if( mystruct == NULL )
mystruct is not a pointer, so you cannot compare it with NULL.
You have three options:
- Add a status field to
MyStructto indicate whether the struct has been initialized correctly. - Allocate the struct on the heap and return it by pointer.
- Pass the structure as a pointer argument and return a status code (thanks @Potatoswatter).
if( mystruct == NULL )
mystruct is not a pointer, so you cannot compare it with NULL.
You have three options:
- Add a status field to
MyStructto indicate whether the struct has been initialized correctly. - Allocate the struct on the heap and return it by pointer.
- Pass the structure as a pointer argument and return a status code (thanks @Potatoswatter).
A structure is not a pointer. If you want to be able to return NULL, you're going to have to allocate the structure on the heap so you can return a pointer to it, and let the caller clean up afterwards.
That way, you can indicate failure, something like:
MyStruct *init_mystruct (void) {
MyStruct *mystruct = malloc (sizeof (*mystruct));
if (mystruct != NULL)
return NULL;
int is_ok = 1;
/* do something ... */
/* everything is OK */
if( is_ok )
return mystruct;
/* something went wrong */
free (mystruct);
return NULL;
}
int main (void) {
MyStruct *mystruct = init_mystruct();
if (mystruct == NULL) {
/* error handler */
return -1;
}
free (mystruct);
return 0;
}
Videos
Source: https://cs50.harvard.edu/x/2023/labs/5/
Here is the portion that I need help with:
// Create a new individual with `generations`
person *create_family(int generations)
{
// TODO: Allocate memory for new person
person *new_person = malloc(sizeof(person));
if (new_person == NULL)
{
return 1;
}Here is what ChatGPT says on my return 1 usage:
The TODO comment indicates that you need to allocate memory for a new person structure. In the provided code, you're using malloc(sizeof(person))to allocate memory dynamically. However, when memory allocation fails (i.e., malloc returns NULL), you're returning the integer 1 from the create_family function. Typically, a function returning a pointer should return NULL to indicate a failure.
Next, I rephrased the code:
// Create a new individual with `generations`
person *create_family(int generations)
{
// TODO: Allocate memory for new person
person *new_person = malloc(sizeof(person));
if (!new_person)
{
return 1;
}ChatGPT reply:
Your updated code checks if new_person is NULL using the condition if (!new_person).
However, it's better to return NULL*(a pointer) in case of memory allocation failure, rather than the integer* 1.
Here's the modified code:
// Create a new individual with `generations`
person *create_family(int generations)
{
// TODO: Allocate memory for new person
person *new_person = malloc(sizeof(person));
// Check if memory allocation was successful
if (!new_person)
{
fprintf(stderr, "Memory allocation failed.\n");
return NULL;
}
// Rest of the code for creating the family tree
// ...
return new_person;
}In this modification:
-
If memory allocation fails, the function prints an error message to the standard error stream using fprintf(stderr, "Memory allocation failed.\n");
-
return NULL;
is used to indicate that the function failed to create a new person due to a memory allocation failure. -
If memory allocation is successful, the function proceeds with the rest of the code for creating the family tree.
Returning NULL in case of memory allocation failure is a common practice in C, as it allows the calling code to check for errors by examining the returned pointer.
Full project code provided by CS50:
// Simulate genetic inheritance of blood type
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Each person has two parents and two alleles
typedef struct person
{
struct person *parents[2];
char alleles[2];
} person;
const int GENERATIONS = 3;
const int INDENT_LENGTH = 4;
person *create_family(int generations);
void print_family(person *p, int generation);
void free_family(person *p);
char random_allele();
int main(void)
{
// Seed random number generator
srand(time(0));
// Create a new family with three generations
person *p = create_family(GENERATIONS);
// Print family tree of blood types
print_family(p, 0);
// Free memory
free_family(p);
}
// Create a new individual with `generations`
person *create_family(int generations)
{
// TODO: Allocate memory for new person
person *new_person = malloc(sizeof(person));
if (new_person == NULL)
{
return 1;
}
// If there are still generations left to create
if (generations > 1)
{
// Create two new parents for current person by recursively calling create_family
person *parent0 = create_family(generations - 1);
person *parent1 = create_family(generations - 1);
// TODO: Set parent pointers for current person
// TODO: Randomly assign current person's alleles based on the alleles of their parents
}
// If there are no generations left to create
else
{
// TODO: Set parent pointers to NULL
// TODO: Randomly assign alleles
}
// TODO: Return newly created person
return NULL;
}
// Free `p` and all ancestors of `p`.
void free_family(person *p)
{
// TODO: Handle base case
// TODO: Free parents recursively
// TODO: Free child
}
// Print each family member and their alleles.
void print_family(person *p, int generation)
{
// Handle base case
if (p == NULL)
{
return;
}
// Print indentation
for (int i = 0; i < generation * INDENT_LENGTH; i++)
{
printf(" ");
}
// Print person
if (generation == 0)
{
printf("Child (Generation %i): blood type %c%c\n", generation, p->alleles[0], p->alleles[1]);
}
else if (generation == 1)
{
printf("Parent (Generation %i): blood type %c%c\n", generation, p->alleles[0], p->alleles[1]);
}
else
{
for (int i = 0; i < generation - 2; i++)
{
printf("Great-");
}
printf("Grandparent (Generation %i): blood type %c%c\n", generation, p->alleles[0], p->alleles[1]);
}
// Print parents of current generation
print_family(p->parents[0], generation + 1);
print_family(p->parents[1], generation + 1);
}
// Randomly chooses a blood type allele.
char random_allele()
{
int r = rand() % 3;
if (r == 0)
{
return 'A';
}
else if (r == 1)
{
return 'B';
}
else
{
return 'O';
}
}
If the return type is int, you can't return a NULL. To show an error, you could instead return a special value like zero or -1, if you check for that value in any calling function. Lots of functions return nonnegative numbers on success, or -1 on error.
NULL cannot be stored in an int variable, unlike in SQL, for example. If you ignore the warning and return NULL anyway, then NULL will be casted to zero. The calling function won't be able to tell whether you returned NULL or zero.
If your function only needs to indicate success or failure, then it's common to return 1 for success, and zero for failure. Zero means "false" when treated as a boolean value (like in if statements), and non-zero means "true."
It looks like you've misunderstood what NULL means in C. Types are not nullable. NULL is effectively just a shorthand for the pointer with value 0! And int is not a pointer.
NULL is a pointer value - or rather a null-pointer value.
NULL means that the function can't find where your pointer should point to - for example if you want to open a file, but it doesn't work your file pointer is returned as NULL. So you can test the value of a pointer and check to see if it worked or not.
If you are writing a routine
int length()
then you could return a negative value if length is unable to read the length of whatever you send it - this would be a way of indicating an error, because normally lengths can never be negative....
It is a matter of convention and you should clearly have one in your head and document it (at least in comments).
Sometimes a pointer really should always point to a valid address (see this intSwap example, both arguments should be valid pointers). At other times, it should either be such a valid address, or be NULL. Conceptually the pointer type is then by convention a sum type (between genuine pointer addresses and the special NULL value).
Notice that the C language does not have a type (or a notation) which enforces that some given pointer is always valid and non-null. BTW, with GCC specifically, you can annotate a function with __attribute__ using nonnull to express that a given argument is never null.
A typical example is FILE* pointers in <stdio.h>. The fopen function is documented to be able to return NULL (on failure), or some valid pointer. But the fprintf function is expecting a valid pointer (and passing NULL to it as the first argument is some undefined behavior, often a segmentation fault; and UB is really bad).
Some non-portable programs even use several "special" pointer values (which should not be dereferenced), e.g. (on Linux/x86-64) #define SPECIAL_SLOT (void*)((intptr_t)-1) (which we know that on Linux it is never a valid address). Then we could have the convention that a pointer is a valid pointer to a valid memory zone, or NULL or SPECIAL_SLOT (hence, if seen as an abstract data type, it is a sum type of two distinct invalid pointers NULL and SPECIAL_SLOT and the set of valid addresses). Another example is MAP_FAILURE as result of mmap(2) on Linux.
BTW, when using pointers in C to heap allocated data (indirectly obtained with malloc), you also need conventions about who is in charge of releasing the data (by using free, often thru a supplied function to free a data and all its internal stuff).
Good C programming requires many explicit conventions regarding pointers, and it is essential to understand them precisely and document them well. Look for example[s] into GTK. Read also about restrict.
[H]ow do I return NULL from a integer result type function?
You don't. NULL is a macro representing a value of type void *. It is not an int, so a function returning int cannot return NULL.
Now, it is possible to convert NULL or any other pointer to type int, but the result of such a conversion is a valid, ordinary int. You seem to be looking instead for some kind of distinguished value, but there is none available unless you reserve such a value yourself. For example, you might reserve INT_MIN for that purpose. Of the built-in types, only pointer types afford general-purpose distinguished values (null pointers).
To provide for your function to signal failure to the caller, you have a couple of alternatives to reserving a value. The most common one is to use the function's return value only to report on the success or failure of the call, and to deliver any output -- in your case a node's value -- via a pointer argument:
int valueOf(t_node *node, int n, int *result) {
int current = 0;
while (current < n && node != NULL) {
node = node->next;
current += 1;
}
if (node == NULL) {
// no such node -- return a failure code
return 0;
} else {
// current == n
*result = node->value;
return 1;
}
}
In C, NULL is just a synonym for 0. So you really only need to do this...
if (current > n) {
return 0;
}
However, NULL usually refers to a pointer value that is undefined and not an integer. In C, integer values are not references as they are in many interpreted languages. They are scalar and can't be referred to implicitly with a pointer.
If you want to indicate an error condition or undefined behavior when current > n, you will have to provide a separate mechanism for indicating that the value isn't usable. Usually, C functions will return a -1 on an error. Since you are using the integer return for a value, that would mean that a valid value could never be -1.
It looks like you're handling a linked list and you want to limit the number of items to be checked. A possible way around this might be...
int valueOf(t_node *node, int n, int *val){
int current = 0;
int value;
while (node -> next != NULL) {
value = node -> val;
if (current == n) {
// This notation is for dereferencing a pointer.
*val = value;
return 0;
}
if (current > n) {
return -1;
}
node = node -> next;
current += 1;
}
// This method also gives you a way to indicate that
// you came to the end of the list. Your code snippet
// would have returned an undefined value if node->next == null
return -1;
}
Maybe there are some basic things you should rethink:
First, only pointers can be NULL, but not objects. Hence, if you return an object of type struct Stack (which is not a pointer), you cannot return NULL but just an instance of struct Stack.
Second, passing in and returning an object of struct Stack by value will result in copying the respective object; I think that passing references or pointers would be a better choice; and - if you pass in and return a pointer, you could also return NULL to indicate a full stack or some other issue.
The problem is that your function must return a value that has the type Stack.
The code you provided doesn't define the type of NULL, but, since you're not seeing another error and you're assigning it to node, I would guess that the type of NULL is StackNode *... or, at least, something compatible with that.
So, there's your problem. You're trying to return something with the type StackNode * when your function claims to return a Stack.
Your best bet is to redesign this function that returns the result of the push operation and not the Stack itself.
In C, NULL is a macro that expands either to 0 or (void*)0 (or something that has a similar effect).
In the first case, you can not differentiate between NULL and 0, because they are literally the same.
In the second case, your code will cause a compile error, because you can't compare an integer variable with a pointer.
First some background ...
The macros are
NULLwhich expands to an implementation-defined null pointer constant; C11 §7.19 3
NULL typically is an integer constant 0 or (void*)0 or the like. It may have a different implementation or type - It could be ((int*) 0xDEADBEEF) as strange as that may be.
NULL might be type int. It might be type void * or something else. The type of NULL is not defined.
When the null pointer constant NULL is cast to any pointer, is is a null pointer. An integer 0 cast to a pointer is also a null pointer. A system could have many different (bit-wise) null pointers. They all compare equally to each other. They all compare unequally to any valid object/function. Recall this compare is done as pointers, not integers.
An integer constant expression with the value 0, or such an expression cast to type
void *, is called a null pointer constant. If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function. C11 §6.3.2.3 3
int x;
if (&x == NULL) ... // this is false
So after all that chapter and verse how to distinguish NULL from 0?
If the macro NULL is defined as an int 0 - it is game over - there is no difference between 0 and NULL.
If NULL is not an int, then code can use _Generic() to differentiate NULL and 0. This does not help OP's "Any change made can only be made within the function itself." requirement as that function accepts an int augment.
If NULL is an int that has a different bit-pattern than 0, then a simple memcmp() can differentiate.
I suspect the whole reason for this exercise is to realize there is no portable method to distinguish NULL from 0.
Returning NULL in case of success (the most simple case for success at that) is definitely contrary to what most people will expect.
Returning a pointer that the user has to manually destruct isn’t super great either. I’d suggest using C++11 unique_ptr but using C++11 may not be feasible in your case.
One thought I had is to make TestRoute private and have the Planner call it whenever it computes a new route.
If the test fails, return NULL, otherwise return the route.
What’s nice about this approach is that you can implement TestRoute however you (or your colleague) please, and the user of the class won’t need to know the details of how it is implemented. The user will just ask for a route from point A to point B and will be guaranteed it’s a valid route with refueling points so long as they don’t receive NULL.
You could also split your method into a few different methods if the performance hit isn’t too great.
For example, for TestRoute, have it return true if the route is possible, false if not.
bool TestRoute(const Route* r)
Have another method TestRouteNeedRefuel that returns true if the route will require refueling, false if not
bool TestRouteNeedRefuel(const Route* r)
Then have a final method, GenerateRefuelRoute that returns a new route with the proper refuel points
Route* TestRoute(const Route* r)
//use this if at all possible
std::unique_ptr<Route> TestRoute(const Route* r)
As far as performance goes, remember to profile before making assumptions. If your colleague is worried about copying Route more than needed (and he may have good reason to, as we don't know how expensive it is or what the target platform is) then clearly performance is an important requirement. I would suggest first implementing as clean an interface as can be done, profiling to find where the bottlenecks REALLY are, and then implementing a few speed hacks where necessary.
I would generally consider returning a pointer from a method in C++ a bad design, and mixing error states and payload data in the return value, too; this is a recipe for unmaintainable code.
Suggested change: Return the fail/success status as int value (or use ternary logic, e. g. boost::tribool), and pass the argument as non-const reference:
/** @returns
- 1 if a solution has been found. The argument will be updated.
- 0 if the request has been processed sucessfully,
but no (immediate) solution has been found.
The argument is not modified in this case.
- -1 if the request failed. The argument is not modified. */
int findSolution(MyClass& argument);
Usage example, leaving out premature optimization to avoid "unnecessary" copies:
MyClass objectToTest(originalUnmutableObject);
switch(findSolution(objectToTest))
{
case 1:
//Replace original with updated object, or whatever
break;
case 0:
//Nothing to do (?)
break;
case -1:
//Error handling
break;
default:
//Unexpected return value
assert(false);
}
An alternative, more sophisticated and reusable approach could be to bundle error state and object into a generic result class; this pattern was inspired by Rust. I leave the implementation of Result to you.
template<typename T>
class Result
{
public:
Result() = delete;
Result(int error);
Result(const T& data);
Result(T&& data);
//Methods
bool isOk() const;
bool isError(int error) const;
int error() const;
const T& data() const;
private:
//Variables
int m_error = 0;
T m_data;
};
...
Result<MyClass> findSolution(const MyClass& argument)
{
int errorCode = 0;
...
if(errorCode != 0)
return Result(errorCode);
else if(solutionFound)
//Error code of result will be 0, Result::isOk() == true
return Result(update(argument, solution));
else
//Error code of result will be 1, Result::isOk() == false
return Result(1);
}