free

I am working on a program where I am using large but limited amount of memory. Memory is allocated and freed on run time on different threads. However, I noticed that the memory usage of the program would not remain within specified bounds. It would increase as time passed on. I wrote the following sample program to check whether memory is being freed back to the OS. Half the allocated memory was freed to check if the memory usage went down. int main() { char *p[COUNT]; for(int i = 0; i < COUNT; i++) { p[i] = new char[1048576]; memset (p[i], 0, 1048576); printf("%p\n", p[i]); } printf("done

I'm trying to free up the memory I've allocated with malloc , but the free command doesn't seem to do its job properly according to Eclipse's debugger. How's this possible? Below is a screenshot of my debugger after it supposedly freed up seCurrent->student->year , which is clearly not the case. year was allocated using malloc . alt text http://img693.imageshack.us/img693/7840/codeo.png free() does not normally change any values in your program - it just makes adjustments to the C runtime heap. This means that the values in the memory that was just freed are retained. However, attempts to

The title maybe confusing. Suppose str is a pointer allocated by malloc . ptr , of type int* , is assigned to it and is freed as shown by the code snippet below: char* str = malloc(64); int* ptr = str; free(ptr); I've tried to compile the above code. It just gives a warning: source_file.c: In function ‘main’: source_file.c:10:16: warning: initialization from incompatible pointer type int* ptr = str; ^ Does the above code invoke Undefined Behavior? Does the above code snippet free the memory allocated by malloc for str ? alk Does the above code invoke Undefined Behavior? It depends. From C11

Will the free() routine work if I dynamically allocate an array and then pass, not the initial pointer, but a pointer to the middle of the array? Example: int* array = malloc(10 * sizeof *array); if(array) { array += 5; // adjusting the indicies free(array); } Or do I need to set the pointer back to the start of the array before calling free() ? Absolutely not. The value passed to free() must be exactly the same value returned by malloc() . In fact, to ensure this is the case, I would recommend you use a copy of the pointer if you need a pointer you can increment or otherwise modify. No (And

What I am trying to do is create an 2-d array of character strings The following seg faults instantly, what is wrong? void add2(char***b, char *i) { if (!i) { b[0][0] = (char*) malloc(120); sprintf(b[0][0], "%s", "hithere"); b[0][1] = (char*) malloc(120); sprintf(b[0][1], "%s", "bithere"); } else { strcat(b[0][0], "\\\\"); strcat(b[0][0], i); strcat(b[0][1], "\\\\"); strcat(b[0][1], i); } } void add1(char ***b) { add2(b,NULL); add2(b,"one"); add2(b,"two"); add2(b,"three"); } int main() { char **keys[2] = {0}; int i,j; add1(keys); for (i = 0; keys[i]; i++) for (j = 0; keys[j]; j++) { fprintf

Normally, if a pointer is freed twice, it's a double free. For example, char *ptr; ptr=malloc(5 * sizeof(*ptr)); free(ptr); free(ptr); The above code is considered as double free. Is the following considered as double free as well? char *ptr; char *ptr1; ptr=malloc(5 * sizeof(*ptr)); ptr1=ptr; free(ptr); free(ptr1); Thank you. Yes. The library doesn't care what name you gave a varaible in your source code (it's long gone by the time the code is executed). All that matters is the value, and in this case the values passed to free() would be the same. Yes. You're freeing the same memory twice.