You want to modify the value of an int* (your array) so need to pass a pointer to it into your increase function: void increase(int** data) { *data = realloc(*data, 5 * sizeof int); } Calling code would then look like: int *data = malloc(4 * sizeof *data); /* do stuff with data */ increase(&data); … 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
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
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
TL;DR Your code provokes Undefined Behavior, as already mentioned in Morlacke’s Answer. Other than that, it seems that you’re having problems on understanding how pointers work. See references for tutorials. First, From your comments When you say that there’s memory allocated for ip in this case: int i = 10; int *ip; ip = &i; … Read more
First of all, 1<<10 is an expression that can be calculated at compile time, and it equals 1024. So your code is equivalent to #include <stdlib.h> int main() { int num=0; while(malloc(1024)) ++num; } So what it does is to allocate chunks of 1024 bytes of memory until it fails to do so. Each time, … Read more
I am not sure what you are asking, But the following program meets your conditions, 1. there is array of pointers 2. there is malloc 3. there is no character array #include <stdio.h> #include <stdlib.h> #define STR_MAX_SIZE 256 int main() { char *str; char *pos[2]; char c; if((str = malloc(STR_MAX_SIZE)) ==NULL) { return -1; } … Read more
You don’t have an array; you have a pointer. The size of the pointer is measured in bytes, not bits. sizeof is evaluated at compile time and is constant for any given expression or type. It does not depend on the number of “filled” elements in an array (or pointer to some space that holds … Read more
Woo, it’s been a long time since I dealt with C pointers, but let me see if I can help. Coordinate_T *p; declares a pointer to a Coordinate_T struct. We don’t have any memory available yet, we just have a pointer to… nothing. p = (Coordinate_T *)malloc(sizeof(Coordinate_T)) actually allocates the memory for us. Now p … 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
In the first code snippet you’re passing the value of the numbers variable. In the function you change a local variable, and doing so has no effect on the variable is the calling function. In the second snippet you correctly pass the address of numbers so it can be set in the function, and the … Read more
head=malloc(sizeof(node*)); Will allocate a space in memory to hold a POINTER to a object of type node. head=malloc(sizeof(node)); Will allocate space for the structure itself, and return to you the address of that memory space, and you will hold that in the head variable For more info, check this other question about Malloc. How do … Read more
“Freeing” means “making available for allocation again”. There is no automatic deleting / overwriting of memory contents because it would negatively impact performance. If you want the area to be set to a value, you have to do it yourself before you call free(). That would be bad practice in release code, though (for anything … Read more