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Why does the arrow (->) operator in C exist?
(3 answers)
Closed 1 year ago.
I am trying to make all the names in the Struct lowercase so that I can compare them and remove them.
int removeNameCard(NameCard *idCard, int *size){
char name[80];
char *ptr;
char rubbish;
int a = 0, c = 0;
printf("removeNameCard():\n");
printf("Enter personName:\n");
scanf("%c", &rubbish); // Why is there a '\n' char here??
fgets(name, 80, stdin);
if((ptr = strchr(name, '\n'))){
*ptr = '\0';
}
if((*size) == 0){
printf("The name card holder is empty\n");
return 0;
}
// Convert everything to Lower Case first
while(name[a]){
name[a] = tolower(name[a]);
a += 1;
}
printf("tolower(): %s", name);
for(int b = 0; b < *size; b += 1){
// Why is this Dot Notation when I passed in a pointer to the Struct?
while (idCard[b].personName)[c]){
(idCard[b].personName)[c] = tolower((idCard[b].personName)[c]);
c += 1;
}
}
for(int i = 0; i < *size; i += 1){
if((idCard[i].personName) == name){
printf("%d. This is from Holder: %s, This is from User: %s", i,(idCard[i].personName),name);
printf("The name card is removed\n");
printf("nameCardID: %d\n", idCard[i].nameCardID);
printf("personName: %s\n", idCard[i].personName);
printf("companyName: %s\n", idCard[i].companyName);
int k = 0;
do{
idCard[i+k].nameCardID = idCard[i+k+1].nameCardID;
strcpy((idCard[i+k].personName),(idCard[i+k+1].personName));
strcpy((idCard[i+k].companyName),(idCard[i+k+1].companyName));
}while((i+k+1) != (*size + 1));
}
}
return 0;
}
However, I am rather confused why the Compiler asked me to use Dot Notation instead of Pointer Notation as I thought I passed in the address of the Struct into *idCard so it should be a pointer if I am not wrong?
And am I wrong for trying to access each individual character in every name of the Struct like this?:
(idCard[b].personName)[c]
Thank you
However, I am rather confused why the Compiler asked me to use Dot Notation instead of Pointer Notation…
idCard[i] is a structure, not a pointer to a structure, so its members are accessed as idCard[i].member, not idCard[i]->member.
idCard[i] is a structure because whenever x is a pointer, x[i] is one of the objects that x points to. It is not the address of the object. You could calculate the address of the object with x+i, and then you could reference the object with *(x+i). And x[i] is actually defined in this way; x[i] is defined to be *(x+i). (In case of expressions generally, E1[E2] is defined to be (*((E1)+(E2))).)
And am I wrong for trying to access each individual character in every name of the Struct like this?: (idCard[b].personName)[c]
This will work, but the parentheses are unnecessary. You can use idCard[b].personName[c]. Due to the C grammar, it is already grouped as (idCard[b].personName)[c].
However, I am rather confused why the Compiler asked me to use Dot Notation instead of Pointer Notation as I thought I passed in the address of the Struct into *idCard so it should be a pointer if I am not wrong?
Arrays are mostly just pointers to the first element in the array (except that the compiler may be aware of the array size).
Because arrays are mostly just pointers; for arrays of integers myInt = myIntArray[x]; is like myInt = *(myIntArray + x); - the pointer dereferencing is implied by array indexing. Note that to access a char in the middle of an int in an array, you could (not portably) use something like myChar = *((char *)(myIntArray + x)) + offset_of_char_in_int);; and this is a little bit like accessing a field inside an array of structures (in that they're both accessing something smaller within an array of larger things).
For arrays of structures; indexing the array causes dereferencing (just like it does for the array of integers); so myIDcardStruct = idCard[i]; is like myIDcardStruct = *(idcard + i);. Because array indexing has implied dereferencing, myIDcardStruct is not a pointer.
The -> operator is like adding the offset of the requested structure's field to the address, casting the address to the requested field's type, and then dereferencing. In other words myInt = myStructPointer->myIntField; is like myInt = (*myStructPointer).myIntField; which is like tempAddress = (void *)myStructPointer + offset_of_myIntField; myInt = *((int *)tempAddress);.
What this means is that if you have an array of structures (which is mostly a pointer to the first structure in the array), indexing the array causes the pointer to be implicitly dereferenced, and using -> also causes implicit dereferencing; and if you do both then you've (implicitly) dereferenced a pointer twice, which is too much dereferencing (because it's not a pointer to a pointer to a structure, or an array of pointers to structures). Because you only want it dereferenced once you have to choose between one implicit deference (array indexing) or the other implicit dereference (->); e.g. you can choose between myInt = idCard[i].nameCardID; or myInt = (idCard + i)->nameCardID;.
Of course the important thing is making code easy to read, and myInt = idCard[i].nameCardID; is easier to read than myInt = (idCard + i)->nameCardID;.
Related
I have currently trouble understanding the following scenario:
I have a multidimensional array of Strings and I want to address it by using pointers only but I always get a Segmentation Fault when using the array annotation on the pointer. This is just an example code I want to use the 3D array in a pthread so I want to pass it in via a structure as a pointer but it just doesn't work and I would like to know why? I thought pointers and arrays are functionally equivalent? Here is the sample code:
#include <stdio.h>
void func(unsigned char ***ptr);
int main() {
// Image of dimension 10 times 10
unsigned char image[10][10][3];
unsigned char ***ptr = image;
memcpy(image[0][0], "\120\200\12", 3);
// This works as expected
printf("Test: %s", image[0][0]);
func(image);
return 0;
}
void func(unsigned char ***ptr) {
// But here I get a Segmentation Fault but why??
printf("Ptr: %s", ptr[0][0]);
}
Thanks in advance for your help :)
I think maybe strdup confuses the issue. Pointers and arrays are not always equivalent. Let me try to demonstrate. I always avoid actual multi-dimension arrays, so I may make a mistake here, but:
int main()
{
char d3Array[10][10][4]; //creates a 400-byte contiguous memory area
char ***d3Pointer; //a pointer to a pointer to a pointer to a char.
int i,j;
d3Pointer = malloc(sizeof(char**) * 10);
for (i = 0; i < 10; ++i)
{
d3Pointer[i] = malloc(sizeof(char*) * 10);
for (j = 0; j < 4; ++j)
{
d3Pointer[i][j] = malloc(sizeof(char) * 4);
}
}
//this
d3Pointer[2][3][1] = 'a';
//is equivalent to this
char **d2Pointer = d3Pointer[2];
char *d1Pointer = d2Pointer[3];
d1Pointer[1] = 'a';
d3Array[2][3][1] = 'a';
//is equivalent to
((char *)d3Array)[(2 * 10 * 4) + (3 * 4) + (1)] = 'a';
}
Generally, I use the layered approach. If I want contiguous memory, I handle the math myself..like so:
char *psuedo3dArray = malloc(sizeof(char) * 10 * 10 * 4);
psuedo3dArray[(2 * 10 * 4) + (3 * 4) + (1)] = 'a';
Better yet, I use a collection library like uthash.
Note that properly encapsulating your data makes the actual code incredibly easy to read:
typedef unsigned char byte_t;
typedef struct
{
byte_t r;
byte_t g;
byte_t b;
}pixel_t;
typedef struct
{
int width;
int height;
pixel_t * pixelArray;
}screen_t;
pixel_t *getxyPixel(screen_t *pScreen, int x, int y)
{
return pScreen->pixelArray + (y*pScreen->width) + x;
}
int main()
{
screen_t myScreen;
myScreen.width = 1024;
myScreen.height = 768;
myScreen.pixelArray = (pixel_t*)malloc(sizeof(pixel_t) * myScreen.height * myScreen.width);
getxyPixel(&myScreen, 150, 120)->r = 255;
}
In C, you should allocate space for your 2D array one row at a time. Your definition of test declares a 10 by 10 array of char pointers, so you don't need to call malloc for it. But to store a string you need to allocate space for the string. Your call to strcpy would crash. Use strdup instead. One way to write your code is as follows.
char ***test = NULL;
char *ptr = NULL;
test = malloc(10 * sizeof(char **));
for (int i = 0; i < 10; i++) {
test[i] = malloc(10 * sizeof(char *));
}
test[0][0] = strdup("abc");
ptr = test[0][0];
printf("%s\n", ptr);
test[4][5] = strdup("efg");
ptr = test[4][5];
printf("%s\n", ptr);
Alternatively, if you want to keep your 10 by 10 definition, you could code it like this:
char *test[10][10];
char *ptr = NULL;
test[0][0] = strdup("abc");
ptr = test[0][0];
printf("%s\n", ptr);
test[4][5] = strdup("efg");
ptr = test[4][5];
printf("%s\n", ptr);
Your problem is, that a char[10][10][3] is something very different from a char***: The first is an array of arrays of arrays, the later is a pointer to a pointer to a pointer. The confusions arises because both can be dereferenced with the same syntax. So, here is a bit of an explanation:
The syntax a[b] is nothing but a shorthand for *(a + b): First you perform pointer arithmetic, then you dereference the resulting pointer.
But, how come you can use a[b] when a is an array instead of a pointer? Well, because...
Arrays decay into pointers to their first element: If you have an array declared like int array[10], saying array + 3 results in array decaying to a pointer of type int*.
But, how does that help to evaluate a[b]? Well, because...
Pointer arithmetic takes the size of the target into account: The expression array + 3 triggers a calculation along the lines of (size_t)array + 3*sizeof(*array). In our case, the pointer that results from the array-pointer-decay points to an int, which has a size, say 4 bytes. So, the pointer is incremented by 3*4 bytes. The result is a pointer that points to the fourths int in the array, the first three elements are skipped by the pointer arithmetic.
Note, that this works for arrays of any element type. Arrays can contain bytes, or integers, or floats, or structs, or other arrays. The pointer arithmetic is the same.
But, how does that help us with multidimensional arrays? Well, because...
Multidimensional arrays are just 1D arrays that happen to contain arrays as elements: When you declare an array with char image[256][512]; you are declaring a 1D array of 256 elements. These 256 elements are all arrays of 512 characters, each. Since the sizeof(char) == 1, the size of an element of the outer array is 512*sizeof(char) = 512, and, since we have 256 such arrays, the total size of image is 256*512. Now, I can declare a 3D array with char animation[24][256][512];...
So, going back to your example that uses
char image[10][10][3]
what happens when you say image[1][2][1] is this: The expression is equivalent to this one:
*(*(*(image + 1) + 2) + 3)
image being of type char[10][10][3] decays into a pointer to its first element, which is of type char(*)[10][3] The size of that element is 10*3*1 = 30 bytes.
image + 1: Pointer arithmetic is performed to add 1 to the resulting pointer, which increments it by 30 bytes.
*(image + 1): The pointer is dereferenced, we are now talking directly about the element, which is of type char[10][3].
This array again decays into a pointer to its first element, which is of type char(*)[3]. The size of the element is 3*1 = 3. This pointer points at the same byte in memory as the pointer that resulted from step 2. The only difference is, that it has a different type!
*(image + 1) + 2: Pointer arithmetic is performed to add 2 to the resulting pointer, which increments it by 2*3 = 6 bytes. Together with the increment in step 2, we now have an offset of 36 bytes, total.
*(*(image + 1) + 2): The pointer is dereferenced, we are now talking directly about the element, which is of type char[3].
This array again decays into a pointer to its first element, which is of type char*. The size of the element is now just a single byte. Again, this pointer has the same value as the pointer resulting from step 5, but a different type.
*(*(image + 1) + 2) + 1: Pointer arithmetic again, adding 1*1 = 1 bytes to the total offset, which increases to 37 bytes.
*(*(*(image + 1) + 2) + 1): The pointer is dereferenced the last time, we are now talking about the char at an offset of 37 bytes into the image.
So, what's the difference to a char***? When you dereference a char***, you do not get any array-pointer-decay. When you try to evaluate the expression pointers[1][2][1] with a variable declared as
char*** pointers;
the expression is again equivalent to:
*(*(*(pointers + 1) + 2) + 3)
pointers is a pointer, so no decay happens. Its type is char***, and it points to a value of type char**, which likely has a size of 8 bytes (assuming a 64 bit system).
pointers + 1: Pointer arithmetic is performed to add 1 to the resulting pointer, which increments it by 1*8 = 8 bytes.
*(pointers + 1): The pointer is dereferenced, we are now talking about the pointer value that is found in memory at an offset of 8 bytes of where pointers points.
Further steps depending on what actually happened to be stored at pointers[1]. These steps do not involve any array-pointer-decay, and thus load pointers from memory instead.
You see, the difference between a char[10][10][3] and a char*** is profound. In the first case, the array-pointer-decay transforms the process into a pure offset computation into a multidimensional array. In the later case, we repeatedly load pointers from memory when accessing elements, all we ever have are 1D arrays of pointers. And it's all down to the types of pointers!
In general, i'm trying to allocate values of first.a and first.b
to a array's in struct secon.
typedef struct {
int a;
int b;
} firs;
//secon is my struct which contains dynamic array
//can i use int here ?
typedef struct {
int *aa;
int *bb;
} secon;
//pointer to secon intialised to NULL;
secon* sp=NULL;
int main()
{
firs first;
//plz assume 2 is coming from user ;
sp=malloc(sizeof(secon)*2);
//setting values
first.a=10;
first.b=11;
/* what i'm trying to do is assign values of first.a and first.b to my
dynamically created array*/
/* plz assume first.a and first.b are changing else where .. that means ,not
all arrays will have same values */
/* in general , i'm trying to allocate values of first.a and first.b
to a array's in struct second. */
for(int i=0; i<2; i++) {
*( &(sp->aa ) + (i*4) ) = &first.a;
*( &(sp->bb ) + (i*4) ) = &first.b;
}
for(int i=0; i<2; i++) {
printf("%d %d \n", *((sp->aa) + (i*4) ),*( (sp->bb) +(i*4) ) );
}
return 0;
}
MY output :
10 11
4196048 0
Problems with my code:
1. whats wrong with my code?
2. can i use int inside struct for dynamic array?
3. what are the alternatives?
4. why am i not getting correct answer?
Grigory Rechistov has done a really good job of untangling the code and you should probably accept his answer, but I want to emphasize one particular point.
In C pointer arithmetic, the offsets are always in units of the size of the type pointed to. Unless the type of the pointer is char* or void* if you find yourself multiplying by the size of the type, you are almost certainly doing it wrong.
If I have
int a[10];
int *p = &(a[5]);
int *q = &(a[7]);
Then a[6] is the same as *(p + 1) not *(p + 1 * sizeof(int)). Likewise a[4] is *(p - 1)
Furthermore, you can subtract pointers when they both point to objects in the same array and the same rule applies; the result is in the units of the size of the type pointed to. q - p is 2, not 2 * sizeof(int). Replace the type int in the example with any other type and the p - q will always be 2. For example:
struct Foo { int n ; char x[37] ; };
struct Foo a[10];
struct Foo *p = &(a[5]);
struct Foo *q = &(a[7]);
q - p is still 2. Incidentally, never be tempted to hard code a type's size anywhere. If you are tempted to malloc a struct like this:
struct Foo *r = malloc(41); // int size is 4 + 37 chars
Don't.
Firstly, sizeof(int) is not guaranteed to be 4. Secondly, even if it is, sizeof(struct Foo) is not guaranteed to be 41. Compilers often add padding to struct types to ensure that the members are properly aligned. In this case it is almost a certainty that the compiler will add 3 bytes (or 7 bytes) of padding to the end of struct Foo to ensure that, in arrays, the address of the n member is aligned to the size of an int. always always always use sizeof.
It looks like your understanding how pointer arithmetic works in C is wrong. There is also a problem with data layout assumptions. Finally, there are portability issues and a bad choice of syntax that complicates understanding.
I assume that wit this expression: *( &(sp->aa ) + (i*4) ) you are trying to access the i-th item in the array by taking address of the 0-th item and then adding a byte offset to it. This is wrong of three reasons:
You assume that after sp[0].aa comes sp[1].aa in memory, but you forget that there is sp[0].bb in between.
You assume that size of int is always 4 bytes, which is not true.
You assume that adding an int to secon* will give you a pointer that is offset by specified number of bytes, while in fact it will be offset in specified number of records of size secon.
The second line of output that you see is random junk from unallocated heap memory because when i == 1 your constructions reference memory that is outside of limits allocated for *secon.
To access an i-th item of array referenced by a pointer, use []:
secon[0].aa is the same as (secon +0)->aa, and secon[1].aa is equal to (secon+1)->aa.
This is a complete mess. If you want to access an array of secons, use []
for(int i=0;i<2;i++)
{
sp[i].aa = &first.a; // Same pointer both times
sp[i].bb = &first.b;
}
You have two copies of pointers to the values in first, they point to the same value
for(int i=0;i<2;i++)
{
sp[i].aa = malloc(sizeof(int)); // new pointer each time
*sp[i].aa = first.a; // assigned with the current value
sp[i].bb = malloc(sizeof(int));
*sp[i].bb = first.b;
}
However the compiler is allowed to assume that first does not change, and it is allowed to re-order these expressions, so you are not assured to have different values in your secons
Either way, when you read back the values in second, you can still use []
for(int i=0;i<2;i++)
{
printf("%d %d \n",*sp[i].aa ),*sp[i].bb );
}
Is there a way to get the length of an Array when I only know a pointer pointing to the Array?
See the following example
int testInt[3];
testInt[0] = 0;
testInt[1] = 1;
testInt[2] = 1;
int* point;
point = testInt;
Serial.println(sizeof(testInt) / sizeof(int)); // returns 3
Serial.println(sizeof(point) / sizeof(int)); // returns 1
(This is a snipplet from Arduino Code - I'm sorry, I don't "speak" real C).
The easy answer is no, you cannot. You'll probably want to keep a variable in memory which stores the amount of items in the array.
And there's a not-so-easy answer. There's a way to determine the length of an array, but for that you would have to mark the end of the array with another element, such as -1. Then just loop through it and find this element. The position of this element is the length. However, this won't work with your current code.
Pick one of the above.
Also doing an Arduino project here...
Everybody on the internet seems to insist it's impossible to do this...
and yet the oldest trick in the book seems to work just fine with null terminated arrays...
example for char pointer:
int getSize(char* ch){
int tmp=0;
while (*ch) {
*ch++;
tmp++;
}return tmp;}
magic...
You can infer the length of an array if you have an array variable.
You cannot infer the length of an array if you have just a pointer to it.
You cannot and you should not attempt deduce array length using pointer arithmetic
if in C++ use vector class
You can if you point the the whole array and NOT point to the first element like:
int testInt[3];
int (*point)[3];
point = testInt;
printf( "number elements: %lu", (unsigned long)(sizeof*point/sizeof**point) );
printf( "whole array size: %lu", (unsigned long)(sizeof*point) );
Is there a way to get the length of an Array when I only know a pointer pointing to the Array?
Technically yes, there is a way when code has a true pointer to an array as the array size is in the type as with int (*array_pointer)[3].
This differs from OP's code as the pointer point is not a pointer to an array, but a pointer to an int.
The line point = testInt; converts the array testInt to the address of the first element of the array (which is an int *) and assigns that to point. Thus the array size info is lost.
int testInt[3];
testInt[0] = 0;
testInt[1] = 1;
testInt[2] = 1;
int* point;
point = testInt; // Get the address of testInt[0]
int (*array_pointer)[3] = &testInt; // Get the address of the array
printf("%zu\n", sizeof(testInt) / sizeof(int));
printf("%zu\n", sizeof(point) / sizeof(int));
printf("%zu\n", sizeof(*point) / sizeof(int));
printf("%zu\n", sizeof(*array_pointer) / sizeof(int));
printf("%p\n", (void *) testInt);
printf("%p\n", (void *) point);
printf("%p\n", (void *) array_pointer);
Sample output
3
2
1
3
0xffffcbc4
0xffffcbc4
0xffffcbc4
Pointers point and array_pointer both have values that point to the same location in memory, but the pointers differ in type.
With C99 or later that support variable length arrays, code could have been the below and achieved similar results without explicitly coding a 3 in the pointer definition.
int (*array_pointer_vla)[sizeof testInt/sizeof testInt[0]] = &testInt;
printf("%zu\n", sizeof(*array_pointer_vla) / sizeof(int));
Output
3
I see now see similarities to #user411313 answer. Perhaps the deeper explanation and VLA discussion will be useful.
I think I will understand this concept better if someone can assist me in a current project that I'm working on. I want to use C to edit data at specific memory addresses, using pointers. Specifically, I have two character arrays (strings) which I need to both read data from at specific locations, and also write to at specific locations.
I'm mostly confused about syntax of pointers, such as * and -> and &.
From my understanding, the * refers to the data kept at the current memory address of the pointer. So, for example, if I wanted to edit the data at the beginning memory address of a char *p, I would do something like: (*p) = 'c';
Now, what if I wanted do to alter a character at the 2nd memory address from the beginning of p?
Also, I understand that & refers to the location of the pointer. But I don't know how to use this syntax.
Here is my example:
int orig_length = strlen(original_string); //-1 for \0?
char *poriginal, *pnew_string;
poriginal = &original_string;
while(orig_length>0) {
k = 0;
j = 0;
while(isalpha(*(poriginal+j))) {
j++;
k++;
}
while(k > 0) {
*(pnew_string+(j-k)) = toupper(*(poriginal+k-1)); //toupper
k--;
}
if(*(poriginal+(j)) == '_') {
*(pnew_string+(j)) = ' ';
}
else {
*(pnew_string+(j)) = *(poriginal+(j));
}
orig_length = orig_length - j;
}
*(pnew_string+strlen(pnew_string)) = '\0'; //Syn? Is this actually necessary?
... //program continues...
By the way, this program is meant to take one string "now_i_understand!" and reverse each word, capitalize each word, switch _ to ' ', and leave other punctuation alone: "WON I DNATSREDNU!"
If what you are dealing with is an array of characters (and it is), use array syntax:
pnew_string[j+1] = poriginal[j+1];
Note that this syntax is equivalent to:
*(pnew_string + j + 1) = *(poriginal + j + 1);
but is more readable.
Dealing with most of the other cases you've got should be obvious given this example.
The * and & operators are inverses of each other. A pointer object hold the address of some other object in memory (or it old a null pointer, which doesn't point to any object).
The unary * operator takes a pointer operand, and gives you the object that it points to; this is called dereferencing.
The unary & operator takes an operand that refers to an object of any type, and gives you a pointer to that object. & is the address-of operator.
For example:
int obj = 42; /* obj is an object of type int; it currently holds the value 42 */
int *ptr; /* ptr is a pointer to an int */
ptr = &obj; /* ptr now holds the address of obj */
printf("obj = %d\n", obj); /* prints 42 */
printf("*ptr = %d\n", *ptr); /* also prints 42; *ptr is another name for obj */
The -> operator is shorthand for dereferencing a pointer and accessing a member of what it points to. The prefix must be a pointer to a struct or union. foo->bar means the same thing as (*foo).bar, where foo is a pointer and bar is the name of a member of what foo points to.
You can also perform arithmetic on pointers. If ptr is a pointer pointing to an element of an array, then ptr + 1 points to the next element of the array, ptr + 2 points to the element after that, and so forth.
The [] array indexing operator is actually defined in terms of pointer arithmetic. ptr[2] means exactly the same thing as *(ptr+2). Combine that with the fact that an array name, in most contexts, decays to a pointer to the array's first element, and with a little thought you'll see how arr[2] refers to the third element of the array arr (third because indexing starts at 0).
I strongly recommend sections 4 (Pointers) and 6 (Arrays and Pointers) of the comp.lang.c FAQ; it will likely explain this stuff better than I have.
Is there a way to get the length of an Array when I only know a pointer pointing to the Array?
See the following example
int testInt[3];
testInt[0] = 0;
testInt[1] = 1;
testInt[2] = 1;
int* point;
point = testInt;
Serial.println(sizeof(testInt) / sizeof(int)); // returns 3
Serial.println(sizeof(point) / sizeof(int)); // returns 1
(This is a snipplet from Arduino Code - I'm sorry, I don't "speak" real C).
The easy answer is no, you cannot. You'll probably want to keep a variable in memory which stores the amount of items in the array.
And there's a not-so-easy answer. There's a way to determine the length of an array, but for that you would have to mark the end of the array with another element, such as -1. Then just loop through it and find this element. The position of this element is the length. However, this won't work with your current code.
Pick one of the above.
Also doing an Arduino project here...
Everybody on the internet seems to insist it's impossible to do this...
and yet the oldest trick in the book seems to work just fine with null terminated arrays...
example for char pointer:
int getSize(char* ch){
int tmp=0;
while (*ch) {
*ch++;
tmp++;
}return tmp;}
magic...
You can infer the length of an array if you have an array variable.
You cannot infer the length of an array if you have just a pointer to it.
You cannot and you should not attempt deduce array length using pointer arithmetic
if in C++ use vector class
You can if you point the the whole array and NOT point to the first element like:
int testInt[3];
int (*point)[3];
point = testInt;
printf( "number elements: %lu", (unsigned long)(sizeof*point/sizeof**point) );
printf( "whole array size: %lu", (unsigned long)(sizeof*point) );
Is there a way to get the length of an Array when I only know a pointer pointing to the Array?
Technically yes, there is a way when code has a true pointer to an array as the array size is in the type as with int (*array_pointer)[3].
This differs from OP's code as the pointer point is not a pointer to an array, but a pointer to an int.
The line point = testInt; converts the array testInt to the address of the first element of the array (which is an int *) and assigns that to point. Thus the array size info is lost.
int testInt[3];
testInt[0] = 0;
testInt[1] = 1;
testInt[2] = 1;
int* point;
point = testInt; // Get the address of testInt[0]
int (*array_pointer)[3] = &testInt; // Get the address of the array
printf("%zu\n", sizeof(testInt) / sizeof(int));
printf("%zu\n", sizeof(point) / sizeof(int));
printf("%zu\n", sizeof(*point) / sizeof(int));
printf("%zu\n", sizeof(*array_pointer) / sizeof(int));
printf("%p\n", (void *) testInt);
printf("%p\n", (void *) point);
printf("%p\n", (void *) array_pointer);
Sample output
3
2
1
3
0xffffcbc4
0xffffcbc4
0xffffcbc4
Pointers point and array_pointer both have values that point to the same location in memory, but the pointers differ in type.
With C99 or later that support variable length arrays, code could have been the below and achieved similar results without explicitly coding a 3 in the pointer definition.
int (*array_pointer_vla)[sizeof testInt/sizeof testInt[0]] = &testInt;
printf("%zu\n", sizeof(*array_pointer_vla) / sizeof(int));
Output
3
I see now see similarities to #user411313 answer. Perhaps the deeper explanation and VLA discussion will be useful.