Mixing
Update cell in struct list and return the list
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0
I'm looking for a way to update a cell in a struct list.
I have a struct list:
struct _list {
int startNode,
targetNode,
substitute;
unsigned char letter;
struct _list *next;
}
And a function that enables me to update a cell.
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
return list;
}
list = list -> next;
}
return NULL;
}
My problem is that this function do not return the whole list. How could I proceed to do so?
c
asked Jan 3 at 10:15
Thomas DI GREGORIOThomas DI GREGORIO
31
add a comment |
0
I'm looking for a way to update a cell in a struct list.
I have a struct list:
struct _list {
int startNode,
targetNode,
substitute;
unsigned char letter;
struct _list *next;
}
And a function that enables me to update a cell.
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
return list;
}
list = list -> next;
}
return NULL;
}
My problem is that this function do not return the whole list. How could I proceed to do so?
c
asked Jan 3 at 10:15
Thomas DI GREGORIOThomas DI GREGORIO
31
insertSubstitutegets a pointer to a struct, so you are modifying the source. Why do you want to return it?
– Jose
Jan 3 at 10:21
Changed your function prototype from 'struct _list*' list to 'struct _list**', you cannot modify a pointer and return it back but you can modify a pointer when you pass a pointer to the pointer, you will need to change internal references to list to '(*list)->, if you do this then you can get rid of the return and change the function return to void.
– SPlatten
Jan 3 at 10:29
add a comment |
0
0
0
I'm looking for a way to update a cell in a struct list.
I have a struct list:
struct _list {
int startNode,
targetNode,
substitute;
unsigned char letter;
struct _list *next;
}
And a function that enables me to update a cell.
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
return list;
}
list = list -> next;
}
return NULL;
}
My problem is that this function do not return the whole list. How could I proceed to do so?
c
asked Jan 3 at 10:15
Thomas DI GREGORIOThomas DI GREGORIO
31
I'm looking for a way to update a cell in a struct list.
I have a struct list:
struct _list {
int startNode,
targetNode,
substitute;
unsigned char letter;
struct _list *next;
}
And a function that enables me to update a cell.
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
return list;
}
list = list -> next;
}
return NULL;
}
My problem is that this function do not return the whole list. How could I proceed to do so?
c
c
asked Jan 3 at 10:15
Thomas DI GREGORIOThomas DI GREGORIO
31
asked Jan 3 at 10:15
Thomas DI GREGORIOThomas DI GREGORIO
31
asked Jan 3 at 10:15
Thomas DI GREGORIOThomas DI GREGORIO
31
asked Jan 3 at 10:15
Thomas DI GREGORIOThomas DI GREGORIO
31
asked Jan 3 at 10:15
Thomas DI GREGORIOThomas DI GREGORIO
31
31
insertSubstitutegets a pointer to a struct, so you are modifying the source. Why do you want to return it?
– Jose
Jan 3 at 10:21
Changed your function prototype from 'struct _list*' list to 'struct _list**', you cannot modify a pointer and return it back but you can modify a pointer when you pass a pointer to the pointer, you will need to change internal references to list to '(*list)->, if you do this then you can get rid of the return and change the function return to void.
– SPlatten
Jan 3 at 10:29
add a comment |
insertSubstitutegets a pointer to a struct, so you are modifying the source. Why do you want to return it?
– Jose
Jan 3 at 10:21
Changed your function prototype from 'struct _list*' list to 'struct _list**', you cannot modify a pointer and return it back but you can modify a pointer when you pass a pointer to the pointer, you will need to change internal references to list to '(*list)->, if you do this then you can get rid of the return and change the function return to void.
– SPlatten
Jan 3 at 10:29
insertSubstitute gets a pointer to a struct, so you are modifying the source. Why do you want to return it?– Jose
Jan 3 at 10:21
insertSubstitute gets a pointer to a struct, so you are modifying the source. Why do you want to return it?– Jose
Jan 3 at 10:21
Changed your function prototype from 'struct _list*' list to 'struct _list**', you cannot modify a pointer and return it back but you can modify a pointer when you pass a pointer to the pointer, you will need to change internal references to list to '(*list)->, if you do this then you can get rid of the return and change the function return to void.
– SPlatten
Jan 3 at 10:29
Changed your function prototype from 'struct _list*' list to 'struct _list**', you cannot modify a pointer and return it back but you can modify a pointer when you pass a pointer to the pointer, you will need to change internal references to list to '(*list)->, if you do this then you can get rid of the return and change the function return to void.
– SPlatten
Jan 3 at 10:29
add a comment |
3 Answers
3
active
oldest
votes
1
Add a new var whole_list.
The following code could work:
struct _list *insertSubstitute(struct _list *list,
int startNode,
unsigned char letter,
int substitute) {
struct _list* whole_list = list;
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
break;
}
list = list -> next;
}
return whole_list;
}
answered Jan 3 at 10:24
Yunbin LiuYunbin Liu
1,2762515
add a comment |
0
you need just to use a temporary pointer to your list to check the elements since you are using the list it self its causing you the problem
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
struct _list *temp=list;
while (temp!= NULL) {
if (temp-> letter == letter && temp-> startNode == startNode) {
temp -> substitute = substitute;
break;
}
temp= temp-> next;
}
return list;
}
answered Jan 3 at 10:25
SpinkooSpinkoo
1,7701216
add a comment |
0
Try this:
void insertSubstitute(struct _list** list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
while (*list != NULL) {
if ((*list) -> letter == letter && (*list) -> startNode == startNode) {
(*list) -> substitute = substitute;
return;
}
(*list) = (*list) -> next;
}
}
When you call this version of the function, you need to pass the address of your list.
Or another version of the original:
struct _list*insertSubstitute(struct _list* list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
struct _list* rc = list;
while (rc != NULL) {
if (rc -> letter == letter && rc -> startNode == startNode) {
rc -> substitute = substitute;
return rc;
}
rc = rc -> next;
}
return NULL;
}
answered Jan 3 at 10:33
SPlattenSPlatten
2,23531953
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Add a new var whole_list.
The following code could work:
struct _list *insertSubstitute(struct _list *list,
int startNode,
unsigned char letter,
int substitute) {
struct _list* whole_list = list;
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
break;
}
list = list -> next;
}
return whole_list;
}
answered Jan 3 at 10:24
Yunbin LiuYunbin Liu
1,2762515
add a comment |
1
Add a new var whole_list.
The following code could work:
struct _list *insertSubstitute(struct _list *list,
int startNode,
unsigned char letter,
int substitute) {
struct _list* whole_list = list;
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
break;
}
list = list -> next;
}
return whole_list;
}
answered Jan 3 at 10:24
Yunbin LiuYunbin Liu
1,2762515
add a comment |
1
1
1
Add a new var whole_list.
The following code could work:
struct _list *insertSubstitute(struct _list *list,
int startNode,
unsigned char letter,
int substitute) {
struct _list* whole_list = list;
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
break;
}
list = list -> next;
}
return whole_list;
}
answered Jan 3 at 10:24
Yunbin LiuYunbin Liu
1,2762515
Add a new var whole_list.
The following code could work:
struct _list *insertSubstitute(struct _list *list,
int startNode,
unsigned char letter,
int substitute) {
struct _list* whole_list = list;
while (list != NULL) {
if (list -> letter == letter && list -> startNode == startNode) {
list -> substitute = substitute;
break;
}
list = list -> next;
}
return whole_list;
}
answered Jan 3 at 10:24
Yunbin LiuYunbin Liu
1,2762515
answered Jan 3 at 10:24
Yunbin LiuYunbin Liu
1,2762515
answered Jan 3 at 10:24
Yunbin LiuYunbin Liu
1,2762515
answered Jan 3 at 10:24
Yunbin LiuYunbin Liu
1,2762515
1,2762515
add a comment |
add a comment |
0
you need just to use a temporary pointer to your list to check the elements since you are using the list it self its causing you the problem
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
struct _list *temp=list;
while (temp!= NULL) {
if (temp-> letter == letter && temp-> startNode == startNode) {
temp -> substitute = substitute;
break;
}
temp= temp-> next;
}
return list;
}
answered Jan 3 at 10:25
SpinkooSpinkoo
1,7701216
add a comment |
0
you need just to use a temporary pointer to your list to check the elements since you are using the list it self its causing you the problem
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
struct _list *temp=list;
while (temp!= NULL) {
if (temp-> letter == letter && temp-> startNode == startNode) {
temp -> substitute = substitute;
break;
}
temp= temp-> next;
}
return list;
}
answered Jan 3 at 10:25
SpinkooSpinkoo
1,7701216
add a comment |
0
0
0
you need just to use a temporary pointer to your list to check the elements since you are using the list it self its causing you the problem
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
struct _list *temp=list;
while (temp!= NULL) {
if (temp-> letter == letter && temp-> startNode == startNode) {
temp -> substitute = substitute;
break;
}
temp= temp-> next;
}
return list;
}
answered Jan 3 at 10:25
SpinkooSpinkoo
1,7701216
you need just to use a temporary pointer to your list to check the elements since you are using the list it self its causing you the problem
struct _list *insertSubstitute(struct _list *list, int startNode,
unsigned char letter, int substitute) {
struct _list *temp=list;
while (temp!= NULL) {
if (temp-> letter == letter && temp-> startNode == startNode) {
temp -> substitute = substitute;
break;
}
temp= temp-> next;
}
return list;
}
answered Jan 3 at 10:25
SpinkooSpinkoo
1,7701216
answered Jan 3 at 10:25
SpinkooSpinkoo
1,7701216
answered Jan 3 at 10:25
SpinkooSpinkoo
1,7701216
answered Jan 3 at 10:25
SpinkooSpinkoo
1,7701216
1,7701216
add a comment |
add a comment |
0
Try this:
void insertSubstitute(struct _list** list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
while (*list != NULL) {
if ((*list) -> letter == letter && (*list) -> startNode == startNode) {
(*list) -> substitute = substitute;
return;
}
(*list) = (*list) -> next;
}
}
When you call this version of the function, you need to pass the address of your list.
Or another version of the original:
struct _list*insertSubstitute(struct _list* list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
struct _list* rc = list;
while (rc != NULL) {
if (rc -> letter == letter && rc -> startNode == startNode) {
rc -> substitute = substitute;
return rc;
}
rc = rc -> next;
}
return NULL;
}
answered Jan 3 at 10:33
SPlattenSPlatten
2,23531953
add a comment |
0
Try this:
void insertSubstitute(struct _list** list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
while (*list != NULL) {
if ((*list) -> letter == letter && (*list) -> startNode == startNode) {
(*list) -> substitute = substitute;
return;
}
(*list) = (*list) -> next;
}
}
When you call this version of the function, you need to pass the address of your list.
Or another version of the original:
struct _list*insertSubstitute(struct _list* list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
struct _list* rc = list;
while (rc != NULL) {
if (rc -> letter == letter && rc -> startNode == startNode) {
rc -> substitute = substitute;
return rc;
}
rc = rc -> next;
}
return NULL;
}
answered Jan 3 at 10:33
SPlattenSPlatten
2,23531953
add a comment |
0
0
0
Try this:
void insertSubstitute(struct _list** list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
while (*list != NULL) {
if ((*list) -> letter == letter && (*list) -> startNode == startNode) {
(*list) -> substitute = substitute;
return;
}
(*list) = (*list) -> next;
}
}
When you call this version of the function, you need to pass the address of your list.
Or another version of the original:
struct _list*insertSubstitute(struct _list* list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
struct _list* rc = list;
while (rc != NULL) {
if (rc -> letter == letter && rc -> startNode == startNode) {
rc -> substitute = substitute;
return rc;
}
rc = rc -> next;
}
return NULL;
}
answered Jan 3 at 10:33
SPlattenSPlatten
2,23531953
Try this:
void insertSubstitute(struct _list** list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
while (*list != NULL) {
if ((*list) -> letter == letter && (*list) -> startNode == startNode) {
(*list) -> substitute = substitute;
return;
}
(*list) = (*list) -> next;
}
}
When you call this version of the function, you need to pass the address of your list.
Or another version of the original:
struct _list*insertSubstitute(struct _list* list
,int startNode
,unsigned char letter
,int substitute) {
if ( list == NULL ) {
//Do nothing list is invalid
return;
}
struct _list* rc = list;
while (rc != NULL) {
if (rc -> letter == letter && rc -> startNode == startNode) {
rc -> substitute = substitute;
return rc;
}
rc = rc -> next;
}
return NULL;
}
answered Jan 3 at 10:33
SPlattenSPlatten
2,23531953
answered Jan 3 at 10:33
SPlattenSPlatten
2,23531953
answered Jan 3 at 10:33
SPlattenSPlatten
2,23531953
answered Jan 3 at 10:33
SPlattenSPlatten
2,23531953
2,23531953
add a comment |
add a comment |
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insertSubstitutegets a pointer to a struct, so you are modifying the source. Why do you want to return it?– Jose
Jan 3 at 10:21
Changed your function prototype from 'struct _list*' list to 'struct _list**', you cannot modify a pointer and return it back but you can modify a pointer when you pass a pointer to the pointer, you will need to change internal references to list to '(*list)->, if you do this then you can get rid of the return and change the function return to void.
– SPlatten
Jan 3 at 10:29