that was actually a mistake , i should have taken the value before i declared it BUT as for arrays goes, we know we can access the a[10] even though we may just declare a[5] ! may be its the same reason i was able to read in "for" loop.... besides here having the size of column matters more than the row size..(see passing 2d arrays as parms)
On Aug 24, 8:30 pm, Anurag Gupta <[email protected]> wrote: > @ Mehnaaz :the variable no_p will be equal to 0,since it is an > external one > so the declaration char p[no_p][max] is equivalent to p[0][max]; > and size of p is zero. > then how you can insert anything into it > as you are doing in for loop? > > no_p may receive some non zero value afterwards but array p is not re- > declared so > it doesnt has any space alloted to it ,then how for loop is working > without any glitch? > > On Aug 24, 5:58 pm, Mehnaaz <[email protected]> wrote: > > > > > > > > > #include <stdio.h> > > #define max 30 > > int no_p; > > int main() { > > > char p[no_p][max]; > > char x; > > int i=0; > > printf("enter the no of productions..\n"); > > scanf("%d", &no_p); > > printf("you have entered :%d\n", no_p); > > printf("Variable who's FOLLOW you want\n"); > > scanf("%c", &x); > > printf("you have entered :%c",X); > > printf("\n"); > > for( i =0 ; i< no_p ; i++){ > > printf("enter the production # %d",i+1); > > scanf("%s",p[i]); > > } > > //follow(p,x); > > return 0; > > > } > > > the output i am getting goes like this > > ...................................... > > enter the no of productions.. > > 3 > > you have entered :3 > > Variable who's FOLLOW you want > > you have entered : > > > enter the production # 1 > > ........................................... > > at the line 4 of the output its supposed to wait for my scanf() entry > > value right??..but it executes the printf after that giving "you have > > entered" > > > i'm using gcc to compile this -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
