_USER,
NULL,
NULL,
& hSC);
if (SCARD_S_SUCCESS!= lReturn)
printf( 失败的SCardEstablishContext \ n);
else
{
SCARDHANDLE hCardHandle;
LONG lReturn1;
DWORD dwAP;
lReturn1 = SCardConnect(hSC,
(LPCTSTR) Rainbow Technologies SCR3531 0 ,
SCARD_SHARE_SHARED,
SCARD_PROTOCOL_T0 | SCARD_PROTOCOL_T1,
& hCardHandle,
& dwAP);
if (SCARD_S_SUCCESS!= lReturn1)
{
printf( SCardConnect\ n失败);
退出( 1 ); // 或其他适当的操作。
}
// 使用连接。
// 显示活动协议。
switch (dwAP)
{
case SCARD_PROTOCOL_T0:
printf( 有效协议T0 \ n);
break ;
case SCARD_PROTOCOL_T1:
printf( 活动协议T1 \ n);
break ;
case SCARD_PROTOCOL_UNDEFINED:
默认:
printf ( 未协商或未知的活动协议);
break ;
}
}
解决方案
您必须将您的应用程序链接到Winscard库。
在一个源文件的顶部添加此行:
#pragma comment(lib,winscard)
i have this attached with a code. undefined reference to 'SCardEstablishContext@16' and 'SCardConnectA@24' keeps coming. what to do?
SCARDCONTEXT hSC;
LONG lReturn;
// Establish the context.
lReturn = SCardEstablishContext(SCARD_SCOPE_USER,
NULL,
NULL,
&hSC);
if ( SCARD_S_SUCCESS != lReturn )
printf("Failed SCardEstablishContext\n");
else
{
SCARDHANDLE hCardHandle;
LONG lReturn1;
DWORD dwAP;
lReturn1 = SCardConnect( hSC,
(LPCTSTR)"Rainbow Technologies SCR3531 0",
SCARD_SHARE_SHARED,
SCARD_PROTOCOL_T0 | SCARD_PROTOCOL_T1,
&hCardHandle,
&dwAP );
if ( SCARD_S_SUCCESS != lReturn1 )
{
printf("Failed SCardConnect\n");
exit(1); // Or other appropriate action.
}
// Use the connection.
// Display the active protocol.
switch ( dwAP )
{
case SCARD_PROTOCOL_T0:
printf("Active protocol T0\n");
break;
case SCARD_PROTOCOL_T1:
printf("Active protocol T1\n");
break;
case SCARD_PROTOCOL_UNDEFINED:
default:
printf("Active protocol unnegotiated or unknown\n");
break;
}
}
解决方案
You must link your application to the Winscard library.
Add this line on top of one source file:
#pragma comment(lib, "winscard")
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