How to write to a void pointer without knowing the Data Type?
I am trying to write the code for a generic Stack using Singly Linked List in C. I am trying to use (void *) as the data type in each of its functions like:
node* getNode(void *, size_t);
void append(node **, size_t); etc.
The structure for each node is:
typedef struct linked_list{
void *data;
struct linked_list *next; }node;
In the getNode function mentioned above I am trying to do this:
node* getNode(void *data, size_t data_size) {
node *newNode;
newNode = (node *)malloc(sizeof(node));
newNode->next = NULL;
// this line should assign the void *data to the data part of the node
return newNode;}
I could not find how to assign the data part without knowing the data type, but only using the data_size variable. I found out about the memcopy function, but even that requires the data type.
Please help.
c adt dynamic-memory-allocation void-pointers
asked Nov 13 '18 at 2:59
The ViperThe Viper
7810
add a comment |
I am trying to write the code for a generic Stack using Singly Linked List in C. I am trying to use (void *) as the data type in each of its functions like:
node* getNode(void *, size_t);
void append(node **, size_t); etc.
The structure for each node is:
typedef struct linked_list{
void *data;
struct linked_list *next; }node;
In the getNode function mentioned above I am trying to do this:
node* getNode(void *data, size_t data_size) {
node *newNode;
newNode = (node *)malloc(sizeof(node));
newNode->next = NULL;
// this line should assign the void *data to the data part of the node
return newNode;}
I could not find how to assign the data part without knowing the data type, but only using the data_size variable. I found out about the memcopy function, but even that requires the data type.
Please help.
c adt dynamic-memory-allocation void-pointers
asked Nov 13 '18 at 2:59
The ViperThe Viper
7810
add a comment |
I am trying to write the code for a generic Stack using Singly Linked List in C. I am trying to use (void *) as the data type in each of its functions like:
node* getNode(void *, size_t);
void append(node **, size_t); etc.
The structure for each node is:
typedef struct linked_list{
void *data;
struct linked_list *next; }node;
In the getNode function mentioned above I am trying to do this:
node* getNode(void *data, size_t data_size) {
node *newNode;
newNode = (node *)malloc(sizeof(node));
newNode->next = NULL;
// this line should assign the void *data to the data part of the node
return newNode;}
I could not find how to assign the data part without knowing the data type, but only using the data_size variable. I found out about the memcopy function, but even that requires the data type.
Please help.
c adt dynamic-memory-allocation void-pointers
asked Nov 13 '18 at 2:59
The ViperThe Viper
7810
I am trying to write the code for a generic Stack using Singly Linked List in C. I am trying to use (void *) as the data type in each of its functions like:
node* getNode(void *, size_t);
void append(node **, size_t); etc.
The structure for each node is:
typedef struct linked_list{
void *data;
struct linked_list *next; }node;
In the getNode function mentioned above I am trying to do this:
node* getNode(void *data, size_t data_size) {
node *newNode;
newNode = (node *)malloc(sizeof(node));
newNode->next = NULL;
// this line should assign the void *data to the data part of the node
return newNode;}
I could not find how to assign the data part without knowing the data type, but only using the data_size variable. I found out about the memcopy function, but even that requires the data type.
Please help.
c adt dynamic-memory-allocation void-pointers
c adt dynamic-memory-allocation void-pointers
asked Nov 13 '18 at 2:59
The ViperThe Viper
7810
asked Nov 13 '18 at 2:59
The ViperThe Viper
7810
asked Nov 13 '18 at 2:59
The ViperThe Viper
7810
asked Nov 13 '18 at 2:59
The ViperThe Viper
7810
asked Nov 13 '18 at 2:59
The ViperThe Viper
7810
7810
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
If you only want to save a pointer to the data, you can just copy the pointer:
newNode->data = data;
If you want to copy what it points to, you need malloc to get the memory and memcpy to copy the bytes:
newNode->data = malloc(data_size);
memcpy(newNode->data, data, data_size);
The signature of memcpy is:
void *memcpy(void *dest, const void *src, size_t n);
It's meant to copy generic memory, so you can pass it a void * for both the source and destination.
answered Nov 13 '18 at 3:02
dbushdbush
93.9k12101134
Doesn'tmemcpyrequire data type?
– The Viper
Nov 13 '18 at 3:46
@TheViper No, just a pointer.
– dbush
Nov 13 '18 at 3:47
okay, thank you @dbush
– The Viper
Nov 13 '18 at 3:48
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you only want to save a pointer to the data, you can just copy the pointer:
newNode->data = data;
If you want to copy what it points to, you need malloc to get the memory and memcpy to copy the bytes:
newNode->data = malloc(data_size);
memcpy(newNode->data, data, data_size);
The signature of memcpy is:
void *memcpy(void *dest, const void *src, size_t n);
It's meant to copy generic memory, so you can pass it a void * for both the source and destination.
answered Nov 13 '18 at 3:02
dbushdbush
93.9k12101134
Doesn'tmemcpyrequire data type?
– The Viper
Nov 13 '18 at 3:46
@TheViper No, just a pointer.
– dbush
Nov 13 '18 at 3:47
okay, thank you @dbush
– The Viper
Nov 13 '18 at 3:48
add a comment |
If you only want to save a pointer to the data, you can just copy the pointer:
newNode->data = data;
If you want to copy what it points to, you need malloc to get the memory and memcpy to copy the bytes:
newNode->data = malloc(data_size);
memcpy(newNode->data, data, data_size);
The signature of memcpy is:
void *memcpy(void *dest, const void *src, size_t n);
It's meant to copy generic memory, so you can pass it a void * for both the source and destination.
answered Nov 13 '18 at 3:02
dbushdbush
93.9k12101134
Doesn'tmemcpyrequire data type?
– The Viper
Nov 13 '18 at 3:46
@TheViper No, just a pointer.
– dbush
Nov 13 '18 at 3:47
okay, thank you @dbush
– The Viper
Nov 13 '18 at 3:48
add a comment |
If you only want to save a pointer to the data, you can just copy the pointer:
newNode->data = data;
If you want to copy what it points to, you need malloc to get the memory and memcpy to copy the bytes:
newNode->data = malloc(data_size);
memcpy(newNode->data, data, data_size);
The signature of memcpy is:
void *memcpy(void *dest, const void *src, size_t n);
It's meant to copy generic memory, so you can pass it a void * for both the source and destination.
answered Nov 13 '18 at 3:02
dbushdbush
93.9k12101134
If you only want to save a pointer to the data, you can just copy the pointer:
newNode->data = data;
If you want to copy what it points to, you need malloc to get the memory and memcpy to copy the bytes:
newNode->data = malloc(data_size);
memcpy(newNode->data, data, data_size);
The signature of memcpy is:
void *memcpy(void *dest, const void *src, size_t n);
It's meant to copy generic memory, so you can pass it a void * for both the source and destination.
answered Nov 13 '18 at 3:02
dbushdbush
93.9k12101134
answered Nov 13 '18 at 3:02
dbushdbush
93.9k12101134
answered Nov 13 '18 at 3:02
dbushdbush
93.9k12101134
answered Nov 13 '18 at 3:02
dbushdbush
93.9k12101134
93.9k12101134
Doesn'tmemcpyrequire data type?
– The Viper
Nov 13 '18 at 3:46
@TheViper No, just a pointer.
– dbush
Nov 13 '18 at 3:47
okay, thank you @dbush
– The Viper
Nov 13 '18 at 3:48
add a comment |
Doesn'tmemcpyrequire data type?
– The Viper
Nov 13 '18 at 3:46
@TheViper No, just a pointer.
– dbush
Nov 13 '18 at 3:47
okay, thank you @dbush
– The Viper
Nov 13 '18 at 3:48
Doesn't
memcpy require data type?– The Viper
Nov 13 '18 at 3:46
Doesn't
memcpy require data type?– The Viper
Nov 13 '18 at 3:46
@TheViper No, just a pointer.
– dbush
Nov 13 '18 at 3:47
@TheViper No, just a pointer.
– dbush
Nov 13 '18 at 3:47
okay, thank you @dbush
– The Viper
Nov 13 '18 at 3:48
okay, thank you @dbush
– The Viper
Nov 13 '18 at 3:48
add a comment |
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