Acutally, the problem is with the malloc for the char* storage into stc->val. You need to add an extra byte to the malloc to accomodate the null terminator character appended by strcpy. This code: stc->val=(char *) malloc(strlen(var)); needs to be replaced with: stc->val=(char *) malloc(strlen(var) + 1); It occurs in three places. Hope this helps. --- Tony Wetmore Solipsys Corporation mailto:tony.wetmore@xxxxxxxxxxxx http://www.solipsys.com -----Original Message----- From: gcc-help-owner@xxxxxxxxxxx [mailto:gcc-help-owner@xxxxxxxxxxx] On Behalf Of Rupert Wood Sent: Sunday, July 20, 2003 8:14 AM To: 'Uílton O. Dutra' Cc: gcc-help@xxxxxxxxxxx Subject: RE: strcpy(stc->val,var) = Segmention Fault Uílton O. Dutra wrote: > I create simple program to read configurations files. > With Bcc32-Win98 my prog works fine, but on Gcc-Linux I got > "Segmention Fault" in "strcpy(stc->val,var)". This doesn't look linux-specific but lines 49-51 is where it first falls over for me: stc->next=(struct symtable *) malloc (sizeof (symtable)); stc=stc->next; strcpy(stc->var," "); strcpy(stc->var," "); You're allocating a new symtable and then attempting to write to use one of its member pointers without assigning it first or allocating it storage; you'd need a stc->var = (char*)malloc(<something>); before last strcpy, but I can't tell you what because it's not obvious to me what you're doing there! The second strcpy, for example, overwrites the first. (It's possible you meant to use one ->var and one ->val.) It strikes me this structure is going to be a nightmare to free() afterwards. Remember that malloc() doesn't zero the contents of the memory it's allocated; at the very least you should cmalloc() your symtables (or malloc then memset(stc,0,sizeof(symtable)) them) so you can see which pointers you have used and should free() and which you haven't (since they're NULL). It's posslble I've overlooked some things here - I haven't given this a lot of thought. Good luck making it work! Rup.
