Conversion of objects attributes to and from pointers as function argument
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My function in c contains objects node, with two attributes Key* a pointer to a key object and int data. The signature of a method is key_comp(key, key), which requires two keys, however the node object contains a pointer to a key. I have tried using node->key and (Node).key however neither work as arguments for this method, how would you go about converting the key pointer to a key for use in a method?
typedef struct
{
Node tree_nodes;
unsigned char *is_free;
int size;
} BStree_struct;
typedef BStree_struct BStree;
void insert_helper(int i, BStree bst, Node node){
if (i>= (*bst).size){
printf("Out of the range of the tree. n");
return;
}
if ((*bst).is_free[i]==1){
(*bst).tree_nodes[i] = node;
(*bst).is_free[i] = 0;
}else if (key_comp(bst->tree_nodes[i].key, node.key) >0){
insert_helper(2*i,bst,node);
}else if (key_comp(bst->tree_nodes[i].key, node.key) <0){
insert_helper(2*i+1, bst, node);
}
}
typedef struct {char *name; int id;} Key;
typedef struct {Key *key; int data;} Node;
Key *key_construct(char *in_name, int in_id);
int key_comp(Key key1, Key key2);
void print_key(Key *key);
void print_node(Node node);
typedef struct {Node *tree_nodes; unsigned char *is_free;
int size;} BStree_struct;
typedef BStree_struct* BStree;
BStree bstree_ini(int size);
void bstree_insert(BStree bst, Key *key, int data);
void bstree_traversal(BStree bst);
void bstree_free(BStree bst)
c arrays pointers methods
asked Nov 12 at 4:33
Ben French
84
|
show 1 more comment
up vote
0
down vote
favorite
My function in c contains objects node, with two attributes Key* a pointer to a key object and int data. The signature of a method is key_comp(key, key), which requires two keys, however the node object contains a pointer to a key. I have tried using node->key and (Node).key however neither work as arguments for this method, how would you go about converting the key pointer to a key for use in a method?
typedef struct
{
Node tree_nodes;
unsigned char *is_free;
int size;
} BStree_struct;
typedef BStree_struct BStree;
void insert_helper(int i, BStree bst, Node node){
if (i>= (*bst).size){
printf("Out of the range of the tree. n");
return;
}
if ((*bst).is_free[i]==1){
(*bst).tree_nodes[i] = node;
(*bst).is_free[i] = 0;
}else if (key_comp(bst->tree_nodes[i].key, node.key) >0){
insert_helper(2*i,bst,node);
}else if (key_comp(bst->tree_nodes[i].key, node.key) <0){
insert_helper(2*i+1, bst, node);
}
}
typedef struct {char *name; int id;} Key;
typedef struct {Key *key; int data;} Node;
Key *key_construct(char *in_name, int in_id);
int key_comp(Key key1, Key key2);
void print_key(Key *key);
void print_node(Node node);
typedef struct {Node *tree_nodes; unsigned char *is_free;
int size;} BStree_struct;
typedef BStree_struct* BStree;
BStree bstree_ini(int size);
void bstree_insert(BStree bst, Key *key, int data);
void bstree_traversal(BStree bst);
void bstree_free(BStree bst)
c arrays pointers methods
asked Nov 12 at 4:33
Ben French
84
1
Please add your code to the question, it sounds like you can just use*(node->key), but it's hard to say without seeing the code.
– Kingsley
Nov 12 at 4:35
added the code here is the method, the node contains object of a pointer to a key, however I keep getting errors when compiling
– Ben French
Nov 12 at 4:38
What areBStreeandNode? Pointers? Please update your question with this information.
– Kingsley
Nov 12 at 4:39
I tried the code you suggested and recieved error, invalid argument type of '->' I added the header files from the two files to the question for supplementary information, I will also add the key_comp method
– Ben French
Nov 12 at 4:43
This data organisation makes very little sense. Are these types and signatures given to you?
– n.m.
Nov 12 at 5:13
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My function in c contains objects node, with two attributes Key* a pointer to a key object and int data. The signature of a method is key_comp(key, key), which requires two keys, however the node object contains a pointer to a key. I have tried using node->key and (Node).key however neither work as arguments for this method, how would you go about converting the key pointer to a key for use in a method?
typedef struct
{
Node tree_nodes;
unsigned char *is_free;
int size;
} BStree_struct;
typedef BStree_struct BStree;
void insert_helper(int i, BStree bst, Node node){
if (i>= (*bst).size){
printf("Out of the range of the tree. n");
return;
}
if ((*bst).is_free[i]==1){
(*bst).tree_nodes[i] = node;
(*bst).is_free[i] = 0;
}else if (key_comp(bst->tree_nodes[i].key, node.key) >0){
insert_helper(2*i,bst,node);
}else if (key_comp(bst->tree_nodes[i].key, node.key) <0){
insert_helper(2*i+1, bst, node);
}
}
typedef struct {char *name; int id;} Key;
typedef struct {Key *key; int data;} Node;
Key *key_construct(char *in_name, int in_id);
int key_comp(Key key1, Key key2);
void print_key(Key *key);
void print_node(Node node);
typedef struct {Node *tree_nodes; unsigned char *is_free;
int size;} BStree_struct;
typedef BStree_struct* BStree;
BStree bstree_ini(int size);
void bstree_insert(BStree bst, Key *key, int data);
void bstree_traversal(BStree bst);
void bstree_free(BStree bst)
c arrays pointers methods
asked Nov 12 at 4:33
Ben French
84
My function in c contains objects node, with two attributes Key* a pointer to a key object and int data. The signature of a method is key_comp(key, key), which requires two keys, however the node object contains a pointer to a key. I have tried using node->key and (Node).key however neither work as arguments for this method, how would you go about converting the key pointer to a key for use in a method?
typedef struct
{
Node tree_nodes;
unsigned char *is_free;
int size;
} BStree_struct;
typedef BStree_struct BStree;
void insert_helper(int i, BStree bst, Node node){
if (i>= (*bst).size){
printf("Out of the range of the tree. n");
return;
}
if ((*bst).is_free[i]==1){
(*bst).tree_nodes[i] = node;
(*bst).is_free[i] = 0;
}else if (key_comp(bst->tree_nodes[i].key, node.key) >0){
insert_helper(2*i,bst,node);
}else if (key_comp(bst->tree_nodes[i].key, node.key) <0){
insert_helper(2*i+1, bst, node);
}
}
typedef struct {char *name; int id;} Key;
typedef struct {Key *key; int data;} Node;
Key *key_construct(char *in_name, int in_id);
int key_comp(Key key1, Key key2);
void print_key(Key *key);
void print_node(Node node);
typedef struct {Node *tree_nodes; unsigned char *is_free;
int size;} BStree_struct;
typedef BStree_struct* BStree;
BStree bstree_ini(int size);
void bstree_insert(BStree bst, Key *key, int data);
void bstree_traversal(BStree bst);
void bstree_free(BStree bst)
c arrays pointers methods
c arrays pointers methods
asked Nov 12 at 4:33
Ben French
84
asked Nov 12 at 4:33
Ben French
84
edited Nov 12 at 4:45
asked Nov 12 at 4:33
Ben French
84
asked Nov 12 at 4:33
Ben French
84
asked Nov 12 at 4:33
Ben French
84
84
1
Please add your code to the question, it sounds like you can just use*(node->key), but it's hard to say without seeing the code.
– Kingsley
Nov 12 at 4:35
added the code here is the method, the node contains object of a pointer to a key, however I keep getting errors when compiling
– Ben French
Nov 12 at 4:38
What areBStreeandNode? Pointers? Please update your question with this information.
– Kingsley
Nov 12 at 4:39
I tried the code you suggested and recieved error, invalid argument type of '->' I added the header files from the two files to the question for supplementary information, I will also add the key_comp method
– Ben French
Nov 12 at 4:43
This data organisation makes very little sense. Are these types and signatures given to you?
– n.m.
Nov 12 at 5:13
|
show 1 more comment
1
Please add your code to the question, it sounds like you can just use*(node->key), but it's hard to say without seeing the code.
– Kingsley
Nov 12 at 4:35
added the code here is the method, the node contains object of a pointer to a key, however I keep getting errors when compiling
– Ben French
Nov 12 at 4:38
What areBStreeandNode? Pointers? Please update your question with this information.
– Kingsley
Nov 12 at 4:39
I tried the code you suggested and recieved error, invalid argument type of '->' I added the header files from the two files to the question for supplementary information, I will also add the key_comp method
– Ben French
Nov 12 at 4:43
This data organisation makes very little sense. Are these types and signatures given to you?
– n.m.
Nov 12 at 5:13
1
1
Please add your code to the question, it sounds like you can just use
*(node->key), but it's hard to say without seeing the code.– Kingsley
Nov 12 at 4:35
Please add your code to the question, it sounds like you can just use
*(node->key), but it's hard to say without seeing the code.– Kingsley
Nov 12 at 4:35
added the code here is the method, the node contains object of a pointer to a key, however I keep getting errors when compiling
– Ben French
Nov 12 at 4:38
added the code here is the method, the node contains object of a pointer to a key, however I keep getting errors when compiling
– Ben French
Nov 12 at 4:38
What are
BStree and Node? Pointers? Please update your question with this information.– Kingsley
Nov 12 at 4:39
What are
BStree and Node? Pointers? Please update your question with this information.– Kingsley
Nov 12 at 4:39
I tried the code you suggested and recieved error, invalid argument type of '->' I added the header files from the two files to the question for supplementary information, I will also add the key_comp method
– Ben French
Nov 12 at 4:43
I tried the code you suggested and recieved error, invalid argument type of '->' I added the header files from the two files to the question for supplementary information, I will also add the key_comp method
– Ben French
Nov 12 at 4:43
This data organisation makes very little sense. Are these types and signatures given to you?
– n.m.
Nov 12 at 5:13
This data organisation makes very little sense. Are these types and signatures given to you?
– n.m.
Nov 12 at 5:13
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
This is the best I can come up with. It's not clear to me exactly what the code is doing, but having a mix of pointers and object copies is throwing it off a bit.
The key thing to consider is that Node.key is a pointer (to a Key). So to pass these to key_comp(Key a, Key b) you need to de-reference it, thereby converting a Key* to a Key.
void insert_helper(int i, BStree bst, Node node)
{
if (i >= bst.size)
{
printf("Out of the range of the tree. n");
return;
}
if (bst.is_free[i] == 1)
{
bst.tree_nodes[i] = node;
bst.is_free[i] = 0;
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) > 0)
{
insert_helper(2*i, bst, node);
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) < 0)
{
insert_helper(2*i+1, bst, node);
}
}
If it were my code, I'd change key_comp() to take a pair of pointers to Key.
answered Nov 12 at 5:07
Kingsley
2,07811022
great works perfectly now !
– Ben French
Nov 12 at 5:13
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
up vote
0
down vote
accepted
This is the best I can come up with. It's not clear to me exactly what the code is doing, but having a mix of pointers and object copies is throwing it off a bit.
The key thing to consider is that Node.key is a pointer (to a Key). So to pass these to key_comp(Key a, Key b) you need to de-reference it, thereby converting a Key* to a Key.
void insert_helper(int i, BStree bst, Node node)
{
if (i >= bst.size)
{
printf("Out of the range of the tree. n");
return;
}
if (bst.is_free[i] == 1)
{
bst.tree_nodes[i] = node;
bst.is_free[i] = 0;
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) > 0)
{
insert_helper(2*i, bst, node);
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) < 0)
{
insert_helper(2*i+1, bst, node);
}
}
If it were my code, I'd change key_comp() to take a pair of pointers to Key.
answered Nov 12 at 5:07
Kingsley
2,07811022
great works perfectly now !
– Ben French
Nov 12 at 5:13
add a comment |
up vote
0
down vote
accepted
This is the best I can come up with. It's not clear to me exactly what the code is doing, but having a mix of pointers and object copies is throwing it off a bit.
The key thing to consider is that Node.key is a pointer (to a Key). So to pass these to key_comp(Key a, Key b) you need to de-reference it, thereby converting a Key* to a Key.
void insert_helper(int i, BStree bst, Node node)
{
if (i >= bst.size)
{
printf("Out of the range of the tree. n");
return;
}
if (bst.is_free[i] == 1)
{
bst.tree_nodes[i] = node;
bst.is_free[i] = 0;
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) > 0)
{
insert_helper(2*i, bst, node);
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) < 0)
{
insert_helper(2*i+1, bst, node);
}
}
If it were my code, I'd change key_comp() to take a pair of pointers to Key.
answered Nov 12 at 5:07
Kingsley
2,07811022
great works perfectly now !
– Ben French
Nov 12 at 5:13
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This is the best I can come up with. It's not clear to me exactly what the code is doing, but having a mix of pointers and object copies is throwing it off a bit.
The key thing to consider is that Node.key is a pointer (to a Key). So to pass these to key_comp(Key a, Key b) you need to de-reference it, thereby converting a Key* to a Key.
void insert_helper(int i, BStree bst, Node node)
{
if (i >= bst.size)
{
printf("Out of the range of the tree. n");
return;
}
if (bst.is_free[i] == 1)
{
bst.tree_nodes[i] = node;
bst.is_free[i] = 0;
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) > 0)
{
insert_helper(2*i, bst, node);
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) < 0)
{
insert_helper(2*i+1, bst, node);
}
}
If it were my code, I'd change key_comp() to take a pair of pointers to Key.
answered Nov 12 at 5:07
Kingsley
2,07811022
This is the best I can come up with. It's not clear to me exactly what the code is doing, but having a mix of pointers and object copies is throwing it off a bit.
The key thing to consider is that Node.key is a pointer (to a Key). So to pass these to key_comp(Key a, Key b) you need to de-reference it, thereby converting a Key* to a Key.
void insert_helper(int i, BStree bst, Node node)
{
if (i >= bst.size)
{
printf("Out of the range of the tree. n");
return;
}
if (bst.is_free[i] == 1)
{
bst.tree_nodes[i] = node;
bst.is_free[i] = 0;
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) > 0)
{
insert_helper(2*i, bst, node);
}
else if (key_comp(*(bst.tree_nodes[i].key), *(node.key)) < 0)
{
insert_helper(2*i+1, bst, node);
}
}
If it were my code, I'd change key_comp() to take a pair of pointers to Key.
answered Nov 12 at 5:07
Kingsley
2,07811022
edited Nov 12 at 5:12
answered Nov 12 at 5:07
Kingsley
2,07811022
answered Nov 12 at 5:07
Kingsley
2,07811022
answered Nov 12 at 5:07
Kingsley
2,07811022
2,07811022
great works perfectly now !
– Ben French
Nov 12 at 5:13
add a comment |
great works perfectly now !
– Ben French
Nov 12 at 5:13
great works perfectly now !
– Ben French
Nov 12 at 5:13
great works perfectly now !
– Ben French
Nov 12 at 5:13
add a comment |
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1
Please add your code to the question, it sounds like you can just use
*(node->key), but it's hard to say without seeing the code.– Kingsley
Nov 12 at 4:35
added the code here is the method, the node contains object of a pointer to a key, however I keep getting errors when compiling
– Ben French
Nov 12 at 4:38
What are
BStreeandNode? Pointers? Please update your question with this information.– Kingsley
Nov 12 at 4:39
I tried the code you suggested and recieved error, invalid argument type of '->' I added the header files from the two files to the question for supplementary information, I will also add the key_comp method
– Ben French
Nov 12 at 4:43
This data organisation makes very little sense. Are these types and signatures given to you?
– n.m.
Nov 12 at 5:13