Help! segmentation fault
Hi Iam getting a segmentation fault in the following program, it would be great if you can suggest where exactly iam going wrong, here Iam trying to extract substrings from a string and store it in a array of strings, please note the recursive call in the function.
Code:}
Last edited by doubty; 06-23-2009 at 03:39 AM.
What are you doing this for?Code:parser(label,label);Shouldn't you be paying attention to the number of rows you have here, instead of just assuming there's an unlimited supply? Typically when you leave out the size of one of the dimensions, you actually pass something along to let the function know when to stop.Code:void parser(char in_str[][25],char out_str[][25])
You really need to learn to indent if you want people to read your code. No one wants to read that.
Quzah.
Hope is the first step on the road to disappointment.
Hi ,Originally Posted by doubty
Srry abt the indentation, when i pasted it frm the file i lost the indentation, anyways this is what i have been trying to do,
I have a String say G(G(G(xxxx))) > yyyyy
I have to extract all the substings of the string namely {G(G(G(xxxx)))>yyyyy,G(G(G(xxxx))),G(G(xxxx)),G(xx xx),xxxx,yyyyy}
I have in my main program accepted the string and put it as the first element of the array of strings label,
Now I have written the function parser that takes 2 arrays,{the input and the output array} in fact bothe are the same, and calculates all the substrings and stores in it. So upfront i dunno how many rows are present in the array.
I have used recursion coz of the pattern in the string.
I know iam going wrong somewheer, can you please guide me
I hope the above details are sufficient to help me
The logic seems awkward. I suggest using a simple and separate char buffer for your string, and using just *one* array for your output.
label and label both obviously point to the base of the same array, so now you have to dance around the big elephant in the living room, which is that both arrays are actually the same array.
Clear and simple is what you need, not this "I'll use one array, but dance around with it like it's really two array's". Easy to debug, easy to maintain, and modify, and easy to have others understand, etc.
Hi Adak,
I changed it as below:
Iam getting segmentation fault while calling the function, can you please help me nowCode:#include<stdio.h> #include<string.h> #include<time.h> #include<stdlib.h> int k=0; int p=1; void parser(char in_str[][25]); //Main function int main() { char inputexp[500]; char label[50][25]; printf("enter expression"); scanf("%s",inputexp); strcpy(label[0],inputexp); //calling function parser parser(label); return 0; getchar(); } //function parser void parser(char in_str[][25]) { while(in_str[k]!= '\0') { int len=strlen(in_str[k]); int openbrac_count=0,closebrac_count=0; int i=0; int from=0; while(i<len) { while(i<len && in_str[k][i]!=')') { if((in_str[k][i]=='(')) { openbrac_count++; } i++; } while(i<len && in_str[k][i]!='(') { if(in_str[k][i]==')') { closebrac_count++; } if(openbrac_count==closebrac_count & openbrac_count != 0) { strncpy(in_str[k+1],&in_str[k][from],i-from); p=p+1; if (in_str[k][i+1]=='&' || in_str[k][i+1]=='|') { from=i+2; } } i++; } if (in_str[k][0]=='G') { strncpy(in_str[p+1],&in_str[k][2],strlen(in_str[k])-2); parser(in_str); } if (in_str[k][0]=='F') { strncpy(in_str[p+1],&in_str[k][2],strlen(in_str[k])-2); parser(in_str); } if (in_str[k][0]=='X') { strncpy(in_str[p+1],&in_str[k][2],strlen(in_str[k])-2); parser(in_str); } } k++; } //end of function parser }
Ok, so say you're extracting strings. What is it you expect your end string to look like? Given:
a( b( c( d ) ) )
What would your end string look like? Is it just one string all squished together somehow? Is it a series of strings?
Quzah.
Hope is the first step on the road to disappointment.
This is what also bothers me. I believe it should parse out as separate strings to be:
What's left after every recursion/iteration:
a( b( c( d ) ) )
b( c( d ) )
c( d )
d
What's parsed out for every recursion/iteration:
a( )
a( b( ) )
a( b( c( ) ) )
a( b( c( d ) ) )
Which I don't see in his example:
Doubty, can you take Quzah's example and show it being correctly parsed?I have a String say G(G(G(xxxx))) > yyyyy
I have to extract all the substings of the string namely {G(G(G(xxxx)))>yyyyy,G(G(G(xxxx))),G(G(xxxx)),G(xx xx),xxxx,yyyyy}
To me, "all substrings", would mean *all* substrings:
etc.Code:{ G ( G ( ... {G G(
Clearly, that's not what you want.
Hi Adak and quzah,
For the example a( b( c( d ) ) ) my substring would be
a( b( c( d ) ) )
b( c( d ) )
c( d )
d
However said that the strings may not be as simple in the sense that it may also be as
a( b( c( d ) ) ) ^ e(f(g)))
in which case my substring set would be:
a( b( c( d ) ) ) ^ e(f(g)))
a( b( c( d ) ) )
b( c( d ) )
c( d )
d
e(f(g)))
f(g)
g
Another constraint is that i can also have a(b(a(c)))
i.e 'a' repeting more than once ( it is for this i have tried to use recursion)
for the above case my substrings would be:
a(b(a(c)))
b(a(c))
a(c)
c
I hope I am clearer now, Thanks a lot for your efforts to help me out
The first and last examples you gave, were fine, but this one is trouble, imo:
Because you're parsing out the ^e(f(g))), and then you're bringing it back to be re-parsed.However said that the strings may not be as simple in the sense that it may also be as
a( b( c( d ) ) ) ^ e(f(g)))
in which case my substring set would be:
a( b( c( d ) ) ) ^ e(f(g)))
a( b( c( d ) ) )
b( c( d ) )
c( d )
d
e(f(g)))
f(g)
g
IMO you need a block of code (or a function), to "feed" the recursive parsing function. The "feeder" will record the whole string as an answer, then remove any part of the string including and after the ^ , and send it to the parsing function only after the leftmost part before the ^ has been parsed.
That "feeding" function seems like a good candidate for recursion, itself.
It's not recursive, but this is my take on parsing this expression
The string sample you posted had 1 extra right parentheses, which I removed.Code:/* Parsing.c a( b( c( d ) ) ) ^ e(f(g)) in which case my substring set would be: a( b( c( d ) ) ) ^ e(f(g)) a( b( c( d ) ) ) b( c( d ) ) c( d ) d e(f(g)) f(g) g */ #include <stdio.h> #include <string.h> int main(void) { int i, lt, rt, hasFlex = 0; char *ptr = NULL; char s[] = {"a( b( c( d ) ) ) ^ e(f(g))"}; printf("\n\n\nOriginal String: \n%s", s); ptr = strchr(s, '^'); //does the string have a circumflex if(ptr) { rt = ptr - s; hasFlex = rt; } else rt = strlen(s); lt = 0; //printf("\n lt: %d, rt: %d", lt, rt); do { if(s[lt++] == '(') { while(s[rt--] != ')' && lt < rt); putchar('\n'); for(i = lt; i <= rt; ++i) printf("%c", s[i]); //i = getchar(); } if(lt >= rt && hasFlex) { //a circumflex component to deal with lt = hasFlex+1; rt = strlen(s); hasFlex = 0; putchar('\n'); for(i = lt; i <= rt; ++i) printf("%c", s[i]); } }while(lt <= rt); printf("\n\n\t\t\t Press Enter When Ready "); lt = getchar(); return 0; }
Hi thanks a lot for your help, but this is too specific for the case where in I have only one ^ symbol between 2 set od substrings , in my case i can have any number of ^ symbols and substrings in between (i.e A(A(d)) ^ B(B(g)) v C(C(h)) ^ D(D(h)), Also please note that i can also have 'V' symbol as well.
Oh, you're just *full* of good news, aren't you!!
So change the code to handle a ^ or a V, in the same way a ^ is handled now.
And add either logic inside the do while loop to handle multiple ^'s or V's, or you can add another do while loop outside the current do while loop, to basically, keep it looping while the string has more separating ^'s or V's with substrings of their own.
Basically, you just want to move, from left (lt) to right(rt) across the string, and see if there are any ^'s or V's.
If there are, put the rt index at the leftmost one of them.
Now parse the substring from lt (0), to the rt index, in the string s.
Afterward, set lt at rt, and scan the string for any more separators (^ or V). If you find one, set the rt index there, and parse the substring between lt and rt.
Repeat until there are no more separators, then process your last substring.
When your string has no more separators, and no more parenthesis, you're done.
Last edited by Adak; 06-24-2009 at 12:27 AM.
Hi ,
I know you will kill me for this ..
I was also checking out if i had strings in any other particular format and i hit upon this one:
i have a very very peculiar requirement :
i.e my input can also be as G(G(p^q)) ^ G(q)
the problem here is that if I go by finding the first occurence of ^ and demarkate it doesnt work.
i.e
my strings need to be :
G(G(p^q)) ^ G(q)
G(G(p^q))
G(p^q)
p^q
p
q
G(q)
i.e ther's a difference between the ^ inside the bracket and the one outside,
Thanks
How does what you posted indicate a difference between a ^ inside a bracket and one outside? In each case, each subexpression of ^ needs to be in your list (as well as the ^ expression itself).
I don't know the context of why you're doing this, exactly, but given that these are logical expressions is there a reason you don't treat them as such? (I.e., make a tree out of this using the shunting yard algorithm and then traverse the tree to get all your subexpressions?)
