c – sprintf_s缓冲区太小

以下代码会导致错误并杀死我的应用程序.缓冲区只有10个字节长,文本长度为22个字节(缓冲区溢出)是有道理的.

char buffer[10];    
int length = sprintf_s( buffer,10,"1234567890.1234567890." );

我如何抓住这个错误,所以我可以报告它,而不是崩溃我的应用程序？

编辑：

阅读下面的评论后,我用_snprintf_s.如果返回-1值,则缓冲区未更新.

length = _snprintf_s( buffer,9,"123456789" );
printf( "1) Length=%d\n",length ); // Length == 9

length = _snprintf_s( buffer,"1234567890.1234567890." );
printf( "2) Length=%d\n",length ); // Length == -1

length = _snprintf_s( buffer,"1234567890.1234567890." );
printf( "3) Length=%d\n",length ); // Crash,it needs room for the NULL char

解决方法

您可以使用snprintf(Windows上的a.k.a _snprintf)而不是sprintf_s.

#ifdef WIN32
#define snprintf _snprintf
#endif

char buffer[10];    
int length = snprintf( buffer,"1234567890.1234567890." );
// unix snprintf returns length output would actually require;
// windows _snprintf returns actual output length if output fits,else negative
if (length >= sizeof(buffer) || length<0) 
{
    /* error handling */
}

原文地址：https://www.jb51.cc/c/113873.html