free

问题 I've a question about free() behavior. Juste a simple code which use malloc and strcpy a char*. So, all is set on the HEAP : (gdb) x/100b 0x602010 0x602010: 66 111 110 106 111 117 114 32 0x602018: 116 111 117 116 32 108 101 32 0x602020: 109 111 110 100 101 0 0 0 0x602028: 0 0 0 0 0 0 0 0 0x602030: 0 0 0 0 0 0 0 0 0x602038: 33 0 0 0 0 0 0 0 When I free the chunk with free(), the result is : (gdb) x/100b 0x602010 0x602010: 0 0 0 0 0 0 0 0 0x602018: 116 111 117 116 32 108 101 32 0x602020: 109

问题 This question already has answers here : Why freed struct in C still has data? (7 answers) Closed 5 years ago . So, I have a struct: tyepdef struct node { struct node *next; double value; } NodeT, *NodeTP; and I have three functions: int deleteNode(NodeTP p) { free(p); p = NULL; return 1; } int deleteOne(NodeTP list) { if (list->next != NULL) deleteOne(list->next); else deleteNode(list); return 1; } int print(NodeT list) { printf("%lf\n", list.value); if (list.next != NULL) print(*list.next);

问题 Basically, I have a block of memory allocated using malloc that I want to start placing objects into using placement new. I know that the destructors for these objects will have to be explicitly called when I'm deleting them, but I want to make sure that I understand this completely, and that I'm going about it the right way. I'm almost certainly going to scrap this approach and go about things in a more straightforward manner, but I figured that I'd ask just for understanding's sake. I have

问题 I have an array of structs. I am trying to delete a list of elements from that array and shift other elements to the left. After shifting the elements I am trying to delete/free the memory at the end of the array which we don't require anymore. I have the following code: #include <iostream> #include<stdio.h> #include<stdlib.h> void removeelement(int*); void displayelements(); typedef struct { int n; }element; element** array; int numofelements=5; int main() { array = (element**)malloc(5

问题 if I have something like struct Node *root; struct Node *q; root = malloc( sizeof(struct Node)); q = root; free(q); is the node q is pointing to freed?? or would I have to pass root to the free function? 回答1: Both root and q have the same value, that is to say, they point to the same memory location. After the call to free(q) , both of them are dangling pointers because the thing they both point to has been freed but they still point to that location. To de-reference either would invoke

问题 How can I free a const char* ? I allocated new memory using malloc , and when I'm trying to free it I always receive the error "incompatible pointer type" The code that causes this is something like: char* name="Arnold"; const char* str=(const char*)malloc(strlen(name)+1); free(str); // error here 回答1: Several people have posted the right answer, but they keep deleting it for some reason. You need to cast it to a non-const pointer; free takes a void* , not a const void* : free((char*)str);

问题 I am not sure why I am getting the below error. Strangely enough, this error doesn't occur when I use Mac OS X, but it does when I use my Linux (Debian) partition. ----------- Empty Queue: 0 0 0 0 0 0 0 0 0 0 --------------- Populated Queue: 5 3 1 7 6 3 2 1 4 4 ------------- After Dequeue: 3 1 7 6 3 2 1 4 4 0 Datum: 5 *** glibc detected *** ./queue_demo: free(): invalid next size (fast): 0x0000000000c73010 *** ======= Backtrace: ========= /lib/x86_64-linux-gnu/libc.so.6(+0x75b76)

问题 I have a simple program which reads a bunch of ini file settings in memory allocated dynamically (malloc), then does stuff in loops for a long time, then ends. When I run valgrind I see that the memory I malloc'ed for my ini strings is not freed. On the one hand, I think that it shouldn't matter since the program is shutting down (and no memory is leaked in the loops). On the other hand, I like when valgrind gives me a big pat on the back for cleaning up my own mess. Aside from the pat on the

问题 I recoded malloc() free() and realloc() . I have a linked list with the pointers returned by malloc() . The question is : what does free() really do ? Currently, I did a memset() to have the same behavior of free() . But was it better just to set a flag in my list as 'is free' rather than doing a memset() in order to make it faster ? 回答1: Usually free(3) does not do anything to the memory itself. (If security or privacy is a concern, you should clear memory before freeing.) If you want to

问题 I'm trying to set up an array of strings (in C, using Linux). The array will hold 11 strings (static length). I initially had the array set up as: char Answers[10][100]; but in my code I have a portion that calls fgets(input,sizeof(input),stdin). When this fgets() portion is called, the final element of my Answers array was being overwritten with the value of input (something about Answers' location on the stack?). So now I'm trying to "lock-in" the memory I use for my Answers array. Would I