malloc tries to allocate a contiguous memory range, and this will initially be in real memory simply due to how swap memory works (at least as far as I remember). It could easily be that your OS sometimes can’t find a contiguous block of 10gb of memory and still leave all the processes that require … Read more
Some other answers suggest that you malloc something and return it. This is bad practice in the same sense as in C++, when you new something in a function and the caller is supposed to delete it (who has ownership?) There is a reason that many C APIs have the format of: function(buf, length); Meaning … Read more
A proof it probably cannot be usefully done: char * p1 = malloc(1); free( p1 ); char * p2 = malloc(1); // probably allocates same block as first malloc Now both p1 and p2 point at same memory on heap, but only p2 is valid.
You need to cast away the const to initialize the fields of a malloc’ed structure: struct deneme *mydeneme = malloc(sizeof(struct deneme)); *(int *)&mydeneme->a = 15; *(int *)&mydeneme->b = 20; Alternately, you can create an initialized version of the struct and memcpy it: struct deneme deneme_init = { 15, 20 }; struct deneme *mydeneme = malloc(sizeof(struct … Read more
Well, you didn’t give us a complete implementation. I assume that you meant. int **mat = (int **)malloc(rows * sizeof(int*)); for(int i = 0; i < rows; i++) mat[i] = (int *)malloc(cols * sizeof(int)); Here’s another option: int *mat = (int *)malloc(rows * cols * sizeof(int)); Then, you simulate the matrix using int offset = … Read more
You should assign an array of char pointers, and then, for each pointer assign enough memory for the string: char **orderedIds; orderedIds = malloc(variableNumberOfElements * sizeof(char*)); for (int i = 0; i < variableNumberOfElements; i++) orderedIds[i] = malloc((ID_LEN+1) * sizeof(char)); // yeah, I know sizeof(char) is 1, but to make it clear… Seems like a … Read more
Ok, what we have here is a chunk of really poorly written code. What I will do in this post could best be described as software archaeology. Step 1: fix the formatting. The indention and compact format doesn’t do anyone any good. Various spaces and empty rows need to be inserted. The comments could be … Read more
When you malloc memory, you’re given a pointer to some space, and when you free it, you’re giving it back to the system. Often, you can still access this memory, but using memory after you have freed it is VERY BAD. The exact behavior is undefined, but on most systems you can either continue to … Read more
Quoting Linux manuals: By default, Linux follows an optimistic memory allocation strategy. This means that when malloc() returns non-NULL there is no guarantee that the memory really is available. This is a really bad bug. In case it turns out that the system is out of memory, one or more processes will be killed by … Read more
like this : int (*arr)[M] = malloc(sizeof(int[N][M])); arr is pointer to int[M]. use like arr[0][M-1]; and free(arr);