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Incrementing IPv6 address
hello every body.
I am working in address incrementation(+1) using c language.User inputs a value which is in hex address and incremented by 1. Suppose my address is - abc5:5723:2109:ffff:76af:82ea:9256:ffff. So my output becomes - abc5:5723:2109:ffff:76af:82ea:9257:0000 I already parse the entire string and put it in 8 different array but unable to do when the the last 4 digit reaches FFFF..I could not write the logic of that last portion.can any body help me to solve the problem? My code is - #include "stdafx.h" #include "conio.h" #include "stdio.h" #include "string.h" int _tmain(int argc, _TCHAR* argv[]) { char address[] = "abc5:5723:2109:ffff:76af:82ea:9256:ffff"; printf("Your IPv6 address is "); puts(address); char a[10], b[10], c[10], d[10], e[10], f[10], g[10], h[10]; if(sscanf(address, "%10[^:]:%10[^:]:%10[^:]:%10[^:]:%10[^:]: %10[^:]:%10[^:]:%10[^:]:",a,b,c,d,e,f,g,h)==8) { //puts(a); //puts(b); } int m,n,o,p,q,r,s,t; sscanf(a,"%x", &m); sscanf(b,"%x", &n); sscanf(c,"%x", &o); sscanf(d,"%x", &p); sscanf(e,"%x", &q); sscanf(f,"%x", &r); sscanf(g,"%x", &s); sscanf(h,"%x", &t); t++; printf("The final output is %x:%x:%x:%x:%x:%x:%x:%x",m,n,o,p,q,r,s,t); getch(); return 0; } Output - You IPv6 address is abc5:5723:2109:ffff:76af:82ea:9256:ffff The final output is abc5:5723:2109:ffff:76af:82ea:9256:10000 Waiting for your reply. Ayanava |
Incrementing IPv6 address
On 30/11/2011 13:14, Tester Tester wrote:
hello every body. I am working in address incrementation(+1) using c language.User Output - You IPv6 address is abc5:5723:2109:ffff:76af:82ea:9256:ffff The final output is abc5:5723:2109:ffff:76af:82ea:9256:10000 Waiting for your reply. This looks like homework. How do you think you might do it? Hint: quite a few Web scripts fell over the exact same problem at Y2k rollover including the USNO who should have known better. Instead of displaying YYYY as 2000 they showed 19100 instead. Regards, Martin Brown |
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