fopen, calloc, and realloc can all fail, returning NULL. In the case of realloc, the original allocation will remain untouched, so immediately overwriting the original pointer with the newly returned value will leak memory and throw away potentially useful data. Reallocating every iteration is rather costly. A better strategy is to reallocate when the buffer … Read more
Your code exhibits undefined behaviour because you passed a non-static, uninitialised pointer to realloc whose contents were indeterminate. From C11, Memory Management Functions: The realloc function deallocates the old object pointed to by ptr and returns a pointer to a new object that has the size specified by size. ….if ptr does not match a … Read more
Introduction If you’ve been struggling with a document for two days and can’t seem to get it to work, you’re not alone. Many people have difficulty understanding the complexities of document formatting and troubleshooting. Fortunately, there are many resources available to help you solve your problem. In this post, we’ll discuss some tips and tricks … Read more
When processing regular characters (final else clause) you read additional input instead of operating over text, i.e. instead of: scanf(“%c”, &document[parano][sentno][wordno][charno]); printf(“%c\n”, document[parano][sentno][wordno][charno]); charno++; you want to do: document[parano][sentno][wordno][charno++] = text[i]; By the time we hit text[i] == ‘ ‘ for the first time the value of ***document is “3\n1 2\n2\n” but you wanted it … Read more
Invalid assumption: void *my_realloc(void *ptr, size_t size) { unsigned char *old_ptr = (unsigned char *)ptr; You’re breaking the realloc() specification here. realloc() doesn’t assume that the ptr would always be pointing to a string. It is type-agnostic. And the cast is redundant. There’s an implicit conversion from void * to any other pointer type. Undefined … Read more
OP’s mostly has it. It is just printing the first element each time. // printf(“%d”, *arr); printf(“%d”, arr[n-1]); is it possible to do it with an array? No. In C an array cannot change size once it is defined. The memory size allocated and referenced by a pointer can change though. solved How to grow … Read more
What you’re doing will work. Each time you increment contents, you also increment counter, so contents – counter gives you the original pointer that you can free. Of course, a better way of doing this would be to use a temporary pointer to increment through the allocated memory so you can use the original to … Read more
This answer is based on the assumption that you do proper handling of your pointers between the part where you allocate memory for 1 element each and where you try to reallocate the memory. This part // realloc to correct size for(int a = 0; a < word_nb; a++) { *(word_list + (word_nb-1)) = (char … Read more
It is always easier to divide program into logical bits and place them in functions. char *addToBuff(char *buff, int ch, size_t *size) { //working on local variable buff. No need for a temporary variable buff = realloc(buff, *size + 1); if(buff) { buff[*size] = ch; *size += 1; } return buff; } char *readToBuff(FILE *fi, … Read more
You should reduce the number of columsn, not rows. Not mat1 but mat1->array should be reallocated. Not nrow and ncol (not updated) but mat1->nrows and mat1->ncols (updated) should be used for the new size. The elements are size_t, so allocating for int mayn’t be enough. Using the variable for calculating size is safe. In the … Read more
At least this statement array->values = realloc(array->values, array->capacity); shall be rewritten like array->values = realloc(array->values, array->capacity * sizeof( Value ) ); Though it would be more safer to use an intermediate pointer like for example Value *tmp = realloc(array->values, array->capacity * sizeof( Value ) ); if ( tmp != NULL ) { array->values = tmp; … Read more
In your function int huffman(node_t *huff, int n) you try to realloc() the huff pointer but in C we pass variables by value so you only receive a copy of the pointer. This means caller retains the original pointer and the realloc() memory is lost when you return from the function. You need to pass … Read more
The function is not updating the caller’s value of client_details, you need one more level of pointer: void func(struct client_detail **client_details,int *size){ *client_details = realloc(*client_details, (*size + 1) * sizeof(**client_details)); *size = *size + 1; } int main() { struct group_detail *group_details = malloc(0 * sizeof(*group_details)); struct client_detail *client_details = malloc(1 * sizeof(*client_details)); int size … Read more
Introduction The malloc: *** error for object 0x147606ac0: pointer being realloc’d was not allocated is a common error encountered when attempting to reallocate memory in C programming. This error occurs when a pointer is passed to the realloc() function, but the pointer was not allocated with malloc() or calloc() beforehand. This error can be caused … Read more
Let’s look at the first two lines of your code: str = (char *) malloc(15); strcpy(str, “63tczfqV4nqB2YnH9iFJbGvGyyDkvK341rrj0G0mo1PEYniOVHejVIFIQnJzHSSMRbuyGMCZ4M5HFMV4y1q4QgYqyxp2XkTjxaolKTkaw1r25S2Emz061tw1”); At this point, you have broken the rules of the C language. strcpy will write past the end of str which causes undefined behavior. Everything that happens after this point is kinda up in the air. 5 solved … Read more