Related
This is well known code to compute array length in C:
sizeof(array)/sizeof(type)
But I can't seem to find out the length of the array passed as an argument to a function:
#include <stdio.h>
int length(const char* array[]) {
return sizeof(array)/sizeof(char*);
}
int main() {
const char* friends[] = { "John", "Jack", "Jim" };
printf("%d %d", sizeof(friends)/sizeof(char*), length(friends)); // 3 1
}
I assume that array is copied by value to the function argument as constant pointer and reference to it should solve this, but this declaration is not valid:
int length(const char**& array);
I find passing the array length as second argument to be redundant information, but why is the standard declaration of main like this:
int main(int argc, char** argv);
Please explain if it is possible to find out the array length in function argument, and if so, why is there the redundancy in main.
sizeof only works to find the length of the array if you apply it to the original array.
int a[5]; //real array. NOT a pointer
sizeof(a); // :)
However, by the time the array decays into a pointer, sizeof will give the size of the pointer and not of the array.
int a[5];
int * p = a;
sizeof(p); // :(
As you have already smartly pointed out main receives the length of the array as an argument (argc). Yes, this is out of necessity and is not redundant. (Well, it is kind of reduntant since argv is conveniently terminated by a null pointer but I digress)
There is some reasoning as to why this would take place. How could we make things so that a C array also knows its length?
A first idea would be not having arrays decaying into pointers when they are passed to a function and continuing to keep the array length in the type system. The bad thing about this is that you would need to have a separate function for every possible array length and doing so is not a good idea. (Pascal did this and some people think this is one of the reasons it "lost" to C)
A second idea is storing the array length next to the array, just like any modern programming language does:
a -> [5];[0,0,0,0,0]
But then you are just creating an invisible struct behind the scenes and the C philosophy does not approve of this kind of overhead. That said, creating such a struct yourself is often a good idea for some sorts of problems:
struct {
size_t length;
int * elements;
}
Another thing you can think about is how strings in C are null terminated instead of storing a length (as in Pascal). To store a length without worrying about limits need a whopping four bytes, an unimaginably expensive amount (at least back then). One could wonder if arrays could be also null terminated like that but then how would you allow the array to store a null?
The array decays to a pointer when passed.
Section 6.4 of the C FAQ covers this very well and provides the K&R references etc.
That aside, imagine it were possible for the function to know the size of the memory allocated in a pointer. You could call the function two or more times, each time with different input arrays that were potentially different lengths; the length would therefore have to be passed in as a secret hidden variable somehow. And then consider if you passed in an offset into another array, or an array allocated on the heap (malloc and all being library functions - something the compiler links to, rather than sees and reasons about the body of).
Its getting difficult to imagine how this might work without some behind-the-scenes slice objects and such right?
Symbian did have a AllocSize() function that returned the size of an allocation with malloc(); this only worked for the literal pointer returned by the malloc, and you'd get gobbledygook or a crash if you asked it to know the size of an invalid pointer or a pointer offset from one.
You don't want to believe its not possible, but it genuinely isn't. The only way to know the length of something passed into a function is to track the length yourself and pass it in yourself as a separate explicit parameter.
As stated by #Will, the decay happens during the parameter passing. One way to get around it is to pass the number of elements. To add onto this, you may find the _countof() macro useful - it does the equivalent of what you've done ;)
First, a better usage to compute number of elements when the actual array declaration is in scope is:
sizeof array / sizeof array[0]
This way you don't repeat the type name, which of course could change in the declaration and make you end up with an incorrect length computation. This is a typical case of don't repeat yourself.
Second, as a minor point, please note that sizeof is not a function, so the expression above doesn't need any parenthesis around the argument to sizeof.
Third, C doesn't have references so your usage of & in a declaration won't work.
I agree that the proper C solution is to pass the length (using the size_t type) as a separate argument, and use sizeof at the place the call is being made if the argument is a "real" array.
Note that often you work with memory returned by e.g. malloc(), and in those cases you never have a "true" array to compute the size off of, so designing the function to use an element count is more flexible.
Regarding int main():
According to the Standard, argv points to a NULL-terminated array (of pointers to null-terminated strings). (5.1.2.2.1:1).
That is, argv = (char **){ argv[0], ..., argv[argc - 1], 0 };.
Hence, size calculation is performed by a function which is a trivial modification of strlen().
argc is only there to make argv length calculation O(1).
The count-until-NULL method will NOT work for generic array input. You will need to manually specify size as a second argument.
This is a old question, and the OP seems to mix C++ and C in his intends/examples. In C, when you pass a array to a function, it's decayed to pointer. So, there is no way to pass the array size except by using a second argument in your function that stores the array size:
void func(int A[])
// should be instead: void func(int * A, const size_t elemCountInA)
They are very few cases, where you don't need this, like when you're using multidimensional arrays:
void func(int A[3][whatever here]) // That's almost as if read "int* A[3]"
Using the array notation in a function signature is still useful, for the developer, as it might be an help to tell how many elements your functions expects. For example:
void vec_add(float out[3], float in0[3], float in1[3])
is easier to understand than this one (although, nothing prevent accessing the 4th element in the function in both functions):
void vec_add(float * out, float * in0, float * in1)
If you were to use C++, then you can actually capture the array size and get what you expect:
template <size_t N>
void vec_add(float (&out)[N], float (&in0)[N], float (&in1)[N])
{
for (size_t i = 0; i < N; i++)
out[i] = in0[i] + in1[i];
}
In that case, the compiler will ensure that you're not adding a 4D vector with a 2D vector (which is not possible in C without passing the dimension of each dimension as arguments of the function). There will be as many instance of the vec_add function as the number of dimensions used for your vectors.
int arsize(int st1[]) {
int i = 0;
for (i; !(st1[i] & (1 << 30)); i++);
return i;
}
This works for me :)
length of an array(type int) with sizeof:
sizeof(array)/sizeof(int)
Best example is here
thanks #define SIZE 10
void size(int arr[SIZE])
{
printf("size of array is:%d\n",sizeof(arr));
}
int main()
{
int arr[SIZE];
size(arr);
return 0;
}
This is well known code to compute array length in C:
sizeof(array)/sizeof(type)
But I can't seem to find out the length of the array passed as an argument to a function:
#include <stdio.h>
int length(const char* array[]) {
return sizeof(array)/sizeof(char*);
}
int main() {
const char* friends[] = { "John", "Jack", "Jim" };
printf("%d %d", sizeof(friends)/sizeof(char*), length(friends)); // 3 1
}
I assume that array is copied by value to the function argument as constant pointer and reference to it should solve this, but this declaration is not valid:
int length(const char**& array);
I find passing the array length as second argument to be redundant information, but why is the standard declaration of main like this:
int main(int argc, char** argv);
Please explain if it is possible to find out the array length in function argument, and if so, why is there the redundancy in main.
sizeof only works to find the length of the array if you apply it to the original array.
int a[5]; //real array. NOT a pointer
sizeof(a); // :)
However, by the time the array decays into a pointer, sizeof will give the size of the pointer and not of the array.
int a[5];
int * p = a;
sizeof(p); // :(
As you have already smartly pointed out main receives the length of the array as an argument (argc). Yes, this is out of necessity and is not redundant. (Well, it is kind of reduntant since argv is conveniently terminated by a null pointer but I digress)
There is some reasoning as to why this would take place. How could we make things so that a C array also knows its length?
A first idea would be not having arrays decaying into pointers when they are passed to a function and continuing to keep the array length in the type system. The bad thing about this is that you would need to have a separate function for every possible array length and doing so is not a good idea. (Pascal did this and some people think this is one of the reasons it "lost" to C)
A second idea is storing the array length next to the array, just like any modern programming language does:
a -> [5];[0,0,0,0,0]
But then you are just creating an invisible struct behind the scenes and the C philosophy does not approve of this kind of overhead. That said, creating such a struct yourself is often a good idea for some sorts of problems:
struct {
size_t length;
int * elements;
}
Another thing you can think about is how strings in C are null terminated instead of storing a length (as in Pascal). To store a length without worrying about limits need a whopping four bytes, an unimaginably expensive amount (at least back then). One could wonder if arrays could be also null terminated like that but then how would you allow the array to store a null?
The array decays to a pointer when passed.
Section 6.4 of the C FAQ covers this very well and provides the K&R references etc.
That aside, imagine it were possible for the function to know the size of the memory allocated in a pointer. You could call the function two or more times, each time with different input arrays that were potentially different lengths; the length would therefore have to be passed in as a secret hidden variable somehow. And then consider if you passed in an offset into another array, or an array allocated on the heap (malloc and all being library functions - something the compiler links to, rather than sees and reasons about the body of).
Its getting difficult to imagine how this might work without some behind-the-scenes slice objects and such right?
Symbian did have a AllocSize() function that returned the size of an allocation with malloc(); this only worked for the literal pointer returned by the malloc, and you'd get gobbledygook or a crash if you asked it to know the size of an invalid pointer or a pointer offset from one.
You don't want to believe its not possible, but it genuinely isn't. The only way to know the length of something passed into a function is to track the length yourself and pass it in yourself as a separate explicit parameter.
As stated by #Will, the decay happens during the parameter passing. One way to get around it is to pass the number of elements. To add onto this, you may find the _countof() macro useful - it does the equivalent of what you've done ;)
First, a better usage to compute number of elements when the actual array declaration is in scope is:
sizeof array / sizeof array[0]
This way you don't repeat the type name, which of course could change in the declaration and make you end up with an incorrect length computation. This is a typical case of don't repeat yourself.
Second, as a minor point, please note that sizeof is not a function, so the expression above doesn't need any parenthesis around the argument to sizeof.
Third, C doesn't have references so your usage of & in a declaration won't work.
I agree that the proper C solution is to pass the length (using the size_t type) as a separate argument, and use sizeof at the place the call is being made if the argument is a "real" array.
Note that often you work with memory returned by e.g. malloc(), and in those cases you never have a "true" array to compute the size off of, so designing the function to use an element count is more flexible.
Regarding int main():
According to the Standard, argv points to a NULL-terminated array (of pointers to null-terminated strings). (5.1.2.2.1:1).
That is, argv = (char **){ argv[0], ..., argv[argc - 1], 0 };.
Hence, size calculation is performed by a function which is a trivial modification of strlen().
argc is only there to make argv length calculation O(1).
The count-until-NULL method will NOT work for generic array input. You will need to manually specify size as a second argument.
This is a old question, and the OP seems to mix C++ and C in his intends/examples. In C, when you pass a array to a function, it's decayed to pointer. So, there is no way to pass the array size except by using a second argument in your function that stores the array size:
void func(int A[])
// should be instead: void func(int * A, const size_t elemCountInA)
They are very few cases, where you don't need this, like when you're using multidimensional arrays:
void func(int A[3][whatever here]) // That's almost as if read "int* A[3]"
Using the array notation in a function signature is still useful, for the developer, as it might be an help to tell how many elements your functions expects. For example:
void vec_add(float out[3], float in0[3], float in1[3])
is easier to understand than this one (although, nothing prevent accessing the 4th element in the function in both functions):
void vec_add(float * out, float * in0, float * in1)
If you were to use C++, then you can actually capture the array size and get what you expect:
template <size_t N>
void vec_add(float (&out)[N], float (&in0)[N], float (&in1)[N])
{
for (size_t i = 0; i < N; i++)
out[i] = in0[i] + in1[i];
}
In that case, the compiler will ensure that you're not adding a 4D vector with a 2D vector (which is not possible in C without passing the dimension of each dimension as arguments of the function). There will be as many instance of the vec_add function as the number of dimensions used for your vectors.
int arsize(int st1[]) {
int i = 0;
for (i; !(st1[i] & (1 << 30)); i++);
return i;
}
This works for me :)
length of an array(type int) with sizeof:
sizeof(array)/sizeof(int)
Best example is here
thanks #define SIZE 10
void size(int arr[SIZE])
{
printf("size of array is:%d\n",sizeof(arr));
}
int main()
{
int arr[SIZE];
size(arr);
return 0;
}
While working on a CS50 problem set (substitution), I encountered a segmentation fault when running the code. After some searching I found out that assigning memory (malloc) to string "output" fixes the issue. However I wanted to understand why assigning memory is necessary here?
Any explanation would be appreciated.
code extract: -please note i am using the cs50.h library
string cipher(string input, string key) {
string output=malloc(strlen(input)+1);
for (int i=0, len = strlen(input); i<len; i++) {
if(isalpha(input[i]) != 0) {
output[i] = substitute(input[i], key);
}
else {
output[i] = input[i];
}
}
return output;
free(output);
}
As much as I could know about CS50.h string, I came to know that
string output
is just declaring a character pointer named output. So what actually happens in your code is that until you explicitly declare the "strlen(input)+1" contiguous memory locations beloging to output only, they are essentially free locations for the program. Hence, your output pointer will contain only the character at the 0th index. The function returns what "output" actually is, a pointer. Some process within the program meanwhile may make use of all the other memory locations other than output[0], since they never belonged to output string. output only pointed to the first character of some string.
Bound checking in C, C++ is essentially done by the programmer. output[i] for any arbitrary i will never give an error because it's a simple pointer arithmetic for the programmer, i.e. output[i] = *(output+i).
:)
For starters this statement
free(output);
never gets the control because it is placed after the return statement
return output;
and moreover it does not make a sense. It is the caller of the function that is responsible to free the allocated memory.
You need to dynamically allocated memory because otherwise if you will declare a variable length array like
char output[strlen(input)+1];
then after exiting the function it will not be alive and an attempt to access the array outside the function results in undefined behavior.
That is if you will just write
string output;
that is equivalent to
char *output;
then the pointer output has indeterminate value because it was not initialized and neither memory was allocated where the source string will be copied.
You could change the source string input in place without creating one more array.
In this case it would be enough to write
if(isalpha(input[i]) != 0) {
input[i] = substitute(input[i], key);
}
and then you could place the statement
return input;
Pay attention to that using the alias string for the type char * is a bad idea.
The function declaration if to rewrite it like
char * cipher(char *input, char *key);
is confusing. It is unclear whether the strings input and key are being changed within the function or not.
If you want that the function returns a new string build from the source string then the function declaration should look like
char * cipher(const char *input, const char *key);
Thus answering your question
However I wanted to understand why assigning memory is necessary here?
if you want to create a new string from the source string pointed to by the pointer input then it is evident that you need to allocate a new character array where elements of the source string will be copied.
Otherwise if you want to change the source string in place then there is no need to create one more array.
This is well known code to compute array length in C:
sizeof(array)/sizeof(type)
But I can't seem to find out the length of the array passed as an argument to a function:
#include <stdio.h>
int length(const char* array[]) {
return sizeof(array)/sizeof(char*);
}
int main() {
const char* friends[] = { "John", "Jack", "Jim" };
printf("%d %d", sizeof(friends)/sizeof(char*), length(friends)); // 3 1
}
I assume that array is copied by value to the function argument as constant pointer and reference to it should solve this, but this declaration is not valid:
int length(const char**& array);
I find passing the array length as second argument to be redundant information, but why is the standard declaration of main like this:
int main(int argc, char** argv);
Please explain if it is possible to find out the array length in function argument, and if so, why is there the redundancy in main.
sizeof only works to find the length of the array if you apply it to the original array.
int a[5]; //real array. NOT a pointer
sizeof(a); // :)
However, by the time the array decays into a pointer, sizeof will give the size of the pointer and not of the array.
int a[5];
int * p = a;
sizeof(p); // :(
As you have already smartly pointed out main receives the length of the array as an argument (argc). Yes, this is out of necessity and is not redundant. (Well, it is kind of reduntant since argv is conveniently terminated by a null pointer but I digress)
There is some reasoning as to why this would take place. How could we make things so that a C array also knows its length?
A first idea would be not having arrays decaying into pointers when they are passed to a function and continuing to keep the array length in the type system. The bad thing about this is that you would need to have a separate function for every possible array length and doing so is not a good idea. (Pascal did this and some people think this is one of the reasons it "lost" to C)
A second idea is storing the array length next to the array, just like any modern programming language does:
a -> [5];[0,0,0,0,0]
But then you are just creating an invisible struct behind the scenes and the C philosophy does not approve of this kind of overhead. That said, creating such a struct yourself is often a good idea for some sorts of problems:
struct {
size_t length;
int * elements;
}
Another thing you can think about is how strings in C are null terminated instead of storing a length (as in Pascal). To store a length without worrying about limits need a whopping four bytes, an unimaginably expensive amount (at least back then). One could wonder if arrays could be also null terminated like that but then how would you allow the array to store a null?
The array decays to a pointer when passed.
Section 6.4 of the C FAQ covers this very well and provides the K&R references etc.
That aside, imagine it were possible for the function to know the size of the memory allocated in a pointer. You could call the function two or more times, each time with different input arrays that were potentially different lengths; the length would therefore have to be passed in as a secret hidden variable somehow. And then consider if you passed in an offset into another array, or an array allocated on the heap (malloc and all being library functions - something the compiler links to, rather than sees and reasons about the body of).
Its getting difficult to imagine how this might work without some behind-the-scenes slice objects and such right?
Symbian did have a AllocSize() function that returned the size of an allocation with malloc(); this only worked for the literal pointer returned by the malloc, and you'd get gobbledygook or a crash if you asked it to know the size of an invalid pointer or a pointer offset from one.
You don't want to believe its not possible, but it genuinely isn't. The only way to know the length of something passed into a function is to track the length yourself and pass it in yourself as a separate explicit parameter.
As stated by #Will, the decay happens during the parameter passing. One way to get around it is to pass the number of elements. To add onto this, you may find the _countof() macro useful - it does the equivalent of what you've done ;)
First, a better usage to compute number of elements when the actual array declaration is in scope is:
sizeof array / sizeof array[0]
This way you don't repeat the type name, which of course could change in the declaration and make you end up with an incorrect length computation. This is a typical case of don't repeat yourself.
Second, as a minor point, please note that sizeof is not a function, so the expression above doesn't need any parenthesis around the argument to sizeof.
Third, C doesn't have references so your usage of & in a declaration won't work.
I agree that the proper C solution is to pass the length (using the size_t type) as a separate argument, and use sizeof at the place the call is being made if the argument is a "real" array.
Note that often you work with memory returned by e.g. malloc(), and in those cases you never have a "true" array to compute the size off of, so designing the function to use an element count is more flexible.
Regarding int main():
According to the Standard, argv points to a NULL-terminated array (of pointers to null-terminated strings). (5.1.2.2.1:1).
That is, argv = (char **){ argv[0], ..., argv[argc - 1], 0 };.
Hence, size calculation is performed by a function which is a trivial modification of strlen().
argc is only there to make argv length calculation O(1).
The count-until-NULL method will NOT work for generic array input. You will need to manually specify size as a second argument.
This is a old question, and the OP seems to mix C++ and C in his intends/examples. In C, when you pass a array to a function, it's decayed to pointer. So, there is no way to pass the array size except by using a second argument in your function that stores the array size:
void func(int A[])
// should be instead: void func(int * A, const size_t elemCountInA)
They are very few cases, where you don't need this, like when you're using multidimensional arrays:
void func(int A[3][whatever here]) // That's almost as if read "int* A[3]"
Using the array notation in a function signature is still useful, for the developer, as it might be an help to tell how many elements your functions expects. For example:
void vec_add(float out[3], float in0[3], float in1[3])
is easier to understand than this one (although, nothing prevent accessing the 4th element in the function in both functions):
void vec_add(float * out, float * in0, float * in1)
If you were to use C++, then you can actually capture the array size and get what you expect:
template <size_t N>
void vec_add(float (&out)[N], float (&in0)[N], float (&in1)[N])
{
for (size_t i = 0; i < N; i++)
out[i] = in0[i] + in1[i];
}
In that case, the compiler will ensure that you're not adding a 4D vector with a 2D vector (which is not possible in C without passing the dimension of each dimension as arguments of the function). There will be as many instance of the vec_add function as the number of dimensions used for your vectors.
int arsize(int st1[]) {
int i = 0;
for (i; !(st1[i] & (1 << 30)); i++);
return i;
}
This works for me :)
length of an array(type int) with sizeof:
sizeof(array)/sizeof(int)
Best example is here
thanks #define SIZE 10
void size(int arr[SIZE])
{
printf("size of array is:%d\n",sizeof(arr));
}
int main()
{
int arr[SIZE];
size(arr);
return 0;
}
I'm just learning C and was wondering which one of these I should use in my main method. Is there any difference? Which one is more common?
As you are just learning C, I recommend you to really try to understand the differences between arrays and pointers first instead of the common things.
In the area of parameters and arrays, there are a few confusing rules that should be clear before going on. First, what you declare in a parameter list is treated special. There are such situations where things don't make sense as a function parameter in C. These are
Functions as parameters
Arrays as parameters
Arrays as parameters
The second maybe is not immediately clear. But it becomes clear when you consider that the size of an array dimension is part of the type in C (and an array whose dimension size isn't given has an incomplete type). So, if you would create a function that takes an array by-value (receives a copy), then it could do so only for one size! In addition, arrays can become large, and C tries to be as fast as possible.
In C, for these reasons, array-values are not existent. If you want to get the value of an array, what you get instead is a pointer to the first element of that array. And herein actually already lies the solution. Instead of drawing an array parameter invalid up-front, a C compiler will transform the type of the respective parameter to be a pointer. Remember this, it's very important. The parameter won't be an array, but instead it will be a pointer to the respective element type.
Now, if you try to pass an array, what is passed instead is a pointer to the arrays' first element.
Excursion: Functions as parameters
For completion, and because I think this will help you better understand the matter, let's look what the state of affairs is when you try to have a function as a parameter. Indeed, first it won't make any sense. How can a parameter be a function? Huh, we want a variable at that place, of course! So what the compiler does when that happens is, again, to transform the function into a function pointer. Trying to pass a function will pass a pointer to that respective function instead. So, the following are the same (analogous to the array example):
void f(void g(void));
void f(void (*g)(void));
Note that parentheses around *g is needed. Otherwise, it would specify a function returning void*, instead of a pointer to a function returning void.
Back to arrays
Now, I said at the beginning that arrays can have an incomplete type - which happens if you don't give a size yet. Since we already figured that an array parameter is not existent but instead any array parameter is a pointer, the array's size doesn't matter. That means, the compiler will translate all of the following, and all are the same thing:
int main(int c, char **argv);
int main(int c, char *argv[]);
int main(int c, char *argv[1]);
int main(int c, char *argv[42]);
Of course, it doesn't make much sense to be able to put any size in it, and it's just thrown away. For that reason, C99 came up with a new meaning for those numbers, and allows other things to appear between the brackets:
// says: argv is a non-null pointer pointing to at least 5 char*'s
// allows CPU to pre-load some memory.
int main(int c, char *argv[static 5]);
// says: argv is a constant pointer pointing to a char*
int main(int c, char *argv[const]);
// says the same as the previous one
int main(int c, char ** const argv);
The last two lines say that you won't be able to change "argv" within the function - it has become a const pointer. Only few C compilers support those C99 features though. But these features make it clear that the "array" isn't actually one. It's a pointer.
A word of warning
Note that all i said above is true only when you have got an array as a parameter of a function. If you work with local arrays, an array won't be a pointer. It will behave as a pointer, because as explained earlier an array will be converted to a pointer when its value is read. But it should not be confused with pointers.
One classic example is the following:
char c[10];
char **c = &c; // does not work.
typedef char array[10];
array *pc = &c; // *does* work.
// same without typedef. Parens needed, because [...] has
// higher precedence than '*'. Analogous to the function example above.
char (*array)[10] = &c;
You could use either. They're completely equivalent. See litb's comments and his answer.
It really depends how you want to use it (and you could use either in any case):
// echo-with-pointer-arithmetic.c
#include <stdio.h>
int main(int argc, char **argv)
{
while (--argc > 0)
{
printf("%s ", *++argv);
}
printf("\n");
return 0;
}
// echo-without-pointer-arithmetic.c
#include <stdio.h>
int main(int argc, char *argv[])
{
int i;
for (i=1; i<argc; i++)
{
printf("%s ", argv[i]);
}
printf("\n");
return 0;
}
As for which is more common - it doesn't matter. Any experienced C programmer reading your code will see both as interchangeable (under the right conditions). Just like an experienced English speaker reads "they're" and "they are" equally easily.
More important is that you learn to read them and recognize how similar they are. You'll be reading more code than you write, and you'll need to be equally comfortable with both.
It doesn't make a difference, but I use char *argv[] because it shows that is a fixed size array of variable length strings (which are usually char *).
You can use either of the two forms, as in C arrays and pointers are interchangeable in function parameter lists. See http://en.wikipedia.org/wiki/C_(programming_language)#Array-pointer_interchangeability.
It doesn't really make a difference, but the latter is more readable. What you are given is an array of char pointers, like the second version says. It can be implicitly converted to a double char pointer like in the first version however.
you should declare it as char *argv[], because of all the many equivalent ways of declaring it, that comes closest to its intuitive meaning: an array of strings.
char ** → pointer to character pointer and char *argv [] means array of character pointers. As we can use pointer instead of an array, both can be used.
I see no special merit of using either approach instead of the other -- use the convention that is most in line with the rest of your code.
If you'll need a varying or dynamic number of strings, char** might be easier to work with. If you're number of string is fixed though, char* var[] would be preferred.
I know this is outdated, but if you are just learning the C programming language and not doing anything major with it, don't use command-line options.
If you are not using command line arguments, don't use either. Just declare the main function as int main()
If you
Want the user of your program to be able to drag a file onto your program so that you can change the outcome of your program with it or
Want to handle command-line options(-help, /?, or any other thing that goes after program name in terminal or command prompt)
use whichever makes more sense to you.
Otherwise, just use int main()
After all, if you end up wanting to add command-line options, you can easily edit them in later.