{"id":24982,"date":"2022-12-07T03:02:48","date_gmt":"2022-12-06T21:32:48","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-pop-function-for-stack-in-c\/"},"modified":"2022-12-07T03:02:48","modified_gmt":"2022-12-06T21:32:48","slug":"solved-pop-function-for-stack-in-c","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-pop-function-for-stack-in-c\/","title":{"rendered":"[Solved] Pop() function for stack in C"},"content":{"rendered":"

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It would help to see how you push items onto the stack. If you’re really calling pop<\/code> without a push<\/code> first, well, then it’s not supposed<\/em> to do anything, is it?<\/p>\n

This bit makes me nervous:<\/p>\n

Node *aux,*prev;\nprev = *stack;\naux = prev->next;\nif(aux == NULL)\n{\n    free(prev);\n    return;\n}\n<\/code><\/pre>\n
You set prev<\/code> to *stack<\/code>, and if nothing follows prev<\/code>, you free it.  Note that since prev == *stack<\/code>, you’ve also freed *stack<\/code>, so that pointer is now invalid<\/em>.  If you try to access that pointer in your caller, you’ll invoke Undefined Behavior.  <\/p>\n
It looks like you’re making your list tail the top of the stack; I’m going to tell you right now that you will make your life much<\/em> simpler if you make the list head<\/em> the top of the stack, such that your pushes and pops look like the following:<\/p>\n
bool push( Node **l, int val )\n{\n  Node *p = calloc( 1, sizeof *p );\n  if ( p )\n  {\n    p->v = val;\n    p->next = *l;   \/\/ set p to point to the current head of the list\n    *l = p;         \/\/ make p the new head of the list\n  }\n  return p != NULL;  \/\/ will return false if the calloc (and by extenion,\n}                    \/\/ the push operation) fails.  \n\nbool pop( Node **l, int *v )\n{\n  Node *p = *l;      \/\/ p points to head of list\n  if ( !p )\n    return false;\n\n  *v = p->val;     \/\/ get value in current node\n  *l = p->next;    \/\/ make the next element the new list head\n  p->next = NULL;  \/\/ sever the old list head\n  free( p );       \/\/ and deallocate it\n\n  return true;\n}\n<\/code><\/pre>\n
No list traversals, no need to keep track of current and previous nodes.  All you care about is the head node.  The statement p->next = NULL;<\/code> isn’t strictly necessary since we immediately free p<\/code> afterwards.  I like it because it makes it obvious that we have removed<\/em> p<\/code> from the list, but if you don’t want to spare the cycles, you can omit it.  <\/p>\n
Edit<\/strong><\/p>\n
I was right to be nervous about that code. <\/p>\n
So here’s basically what’s happening – when you have exactly one item in the stack, you free the head of the list, but you don’t update the value of the list pointer<\/strong> (*stack<\/code> in the original code, *l<\/code> in the latest edit).  The value of the stack<\/code> variable in main<\/code> is unchanged, and now it’s invalid<\/em> – the memory at that address is no longer allocated.  So the next time you call push<\/code>, it sees that *l<\/code> is not NULL<\/code>, and attempts to traverse down the (non-existent) list.  <\/p>\n
At this point the behavior is undefined; literally anything<\/em> can happen.  On my system after the first push<\/code>, stack<\/code> has the value 0x501010<\/code>.  I do a pop<\/code>, which free<\/code>s that memory, but doesn’t change the value of stack<\/code>.  On the next push<\/code>, *l<\/code> is not NULL<\/code>, so I set (*l)->next<\/code> to whatever malloc<\/code> returns, which in my case is…0x501010<\/code> again.  So *l<\/code> is 0x501010<\/code>, and (*l)->next<\/code> is 0x501010<\/code>.  If I try to push another item, I wind up in an infinite loop (p<\/code> == p->next<\/code>).  <\/p>\n
To fix this, you need to set the list pointer to NULL<\/code> after you free<\/code> it:<\/p>\n
Node *aux,*prev;\nprev = *l;\naux = prev->next;\nif(aux == NULL)\n{\n    free(prev);\n    *l = NULL;\n    return;\n}\n<\/code><\/pre>\n<\/p><\/div>\n
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            5<\/span> <\/p><\/div>\n<\/div>\n
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solved Pop() function for stack in C <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] It would help to see how you push items onto the stack. If you’re really calling pop without a push first, well, then it’s not supposed to do anything, is it? This bit makes me nervous: Node *aux,*prev; prev = *stack; aux = prev->next; if(aux == NULL) { free(prev); return; } You set prev … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[324,712,376],"class_list":["post-24982","post","type-post","status-publish","format-standard","hentry","category-solved","tag-c","tag-pointers","tag-stack"],"yoast_head":"\n[Solved] Pop() function for stack in C - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-pop-function-for-stack-in-c\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Pop() function for stack in C - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] It would help to see how you push items onto the stack. If you’re really calling pop without a push first, well, then it’s not supposed to do anything, is it? This bit makes me nervous: Node *aux,*prev; prev = *stack; aux = prev->next; if(aux == NULL) { free(prev); return; } You set prev ... 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If you’re really calling pop without a push first, well, then it’s not supposed to do anything, is it? This bit makes me nervous: Node *aux,*prev; prev = *stack; aux = prev->next; if(aux == NULL) { free(prev); return; } You set prev ... 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