free

问题 Until today I lived in belief that calling free() on memory space releases it for further allocation without any other modifications. Especially, considering this SO question that clearly states that free() DOESN'T zero out memory. Yet, let's consider this piece of code (test.c): #include<stdlib.h> #include<stdio.h> int main() { int* pointer; if (NULL == (pointer = malloc(sizeof(*pointer)))) return EXIT_FAILURE; *pointer = 1337; printf("Before free(): %p, %d\n", pointer, *pointer); free

问题 If I allocated memory in my C program using malloc and now I want to exit, do I have to free the allocated memory, or can I assume that since my entire program terminates, it will be freed by the OS? I run in Linux environment. 回答1: Any modern operating system will clean up everything after a process terminates, but it's generally not a good practice to rely on this. It depends on the program you are writing. If it's just a command line tool that runs and terminates quickly, you may not

问题 This question already has answers here : Facing an error “*** glibc detected *** free(): invalid next size (fast)” (2 answers) Closed 5 years ago . I am freeing a char* after a concatenation process. But I receive this error: free(): invalid next size (fast): 0x0000000001b86170 Below is my code: void concat(stringList *list) { char *res = (char*)malloc(sizeof(char*)); strcpy(res, list->head->string); list->tmp = list->head->next; while (list->tmp != NULL) { strcat(res, ","); strcat(res, list-

问题 Theoretically I can say that free(ptr); free(ptr); is a memory corruption since we are freeing the memory which has already been freed. But what if free(ptr); ptr=NULL; free(ptr); As the OS will behave in an undefined manner I cannot get an actual theoretical analysis for this about what's happening. Whatever I am doing, is this memory corruption or not? Is freeing a NULL pointer valid? 回答1: 7.20.3.2 The free function Synopsis #include <stdlib.h> void free(void *ptr); Description The free

问题 I have been trying to free the allocated memory of a file loaded into a linked list, I have managed to free the nodes, but I can't figure out how to free the allocated memory of the file's values copies. I have tried something like that: void FreeString(struct node * newNode) { for (int i = 0; i < 5; i++) { free(newNode->string); } } but the compiler would crash with a segmentation fault, and valgrind would still point out to memory leaks. it would be appreciated if anyone can tell me what am

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 7 years ago . void str_cpy(char **des,char* src){ if(*des || *des==src) free(*des); *des = (char*)malloc(sizeof(char)*(strlen(src)+1)); if(!*des) return ; memcpy ( *des, src, (strlen(src)+1)*sizeof(char) ); } Why the free in

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed last year . I get the following error when running a c program: *** Error in `./a.out': double free or corruption (!prev): 0x0000000000bb0470 *** I believe this is due to fclose() being called in the program, It is a Lexical Analyzer for compilers in c language and it uses file pointers . Here is code: #include<stdio.h>

问题 So I have created and allocated memory for a double pointer using the following function: void mallocDoubleArr(double ***arr, int size) { printf("Here: %d", size); int i, j; *arr = malloc(size * sizeof(double*)); for(i = 0; i < size; i++) { (*arr)[i]= malloc(size*sizeof(double)); for (j = 0; j < size; j++) { (*arr)[i][j] = 0; } } } And I called the function using: double **G; //Create double pointer to hold 2d matrix mallocDoubleArr(&G, numNodes); Now my question is how would I write a

问题 gcc (GCC) 4.7.0 c89 Hello, I am wondering if I am thinking correctly here. When I allocate memory using malloc. malloc will return a pointer to a size in memory. So before I allocate my memory all pointers will have the value NULL. Using this code snippet: struct address *db_row = NULL; db_row = malloc(sizeof(struct address)); db_row->name = malloc(sizeof(char) * 10); db_row->email = malloc(sizeof(char) *10); free(db_row->name); free(db_row->email); free(db_row); I have done this in the gdb

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 7 years ago . void str_cpy(char **des,char* src){ if(*des || *des==src) free(*des); *des = (char*)malloc(sizeof(char)*(strlen(src)+1)); if(!*des) return ; memcpy ( *des, src, (strlen(src)+1)*sizeof(char) ); } Why the free in