For pointers, I'm getting confused with declarations and function parameters on when to use char ** or char * or *array[n], etc. Like if a function takes a (*array[n]) parameter, do I pass it a **type?
I try using the Right-Left rule and know that p would be a pointer to a pointer to a char (char **p), and p is an array of n pointers (*p[n]), but someone said that *p[n] and **p are essentially equivalent. Is that true?
In the correct context (namely, arguments to a function), then the following declarations are equivalent:
int main(int argc, char *argv[]);
int main(int argc, char **argv);
int main(int argc, char *argv[12]); // Very aconventional!
Similar comments apply to the function definitions (which have a block enclosed in braces in place of the semi-colon).
In any other context, there are important differences between the notations. For example:
extern char *list1[];
extern char **list2;
extern char *list3[12];
The first says that somewhere there is an array of indeterminate size containing 'char *' values. The second says that somewhere - possibly here - there is a single value containing a pointer to a char pointer. The third says that somewhere - possibly here - there is an array of 12 character pointers.
However, all the three lists can be referenced in somewhat the same way - assuming that they actually have been defined and initialized.
list1[0][0] = '1';
list2[0][0] = '2';
list3[0][0] = '3';
Further, if they are passed into a function like this:
function(list1, list2, list3);
then the function can be declared as:
void function(char **list1, char **list2, char **list3);
The arrays (list1, list3) decay from the array to the pointer to the first element of the array; list2, of course, is already a pointer to a pointer.
One detail to note in a function such as:
void otherfunction(char *list[12])
{
...
}
The C compiler does not treat that declaration any differently from:
void otherfunction(char **list)
{
...
}
or
void otherfunction(char *list[])
{
...
}
In particular, it does no array bounds checking, and as far as the function is concerned, the 12 may as well be absent.
C99 introduces VLA (variable length array) types and also introduces a notation with 'static' and a size in the array bounds. You would need to read the standard to understand those fully.
Suffice to say in a function like the following the size of the array does matter, and is determined at run-time. With two-dimensional arrays in general, all the dimensions except the first need to be specified.
void vla_function(size_t m, int vla[m][m]);
Quoting from the standard (section 6.7.5.3):
void f(double (* restrict a)[5]);
void f(double a[restrict][5]);
void f(double a[restrict 3][5]);
void f(double a[restrict static 3][5]);
(Note that the last declaration also specifies that the argument corresponding to a in any call to f must be a
non-null pointer to the first of at least three arrays of 5 doubles, which the others do not.)
Reading C declarators (that's the part of the variable with the * and []) is fairly nuanced. There are some websites with tips:
http://www.antlr.org/wiki/display/CS652/How+To+Read+C+Declarations
http://www.ericgiguere.com/articles/reading-c-declarations.html
A char** is a pointer to (possible multiple) pointer(s) to (possibly multiple) char(s). For example, it might be a pointer to a string pointer, or a pointer to an array of string pointers.
A char*[] is an array of pointers to char. When you have a function that takes this as a parameter, the C compiler makes it "decay" into a char**. This only happens to the first layer... so, taking a complicated example, char*[4][] becomes char*(*)[4]. Read the links above so you can understand what the heck that means.
Or you can do a (very sensible) thing and make a bunch of typedefs. I don't do this, but until you're good at reading declarators, it's a good idea.
typedef char * stringp;
void func(stringp array[]) { ... }
static stringp FOUR_STRINGS[4] = { ... };
If n==0 then they reference the same memory. Array indexing is basically a pointer plus an offset. *(p[n]) would be the same as **(p+n). You can see for yourself how simple this is in C, because array[4] and 4[array] will give you the same thing.
Related
#include <stdio.h>
void lower_string(char*);
int main()
{
char string[100];
printf("Enter a string to convert it into lower case\n");
gets(string);
lower_string(string);
printf("Entered string in lower case is \"%s\"\n", string);
return 0;
}
void lower_string(char *string)
{
while(*string)
{
if ( *string >= 'A' && *string <= 'Z' )
{
*string = *string + 32;
}
string++;
}
}
In this program what if i replace *string with string[]?
can anyone help me in writing the above program without using any function?
And please explain what does this mean while(*str) ?
void lower_string(char *string);
and
void lower_string(char string[]);
are equivalent in C. A parameter of type char [] is adjusted to type char *.
Of course when string is the operand of the * operator like in:
while (*string)
then you cannot change it to string[] as * here is the indirection operator and not a part of a type name.
There is no difference between the two, because when an array is passed to a function, it decays adjusts to a pointer.
I wouldn't write such code and then tag it C++, because string is a standard library class name.
C-FAQ: Q-6.4:
Since arrays decay immediately into pointers, an array is never actually passed to a function. You can pretend that a function receives an array as a parameter, and illustrate it by declaring the corresponding parameter as an array:
void f(char a[])
{ ... }
Interpreted literally, this declaration would have no use, so the compiler turns around and pretends that you'd written a pointer declaration, since that's what the function will in fact receive:
void f(char *a)
{ ... }
There's nothing particularly wrong with talking about a function as if it ``receives'' an array, if the function is traditionally used to operate on arrays, or if the parameter is naturally treated within the function as an array.
This conversion of array-like declarators into pointers holds only within function formal parameter declarations, nowhere else.
The symbol [] is basically equivalent to * in some way.
You can accept char[] and still treat it as a pointer with * and you can accept a char* as a parameter and access it with [] as a regular array - it's exactly the same.
You're always getting a pointer to that spot in memory, when it's an array defined with [] though, the pointer will be constant (can't deference it, only access where it's pointing to).
When you're declaring an array using char str[100]; you're actually declaring a const char* named str and pointing to a memory area in the stack.
They are pretty much the same thing. For example,
sz[x];
is short for
*(sz + x);
However, when you write char *sz;, all yau are getting is a pointer. Wen you write char sz[size];, you are getting an array also. See haccks' answer.
I am confused as to how the following passage matches up with the code that follows it:
Since argv is a pointer to an array of pointers, we can manipulate the
pointer rather than index the array. This next variant is based on
incrementing argv, which is a pointer to pointer to char, while argc
is counted down:
#include <stdio.h>
/* echo command-line arguments; 2nd version */
main(int argc, char *argv[])
{
while (--argc > 0)
printf("%s%s", *++argv, (argc > 1) ? " " : "");
printf("\n");
return 0;
}
Isn't char *argv[] just an array of pointers? Wouldn't a pointer to an array of pointers be written as char *(*argv[]) or something similar?
As a side note, is it normal that in general I find declarations that mix arrays and pointers rather confusing?
Such terms as "pointer to array" or "to point to an array" are often treated rather loosely in C terminology. They can mean at least two different things.
In the most strict and pedantic sense of the term, a "pointer to array" has to be declared with "pointer to array" type, as in
int a[10];
int (*p)[10] = &a;
In the above example p is declared as a pointer to array of 10 ints and it is actually initialized to point to such an array.
However, the term is also often used is its less formal meaning. In this example
int a[10];
int *p = &a;
p is declared as a mere pointer to int. It is initialized to point to the first element of array a. You can often hear and see people say that p in this case also "points to an array" of ints, even though this situation is semantically different from previous one. "Points to an array" in this case means "provides access to elements of an array through pointer arithmetic", as in p[5] or *(p + 3).
This is exactly what is meant by the phrase "...argv is a pointer to an array of pointers..." you quoted. argv's declaration in parameter list of main is equivalent to char **argv, meaning that argv is actually a pointer to a char * pointer. But since it physically points to the first element of some array of char * pointers (maintained by the calling code), it is correct to say semi-informally that argv points to an array of pointers.
That's exactly what is meant by the text you quoted.
Where C functions claim to accept arrays, strictly they accept pointers instead. The language does not distinguish between void fn(int *foo) {} and void fn(int foo[]). It doesn't even care if you have void fn(int foo[100]) and then pass that an array of int [10].
int main(int argc, char *argv[])
is the same as
int main(int argc, char **argv)
Consequently, argv points to the first element of an array of char pointers, but it is not itself an array type and it does not (formally) point to a whole array. But we know that array is there, and we can index into it to get the other elements.
In more complex cases, like accepting multi-dimensional arrays, it is only the first [] which drops back to a pointer (and which can be left unsized). The others remain as part of the type that is being pointed to, and they have an influence on pointer arithmetic.
The array-pointer equivalence thing only holds true only for function arguments, so while void fn(const char* argv[]) and void fn(const char** argv) are equivalent, it doesn't hold true when it comes to the variables you might want to pass TO the function.
Consider
void fn(const char** argv)
{
...
}
int main(int argc, const char* argv[])
{
fn(argv); // acceptable.
const char* meats[] = { "Chicken", "Cow", "Pizza" };
// "meats" is an array of const char* pointers, just like argv, so
fn(meats); // acceptable.
const char** meatPtr = meats;
fn(meatPtr); // because the previous call actually cast to this,.
// an array of character arrays.
const char vegetables[][10] = { "Avocado", "Pork", "Pepperoni" };
fn(vegetables); // does not compile.
return 0;
}
"vegetables" is not a pointer to a pointer, it points directly to the first character in a 3*10 contiguous character sequence. Replace fn(vegetables) in the above to get
int main(int argc, const char* argv[])
{
// an array of character arrays.
const char vegetables[][10] = { "Avocado", "Pork", "Pepperoni" };
printf("*vegetables = %c\n", *(const char*)vegetables);
return 0;
}
and the output is "A": vegetables itself is pointing directly - without indirection - to the characters, and not intermediate pointers.
The vegetables assignment is basically a shortcut for this:
const char* __vegetablesPtr = "Avocado\0\0\0Pork\0\0\0\0\0\0Pepperoni\0";
vegetables = __vegetablesPtr;
and
const char* roni = vegetables[2];
translates to
const char* roni = (&vegetables[0]) + (sizeof(*vegetables[0]) * /*dimension=*/10 * /*index=*/2);
Since argv is a pointer to an array of pointers.
This is wrong. argv is an array of pointers.
Since argv is a pointer to an array of pointers,
No, not even close.
Isn't char *argv[] just an array of pointers?
No, it's a pointer to pointers.
"Pointer to the first element of an array" is a common construct. Every string function uses it, including stdio functions that input and output strings. main uses it for argv.
"Pointer to an array" is a rare construct. I can't find any uses of it in the C standard library or POSIX. grepping all the headers I have installed locally (for '([^)]*\*[^)]) *\[') I find exactly 2 legitimate instances of pointer-to-array, one in libjpeg and one in gtk. (Both are struct members, not function parameters, but that's beside the point.)
So if we stick to official language, we have a rare thing with a short name and a similar but much more common thing with a long name. That's the opposite of the way human language naturally wants to work, so there's tension, which gets resolved in all but the most formal situations by using the short name "incorrectly".
The reason we don't just say "pointer to pointer" is that there's another common use of pointers as function parameters, in which the parameter points to a single object that's not a member of an array. For example, in
long strtol(const char *nptr, char **endptr, int base);
endptr is exactly the same type as argv is in main, both are pointer-to-pointer, but they're used in different ways. argv points to the first char * in an array of char *s; inside main you're expected to use it with indexes like argv[0], argv[optind], etc., or step through the array by incrementing it with ++argv.
endptr points to a single char *. Inside strtol, it is not useful to increment endptr or to refer to endptr[n] for any value of n other than zero.
That's semantic difference is expressed by the informal usage of "argv is a pointer to an array". The possible confusion with what "pointer to array" means in formal language is ignored, because the natural instinct to use concise language is stronger than the desire to adhere to a formal definition that tells you not to use the most obvious simple phrase because it's reserved for a situation that will almost never happen.
My question is base on the following code:
int myfunct(int ary[], int arysize)
int myfunct2(int *ary, int arysize)
int main(void){
int numary[10];
myfunct(numary, 10)
myfunct2(numary, 10)
return;
}
int myfunct(int ary[], int arysize) {
//Whatever work is done
}
int myfunct2(int *ary, int arysize) {
// Whatever work is done
}
Is there a reason to use one of these over the other? To elaborate, when concerned with numeric arrays, is there any reason one would want to use pointer notation over array notation. If one uses pointer notation then within the function pointer arithmetic would be used etc.. AND if one uses the [] array notation, one could work with the array as usual. I'm new to programming and I currently do not see any benefit to using the pointer notation.
My precise question, is there any reason to pass a numeric array to a function using pointer notation and therefore using pointer manipulations within the function.
When you declare a function parameter as an array, the compiler automatically ignores the array size (if any) and converts it to a pointer. That is, this declaration:
int foo(char p[123]);
is 100% equivalent to:
int foo(char *p);
In fact, this isn't about notation but about the actual type:
typedef char array_t[42];
int foo(array_t p); // still the same function
This has nothing to do with how you access p within the function. Furthermore, the [] operator is not "array notation". [] is a pointer operator:
a[b]
is 100% equivalent to:
*(a + b)
There is no real functional difference between the two notations. In C, when you pass an array variable to a function, it decays to a pointer regardless of the notation. However, in my opinion, the pointer notation is preferable. The problem with [] notation in function definitions is that, in my opinion, it is somewhat misleading:
void foo(int array[])
{
}
A ubiquitous mistake among novice C programmers is to assume that sizeof(array) will give you the number of elements in the array multiplied by sizeof(int), like it would if array were an array variable declared on the stack. But the reality is that array has been decayed to a pointer, despite the misleading [] notation, and so sizeof(array) is going to be sizeof(int*). array is really just a pointer to the first element, or possibly a pointer to a single integer allocated anywhere.
For example, we could call foo like this:
int x = 10;
foo(&x);
In which case the [] notation in the definition of foo is kind of misleading.
Those declarations are absolutely identical. To quote the standard:
A declaration of a parameter as "array of type" shall be adjusted to
"qualified pointer to type"
C99 standard section 6.7.5.3 paragraph 7
In modern C that has variable length arrays since C99, the array notation is preferable if is an array, I think. For one dimensional arrays, you can do things like
int myfunct(size_t size, int array[size]) {
... array[i] ..
}
and for two dimensional
int myfunct(size_t size, int array[size][size]) {
... array[i][j] ..
}
So array notation fits much better in the general picture. A sophisticated compiler could then even do bounds checking, but I don't know of any that does this yet.
In my opinion, the main reason to prefer pointer notation over empty array notation in function prototypes is that the latter is not consistent with structure definitions:
struct person {
char *firstname;
char *lastname;
};
void setperson(struct person *p, char firstname[], char lastname[])
{
p->firstname = firstname;
p->lastname = lastname;
}
In structures you will have to use the pointer notation anyway because empty array notation is only valid, at least since C99, for the last member when you want to make it a flexible array member.
You only need to use the array notation for multidimensional arrays. (You do not have to provide the size of the first dimension).
This might be a bit of a basic question, but what is the difference between writing char * [] and char **? For example, in main,I can have a char * argv[]. Alternatively I can use char ** argv. I assume there's got to be some kind of difference between the two notations.
Under the circumstances, there's no difference at all. If you try to use an array type as a function parameter, the compiler will "adjust" that to a pointer type instead (i.e., T a[x] as a function parameter means exactly the same thing as: T *a).
Under the right circumstances (i.e., not as a function parameter), there can be a difference between using array and pointer notation though. One common one is in an extern declaration. For example, let's assume we have one file that contains something like:
char a[20];
and we want to make that visible in another file. This will work:
extern char a[];
but this will not:
extern char *a;
If we make it an array of pointers instead:
char *a[20];
...the same remains true -- declaring an extern array works fine, but declaring an extern pointer does not:
extern char *a[]; // works
extern char **a; // doesn't work
Depends on context.
As a function parameter, they mean the same thing (to the compiler), but writing it char *argv[] might help make it obvious to programmers that the char** being passed points to the first element of an array of char*.
As a variable declaration, they mean different things. One is a pointer to a pointer, the other is an array of pointers, and the array is of unspecified size. So you can do:
char * foo[] = {0, 0, 0};
And get an array of 3 null pointers. Three char*s is a completely different thing from a pointer to a char*.
You can use cdecl.org to convert them to English:
char *argv[] = declare argv as array of pointer to char
char **argv = declare argv as pointer to pointer to char
I think this is a little bit more than syntactic sugar, it also offers a way to express semantic information about the (voluntary) contract implied by each type of declaration.
With char*[] you are saying that this is intended to be used as an array.
With char**, you are saying that you CAN use this as an array but that's not the way it's intended to be used.
As it was mentioned in the other answers, char*[] declares an array of pointers to char, char** declares a pointer to a pointer to char (which can be used as array).
One difference is that the array is constant, whereas the pointer is not.
Example:
int main()
{
char** ppc = NULL;
char* apc[] = {NULL};
ppc++;
apc++; /* this won't compile*/
return 0;
}
This really depends on the context of where the declarations occur.
Outside of a function parameter definition, the declaration
T a[];
declares a as an unknown-sized array of T; the array type is incomplete, so unless a is defined elsewhere (either in this translation unit or another translation unit that gets linked) then no storage is set aside for it (and you will probably get an "undefined reference" error if you attempt to link, although I think gcc's default behavior is to define the array with 1 element) . It cannot be used as an operand to the sizeof operator. It can be used as an operand of the & operator.
For example:
/**
* module1.c
*/
extern char *a[]; /* non-defining declaration of a */
void foo()
{
size_t i = 0;
for (i = 0; a[i] != NULL; i++)
printf("a[%lu] = %s\n", (unsigned long) i, a[i++]);
}
module1.c uses a non-defining declaration of a to introduce the name so that it can be used in the function foo, but since no size is specified, no storage is set aside for it in this translation unit. Most importantly, the expression a is not a pointer type; it is an incomplete array type. It will be converted to a pointer type in the call to printf by the usual rules.
/**
* module2.c
*/
char *a[] = {"foo", "bar", "bletch", "blurga", NULL}; /* defining declaration of a */
int main(void)
{
void foo();
foo();
return 0;
}
module2.c contains a defining declaration for a (the size of the array is computed from the number of elements in the initializer), which causes storage to be allocated for the array.
Style note: please don't ever write code like this.
In the context of a function parameter declaration, T a[] is synonymous with T *a; in both cases, a is a pointer type. This is only true in the context of a function parameter declaration.
As Paul said in the comment above, it's syntactic sugar. Both char* and char[] are the same data type. In memory, they will both contain the address of a char.
The array/index notation is equivalent to the pointer notation, both in declaration and in access, but sometimes much more intuitive. If you are creating an array of char pointers, you may want to write it one way or another to clarify your intention.
Edit: didn't consider the case Jerry mentioned in the other answer. Take a look at that.
char *ptr[2]={"good","bad"}; //Array of ptr to char
char **str; //Refer ptr to ptr to char
int i;
//str = &ptr[0]; //work
str = ptr;
for(i=0;i<2;i++) printf("%s %s\n",ptr[i],str[i]);
Its o/p same. Using that we can easily understand.
In the C program below, I don't understand why buf[0] = 'A' after I call foo. Isn't foo doing pass-by-value?
#include <stdio.h>
#include <stdlib.h>
void foo(char buf[])
{
buf[0] = 'A';
}
int main(int argc, char *argv[])
{
char buf[10];
buf[0] = 'B';
printf("before foo | buf[0] = %c\n", buf[0]);
foo(buf);
printf("after foo | buf[0] = %c\n", buf[0]);
system("PAUSE");
return 0;
}
output:
before foo | buf[0] = 'B'
after foo | buf[0] = 'A'
void foo(char buf[])
is the same as
void foo(char* buf)
When you call it, foo(buf), you pass a pointer by value, so a copy of the pointer is made.
The copy of the pointer points to the same object as the original pointer (or, in this case, to the initial element of the array).
C does not have pass by reference semantics in the sense that C++ has pass by reference semantics. Everything in C is passed by value. Pointers are used to get pass by reference semantics.
an array is just a fancy way to use a pointer. When you pass buf to the function, you're passing a pointer by value, but when you dereference the pointer, you're still referencing the string it points to.
Array as function parameter is equivalent to a pointer, so the declaration
void foo( char buf[] );
is the same as
void foo( char* buf );
The array argument is then decayed to the pointer to its first element.
Arrays are treated differently than other types; you cannot pass an array "by value" in C.
Online C99 standard (draft n1256), section 6.3.2.1, "Lvalues, arrays, and function designators", paragraph 3:
Except when it is the operand of the sizeof operator or the unary & operator, or is a
string literal used to initialize an array, an expression that has type ‘‘array of type’’ is
converted to an expression with type ‘‘pointer to type’’ that points to the initial element of
the array object and is not an lvalue. If the array object has register storage class, the
behavior is undefined.
In the call
foo(buf);
the array expression buf is not the operand of sizeof or &, nor is it a string literal being used to initialize an array, so it is implicitly converted ("decays") from type "10-element array of char" to "pointer to char", and the address of the first element is passed to foo. Therefore, anything you do to buf in foo() will be reflected in the buf array in main(). Because of how array subscripting is defined, you can use a subscript operator on a pointer type so it looks like you're working with an array type, but you're not.
In the context of a function parameter declaration, T a[] and T a[N] are synonymous with T *a, but this is only case where that is true.
*char buf[] actually means char ** so you are passing by pointer/reference.
That gives you that buf is a pointer, both in the main() and foo() function.
Because you are passing a pointer to buf (by value). So the content being pointed by buf is changed.
With pointers it's different; you are passing by value, but what you are passing is the value of the pointer, which is not the same as the value of the array.
So, the value of the pointer doesn't change, but you're modifying what it's pointing to.
arrays and pointers are (almost) the same thing.
int* foo = malloc(...)
foo[2] is the same as *(foo+2*sizeof(int))
anecdote: you wrote
int main(int argc, char *argv[])
it is also legal (will compile and work the same) to write
int main(int argc, char **argv)
and also
int main(int argc, char argv[][])
they are effectively the same. its slightly more complicated than that, because an array knows how many elements it has, and a pointer doesn't. but they are used the same.
in order to pass that by value, the function would need to know the size of the argument. In this case you are just passing a pointer.
You are passing by reference here. In this example, you can solve the problem by passing a single char at the index of the array desired.
If you want to preserve the contents of the original array, you could copy the string to temporary storage in the function.
edit: What would happen if you wrapped your char array in a structure and passed the struct? I believe that might work too, although I don't know what kind of overhead that might create at the compiler level.
please note one thing,
declaration
void foo(char buf[])
says, that will be using [ ] notation. Not which element of array you will use.
if you would like to point that, you want to get some specific value, then you should declare this function as
void foo(char buf[X]); //where X would be a constant.
Of course it is not possible, because it would be useless (function for operating at n-th element of array?). You don't have to write down information which element of array you want to get. Everything what you need is simple declaration:
voi foo(char value);
so...
void foo(char buf[])
is a declaration which says which notation you want to use ( [ ] - part ), and it also contains pointer to some data.
Moreover... what would you expect... you sent to function foo a name of array
foo(buf);
which is equivalent to &buf[0]. So... this is a pointer.
Arrays in C are not passed by value. They are not even legitimate function parameters. Instead, the compiler sees that you're trying to pass an array and demotes it to pointer. It does this silently because it's evil. It also likes to kick puppies.
Using arrays in function parameters is a nice way to signal to your API users that this thing should be a block of memory segmented into n-byte sized chunks, but don't expect compilers to care if you spell char *foo char foo[] or char foo[12] in function parameters. They won't.