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.NET Forum / Languages / Managed C++ / February 2005Getting File size from network byte order
Hi, Following code shows how a packet is sent from a client to a server's socket. The file size (fileSizeTemp) is converted to network byte order before sending to the server. *((unsigned short *)&dataBuff[6 + filenameLen + 1]) = htons((unsigned short)(fileSizeTemp >> 16)); *((unsigned short *)&dataBuff[6 + filenameLen + 3]) = htons((unsigned short)((fileSizeTemp << 16) >> 16)); if (send(clientSock, ..., ...) == SOCKET_ERROR) { //... } My question is, how to get the size of the file (from the received packet) on the server? Thanks Manu. > Hi, > [quoted text clipped - 3 lines] > > *((unsigned short *)&dataBuff[6 + filenameLen + 1]) = htons((unsigned > short)(fileSizeTemp >> 16)); > *((unsigned short *)&dataBuff[6 + filenameLen + 3]) = htons((unsigned > short)((fileSizeTemp << 16) >> 16)); You could do it more simply with htonl : *((unsigned long *)&dataBuff[6 + filenameLen + 1]) = htonl(fileSizeTemp) > if (send(clientSock, ..., ...) == SOCKET_ERROR) > { [quoted text clipped - 3 lines] > My question is, how to get the size of the file (from the received packet) > on the server? ntohl. Arnaud MVP - VC > Hi, > [quoted text clipped - 3 lines] > > *((unsigned short *)&dataBuff[6 + filenameLen + 1]) = htons((unsigned > short)(fileSizeTemp >> 16)); > *((unsigned short *)&dataBuff[6 + filenameLen + 3]) = htons((unsigned > short)((fileSizeTemp << 16) >> 16)); You could do it more simply with htonl : *((unsigned long *)&dataBuff[6 + filenameLen + 1]) = htonl(fileSizeTemp) > if (send(clientSock, ..., ...) == SOCKET_ERROR) > { [quoted text clipped - 3 lines] > My question is, how to get the size of the file (from the received packet) > on the server? ntohl. Arnaud MVP - VC But there are 4 byes of data. How do I supply the netlong parameter to ntohl from 4 bytes of data? Manu. >> Hi, >> [quoted text clipped - 26 lines] > Arnaud > MVP - VC OK, I got it. long fileSize = ntohl(((ULONG)dataBuff[6 + filenameLen + 4] << 24) + ((ULONG)dataBuff[6 + filenameLen + 3] << 16) + ((ULONG)dataBuff[6 + filenameLen + 2] << 8) + ((ULONG)dataBuff[6 + filenameLen + 1])); That is all that is required. Thanks Arnaud :-) Manu. > But there are 4 byes of data. How do I supply the netlong parameter to > ntohl from 4 bytes of data? [quoted text clipped - 31 lines] >> Arnaud >> MVP - VC Aww... those ULONGs should be long :) > OK, I got it. > [quoted text clipped - 44 lines] >>> Arnaud >>> MVP - VC > OK, I got it. > > long fileSize = ntohl(((ULONG)dataBuff[6 + filenameLen + 4] << 24) > + ((ULONG)dataBuff[6 + filenameLen + 3] << 16) > + ((ULONG)dataBuff[6 + filenameLen + 2] << 8) > + ((ULONG)dataBuff[6 + filenameLen + 1])); long fileSize = ntohl(*(long*)(dataBuff+6+filenameLen+1)); As you've written it above you'll get the wrong answer if you compile the code on a big-endian machine. Similarly, on the server end you should be using htonl as Arnaud pointed out. -cd > OK, I got it. > [quoted text clipped - 4 lines] > > That is all that is required. DO NOT mess with byte shifting and ordering when ntohl is doing exactly that for you (and do it right!) long* ptr=reinterpret_cast<long*>(dataBuff[6 + filenameLen + 1]); long fileSize=ntohl(*ptr); Arnaud MVP - VC >>OK, I got it. >> [quoted text clipped - 10 lines] > long* ptr=reinterpret_cast<long*>(dataBuff[6 + filenameLen + 1]); > long fileSize=ntohl(*ptr); And note that the reinterpret_cast above is potential undefined behaviour due to alignment issues, but is fortunately perfect well defined for x86, which is unfussy about alignment (although you lose atomicity and performance). Beware if you are using a non-x86 platform which may be fussier about alignment, in which case you should use memcpy to copy the 4 bytes into a ULONG before calling ntohl. Tom Oh well... Thanks a lot guys. That is a lot of information I got. Manu. >>>OK, I got it. >>> [quoted text clipped - 19 lines] > > Tom Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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