realloc() problems with the old size
I want to read a lot of information from a file, so I need a dynamic memory.
This is why I use malloc for my struct in the main.
I wanted to realloc with every new line I get from the file, but he says "realloc(): invalid old size", and does not even realloc once.
typedef struct{
int anzahl;
int ANR;
char MHD[10];
char Bezeichnung[20];
char VPE[5];
float Preis;
float gesamtpreis;
}Ware;
int DateiLesen(Ware *Rechnung)
{
FILE *datei_lesen = NULL;
char trennung = " :,;n=";
char zeilen_lesen[256] = {0};
char *formatierer = NULL;
int count = 0;
datei_lesen = fopen("artikel.txt","r");
while(fgets(zeilen_lesen,256,datei_lesen))
{
count++;
}
fclose(datei_lesen);
if(count == 0)
{
return -1;
}
datei_lesen = fopen("artikel.txt","r");
while(fgets(zeilen_lesen,256,datei_lesen))
{
fputs(zeilen_lesen,datei_lesen);
formatierer = strtok(zeilen_lesen,trennung);
if(atoi(formatierer) >= 100000)
{
Rechnung->ANR = atoi(formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->MHD,formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->Bezeichnung,formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->VPE,formatierer);
formatierer = strtok(NULL,trennung);
Rechnung->Preis = atoi(formatierer);
Rechnung = realloc(Rechnung,1*sizeof(Ware));
//Rechnung = (Ware*) realloc(Rechnung,1);
Rechnung++;
}
}
fclose(datei_lesen);
return 0;
}
int main(void) {
Ware *Rechnung = (Ware*) malloc(sizeof(Ware));
int test = 0;
initialisiere(&Rechnung);
test = DateiLesen(&Rechnung);
return EXIT_SUCCESS;
}
c pointers struct malloc realloc
add a comment |
I want to read a lot of information from a file, so I need a dynamic memory.
This is why I use malloc for my struct in the main.
I wanted to realloc with every new line I get from the file, but he says "realloc(): invalid old size", and does not even realloc once.
typedef struct{
int anzahl;
int ANR;
char MHD[10];
char Bezeichnung[20];
char VPE[5];
float Preis;
float gesamtpreis;
}Ware;
int DateiLesen(Ware *Rechnung)
{
FILE *datei_lesen = NULL;
char trennung = " :,;n=";
char zeilen_lesen[256] = {0};
char *formatierer = NULL;
int count = 0;
datei_lesen = fopen("artikel.txt","r");
while(fgets(zeilen_lesen,256,datei_lesen))
{
count++;
}
fclose(datei_lesen);
if(count == 0)
{
return -1;
}
datei_lesen = fopen("artikel.txt","r");
while(fgets(zeilen_lesen,256,datei_lesen))
{
fputs(zeilen_lesen,datei_lesen);
formatierer = strtok(zeilen_lesen,trennung);
if(atoi(formatierer) >= 100000)
{
Rechnung->ANR = atoi(formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->MHD,formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->Bezeichnung,formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->VPE,formatierer);
formatierer = strtok(NULL,trennung);
Rechnung->Preis = atoi(formatierer);
Rechnung = realloc(Rechnung,1*sizeof(Ware));
//Rechnung = (Ware*) realloc(Rechnung,1);
Rechnung++;
}
}
fclose(datei_lesen);
return 0;
}
int main(void) {
Ware *Rechnung = (Ware*) malloc(sizeof(Ware));
int test = 0;
initialisiere(&Rechnung);
test = DateiLesen(&Rechnung);
return EXIT_SUCCESS;
}
c pointers struct malloc realloc
4
your signature for theDateiLesen
function doesn't match the parameters that you pass in. Do you see any compiler warnings?DateiLesen
says it wants a pointer to i.e. a*Ware
type, but you have passed in a**Ware
type.
– bruceg
Nov 19 at 21:36
1
StackOverflow does not really require perfect English, or I myself would be pretty lost. But please try a little harder to avoid the impression that you cannot be bothered with punctuation, capitalisation only where appropriate and basic grammar. You will not actually insist that you really think "wanna" is correct, will you? Giving instead the impression that you at least tried will give you better reactions. Also even in code, using English words for identifiers will make your question/code easier to understand for the majority of potential answerers.
– Yunnosch
Nov 19 at 21:43
add a comment |
I want to read a lot of information from a file, so I need a dynamic memory.
This is why I use malloc for my struct in the main.
I wanted to realloc with every new line I get from the file, but he says "realloc(): invalid old size", and does not even realloc once.
typedef struct{
int anzahl;
int ANR;
char MHD[10];
char Bezeichnung[20];
char VPE[5];
float Preis;
float gesamtpreis;
}Ware;
int DateiLesen(Ware *Rechnung)
{
FILE *datei_lesen = NULL;
char trennung = " :,;n=";
char zeilen_lesen[256] = {0};
char *formatierer = NULL;
int count = 0;
datei_lesen = fopen("artikel.txt","r");
while(fgets(zeilen_lesen,256,datei_lesen))
{
count++;
}
fclose(datei_lesen);
if(count == 0)
{
return -1;
}
datei_lesen = fopen("artikel.txt","r");
while(fgets(zeilen_lesen,256,datei_lesen))
{
fputs(zeilen_lesen,datei_lesen);
formatierer = strtok(zeilen_lesen,trennung);
if(atoi(formatierer) >= 100000)
{
Rechnung->ANR = atoi(formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->MHD,formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->Bezeichnung,formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->VPE,formatierer);
formatierer = strtok(NULL,trennung);
Rechnung->Preis = atoi(formatierer);
Rechnung = realloc(Rechnung,1*sizeof(Ware));
//Rechnung = (Ware*) realloc(Rechnung,1);
Rechnung++;
}
}
fclose(datei_lesen);
return 0;
}
int main(void) {
Ware *Rechnung = (Ware*) malloc(sizeof(Ware));
int test = 0;
initialisiere(&Rechnung);
test = DateiLesen(&Rechnung);
return EXIT_SUCCESS;
}
c pointers struct malloc realloc
I want to read a lot of information from a file, so I need a dynamic memory.
This is why I use malloc for my struct in the main.
I wanted to realloc with every new line I get from the file, but he says "realloc(): invalid old size", and does not even realloc once.
typedef struct{
int anzahl;
int ANR;
char MHD[10];
char Bezeichnung[20];
char VPE[5];
float Preis;
float gesamtpreis;
}Ware;
int DateiLesen(Ware *Rechnung)
{
FILE *datei_lesen = NULL;
char trennung = " :,;n=";
char zeilen_lesen[256] = {0};
char *formatierer = NULL;
int count = 0;
datei_lesen = fopen("artikel.txt","r");
while(fgets(zeilen_lesen,256,datei_lesen))
{
count++;
}
fclose(datei_lesen);
if(count == 0)
{
return -1;
}
datei_lesen = fopen("artikel.txt","r");
while(fgets(zeilen_lesen,256,datei_lesen))
{
fputs(zeilen_lesen,datei_lesen);
formatierer = strtok(zeilen_lesen,trennung);
if(atoi(formatierer) >= 100000)
{
Rechnung->ANR = atoi(formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->MHD,formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->Bezeichnung,formatierer);
formatierer = strtok(NULL,trennung);
strcpy(Rechnung->VPE,formatierer);
formatierer = strtok(NULL,trennung);
Rechnung->Preis = atoi(formatierer);
Rechnung = realloc(Rechnung,1*sizeof(Ware));
//Rechnung = (Ware*) realloc(Rechnung,1);
Rechnung++;
}
}
fclose(datei_lesen);
return 0;
}
int main(void) {
Ware *Rechnung = (Ware*) malloc(sizeof(Ware));
int test = 0;
initialisiere(&Rechnung);
test = DateiLesen(&Rechnung);
return EXIT_SUCCESS;
}
c pointers struct malloc realloc
c pointers struct malloc realloc
edited Nov 19 at 21:39
Yunnosch
11.1k52033
11.1k52033
asked Nov 19 at 21:31
Yondaime Mugiwara
31
31
4
your signature for theDateiLesen
function doesn't match the parameters that you pass in. Do you see any compiler warnings?DateiLesen
says it wants a pointer to i.e. a*Ware
type, but you have passed in a**Ware
type.
– bruceg
Nov 19 at 21:36
1
StackOverflow does not really require perfect English, or I myself would be pretty lost. But please try a little harder to avoid the impression that you cannot be bothered with punctuation, capitalisation only where appropriate and basic grammar. You will not actually insist that you really think "wanna" is correct, will you? Giving instead the impression that you at least tried will give you better reactions. Also even in code, using English words for identifiers will make your question/code easier to understand for the majority of potential answerers.
– Yunnosch
Nov 19 at 21:43
add a comment |
4
your signature for theDateiLesen
function doesn't match the parameters that you pass in. Do you see any compiler warnings?DateiLesen
says it wants a pointer to i.e. a*Ware
type, but you have passed in a**Ware
type.
– bruceg
Nov 19 at 21:36
1
StackOverflow does not really require perfect English, or I myself would be pretty lost. But please try a little harder to avoid the impression that you cannot be bothered with punctuation, capitalisation only where appropriate and basic grammar. You will not actually insist that you really think "wanna" is correct, will you? Giving instead the impression that you at least tried will give you better reactions. Also even in code, using English words for identifiers will make your question/code easier to understand for the majority of potential answerers.
– Yunnosch
Nov 19 at 21:43
4
4
your signature for the
DateiLesen
function doesn't match the parameters that you pass in. Do you see any compiler warnings? DateiLesen
says it wants a pointer to i.e. a *Ware
type, but you have passed in a **Ware
type.– bruceg
Nov 19 at 21:36
your signature for the
DateiLesen
function doesn't match the parameters that you pass in. Do you see any compiler warnings? DateiLesen
says it wants a pointer to i.e. a *Ware
type, but you have passed in a **Ware
type.– bruceg
Nov 19 at 21:36
1
1
StackOverflow does not really require perfect English, or I myself would be pretty lost. But please try a little harder to avoid the impression that you cannot be bothered with punctuation, capitalisation only where appropriate and basic grammar. You will not actually insist that you really think "wanna" is correct, will you? Giving instead the impression that you at least tried will give you better reactions. Also even in code, using English words for identifiers will make your question/code easier to understand for the majority of potential answerers.
– Yunnosch
Nov 19 at 21:43
StackOverflow does not really require perfect English, or I myself would be pretty lost. But please try a little harder to avoid the impression that you cannot be bothered with punctuation, capitalisation only where appropriate and basic grammar. You will not actually insist that you really think "wanna" is correct, will you? Giving instead the impression that you at least tried will give you better reactions. Also even in code, using English words for identifiers will make your question/code easier to understand for the majority of potential answerers.
– Yunnosch
Nov 19 at 21:43
add a comment |
1 Answer
1
active
oldest
votes
This is not how you grow an array.
First,
Rechnung++;
is fine and dandy, but Rechnung
is no longer a pointer returned by a previous call to malloc
or realloc
. So you can neither realloc
nor free
it.
Second,
Rechnung = realloc(Rechnung,1*sizeof(Ware));
is fine if you want to leave the size of the array always at 1 element no matter what. If you want the size to increase, you need to feed in the new size.
A typical array-growing loop often looks like this:
Data *array = NULL; // note it's fine to realloc a NULL
size_t size = 0;
while (fgets(...)) {
size_t new_size = size + 1;
Data *new_array = realloc(array, sizeof(Data) * new_size);
if (new_array == NULL) {
// report an error, exit, abort, try again, whatever
// note having a separate `new_array` variable allows you
// to retain old data in `array` in the case of `realloc` erroring on you
} else {
array = new_array;
array[size].foo = make_foo();
array[size].bar = make_bar();
size = new_size;
}
}
Note you never increment array
because you cannot pass an incremented array
to realloc
. You cannot also have something like this:
Data *array = malloc(sizeof(Data));
Data *current = data;
while (...) {
...
current->foo = make_foo();
current->bar = make_bar();
array = realloc(array, sizeof(Data) * new_size);
current++;
}
because realloc
may and will return a pointer different from what it was passed, and current
will become invalid. So use plain old boring array indexing.
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is not how you grow an array.
First,
Rechnung++;
is fine and dandy, but Rechnung
is no longer a pointer returned by a previous call to malloc
or realloc
. So you can neither realloc
nor free
it.
Second,
Rechnung = realloc(Rechnung,1*sizeof(Ware));
is fine if you want to leave the size of the array always at 1 element no matter what. If you want the size to increase, you need to feed in the new size.
A typical array-growing loop often looks like this:
Data *array = NULL; // note it's fine to realloc a NULL
size_t size = 0;
while (fgets(...)) {
size_t new_size = size + 1;
Data *new_array = realloc(array, sizeof(Data) * new_size);
if (new_array == NULL) {
// report an error, exit, abort, try again, whatever
// note having a separate `new_array` variable allows you
// to retain old data in `array` in the case of `realloc` erroring on you
} else {
array = new_array;
array[size].foo = make_foo();
array[size].bar = make_bar();
size = new_size;
}
}
Note you never increment array
because you cannot pass an incremented array
to realloc
. You cannot also have something like this:
Data *array = malloc(sizeof(Data));
Data *current = data;
while (...) {
...
current->foo = make_foo();
current->bar = make_bar();
array = realloc(array, sizeof(Data) * new_size);
current++;
}
because realloc
may and will return a pointer different from what it was passed, and current
will become invalid. So use plain old boring array indexing.
add a comment |
This is not how you grow an array.
First,
Rechnung++;
is fine and dandy, but Rechnung
is no longer a pointer returned by a previous call to malloc
or realloc
. So you can neither realloc
nor free
it.
Second,
Rechnung = realloc(Rechnung,1*sizeof(Ware));
is fine if you want to leave the size of the array always at 1 element no matter what. If you want the size to increase, you need to feed in the new size.
A typical array-growing loop often looks like this:
Data *array = NULL; // note it's fine to realloc a NULL
size_t size = 0;
while (fgets(...)) {
size_t new_size = size + 1;
Data *new_array = realloc(array, sizeof(Data) * new_size);
if (new_array == NULL) {
// report an error, exit, abort, try again, whatever
// note having a separate `new_array` variable allows you
// to retain old data in `array` in the case of `realloc` erroring on you
} else {
array = new_array;
array[size].foo = make_foo();
array[size].bar = make_bar();
size = new_size;
}
}
Note you never increment array
because you cannot pass an incremented array
to realloc
. You cannot also have something like this:
Data *array = malloc(sizeof(Data));
Data *current = data;
while (...) {
...
current->foo = make_foo();
current->bar = make_bar();
array = realloc(array, sizeof(Data) * new_size);
current++;
}
because realloc
may and will return a pointer different from what it was passed, and current
will become invalid. So use plain old boring array indexing.
add a comment |
This is not how you grow an array.
First,
Rechnung++;
is fine and dandy, but Rechnung
is no longer a pointer returned by a previous call to malloc
or realloc
. So you can neither realloc
nor free
it.
Second,
Rechnung = realloc(Rechnung,1*sizeof(Ware));
is fine if you want to leave the size of the array always at 1 element no matter what. If you want the size to increase, you need to feed in the new size.
A typical array-growing loop often looks like this:
Data *array = NULL; // note it's fine to realloc a NULL
size_t size = 0;
while (fgets(...)) {
size_t new_size = size + 1;
Data *new_array = realloc(array, sizeof(Data) * new_size);
if (new_array == NULL) {
// report an error, exit, abort, try again, whatever
// note having a separate `new_array` variable allows you
// to retain old data in `array` in the case of `realloc` erroring on you
} else {
array = new_array;
array[size].foo = make_foo();
array[size].bar = make_bar();
size = new_size;
}
}
Note you never increment array
because you cannot pass an incremented array
to realloc
. You cannot also have something like this:
Data *array = malloc(sizeof(Data));
Data *current = data;
while (...) {
...
current->foo = make_foo();
current->bar = make_bar();
array = realloc(array, sizeof(Data) * new_size);
current++;
}
because realloc
may and will return a pointer different from what it was passed, and current
will become invalid. So use plain old boring array indexing.
This is not how you grow an array.
First,
Rechnung++;
is fine and dandy, but Rechnung
is no longer a pointer returned by a previous call to malloc
or realloc
. So you can neither realloc
nor free
it.
Second,
Rechnung = realloc(Rechnung,1*sizeof(Ware));
is fine if you want to leave the size of the array always at 1 element no matter what. If you want the size to increase, you need to feed in the new size.
A typical array-growing loop often looks like this:
Data *array = NULL; // note it's fine to realloc a NULL
size_t size = 0;
while (fgets(...)) {
size_t new_size = size + 1;
Data *new_array = realloc(array, sizeof(Data) * new_size);
if (new_array == NULL) {
// report an error, exit, abort, try again, whatever
// note having a separate `new_array` variable allows you
// to retain old data in `array` in the case of `realloc` erroring on you
} else {
array = new_array;
array[size].foo = make_foo();
array[size].bar = make_bar();
size = new_size;
}
}
Note you never increment array
because you cannot pass an incremented array
to realloc
. You cannot also have something like this:
Data *array = malloc(sizeof(Data));
Data *current = data;
while (...) {
...
current->foo = make_foo();
current->bar = make_bar();
array = realloc(array, sizeof(Data) * new_size);
current++;
}
because realloc
may and will return a pointer different from what it was passed, and current
will become invalid. So use plain old boring array indexing.
answered Nov 19 at 22:13
n.m.
71k882166
71k882166
add a comment |
add a comment |
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4
your signature for the
DateiLesen
function doesn't match the parameters that you pass in. Do you see any compiler warnings?DateiLesen
says it wants a pointer to i.e. a*Ware
type, but you have passed in a**Ware
type.– bruceg
Nov 19 at 21:36
1
StackOverflow does not really require perfect English, or I myself would be pretty lost. But please try a little harder to avoid the impression that you cannot be bothered with punctuation, capitalisation only where appropriate and basic grammar. You will not actually insist that you really think "wanna" is correct, will you? Giving instead the impression that you at least tried will give you better reactions. Also even in code, using English words for identifiers will make your question/code easier to understand for the majority of potential answerers.
– Yunnosch
Nov 19 at 21:43