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.
Related
I am trying to create argv for a new process(trying to use execvp), and I checked the execvp manual page which says it needs char *const argv[].
I assume this is an pointer to array of char pointers. So is it possible to pass double pointer of char to this argument?
Basically, what I am trying to do is as following
(argvcounter is number of arguments. ex) cat a -> argvcount = 2)
int argvcount;
char **argv;
...
argv = malloc(sizeof(char*)*(argvcount+1));
for (int i = 0; i<argvcount; i++){
argv[i] = some char pointer;
}
argv[-1] = NULL;
I am not sure about the last line either. I am setting last element to NULL since the last element of array of arguments have to be NULL.
Is it possible to pass this argv to execvp?
Thank you.
According to the C Standard (5.1.2.2.1 Program startup, p.#2)
— argv[argc] shall be a null pointer
So you have to write
argv[argvcount] = NULL;
This statement
argv[-1] = NULL;
does not make sense and results in undefined behavior.
I assume this is an pointer to array of char pointers. So is it
possible to pass double pointer of char to this argument?
An array designater with rare exceptions is implicitly converted to pointer to its first element.
So if for example you have an array like this
char * argv[argvcount];
then passed to a function it is converted to pointer to its first element and has type char **.
On the other hand, these function declarations
void f( char *a[] );
and
void f( char **a );
are equivalent because the compiler adjusts the type of a parameter declared as an arrray to the type of pointer to an object of the array element type.
it needs char *const argv[]. I assume this is an pointer to array of char pointers.
No, it is an array of char* const pointers. It might help reading these declarations from right to left:
[] An array (of unknown size)...
argv ...named argv...
const ... of const...
* ...pointers...
char ...to char.
In plain English: An array (of unknown size) named argv, of read-only pointers to character.
So is it possible to pass double pointer of char to this argument?
Please note the subtle difference between arguments and parameters. Parameter referring to the variable in the function declaration, argument referring to the things you pass to the function on the caller side. It matters here.
Because a function taking a parameter of type char *const argv[], will have that parameter silently "adjusted" by the compiler into a pointer to the first element (sometimes called "array decay"). This is why we don't have to specify the array size - it will "decay" no matter the array size.
The first element is a char*const and a pointer to such an element is of type char*const*, so that's the type that the function will expect. A pointer to a const pointer to char - at the second level of indirection, the pointer itself cannot be modified.
As it happens, char** is a type that may be implicitly converted to char*const*, because the latter is a "qualified" version of the former - it is the same type but with "more const in the right places". Generally, any type* can be converted to type*const.
Had the parameter been const char* argv[], it wouldn't have been possible to use char**, because in that case the const belongs to the pointed-at type and not the pointer.
As already pointed out, note that argv[-1] = NULL; is nonsense, it should be argv[argvcount] = NULL;
Even with int foo(char str[]); which will take in an array initialized to a string literal sizeof doesn't work. I was asked to do something like strlen and the approach I want to take is to use sizeof on the whole string then subtract accordingly depending on a certain uncommon token. Cuts some operations than simply counting through everything.
So yea, I tried using the dereferencing operator on the array(and pointer too, tried it) but I end up getting only the first array element.
How can I sizeof passed arguments. I suppose passing by value might work but I don't really know if that's at all possible with strings.
int foo(char str[]); will take in an array initialized to a string literal
That's not what that does. char str[] here is identical to char* str. When an array type is used as the type of a parameter, it is converted to its corresponding pointer type.
If you need the size of a pointed-to array in a function, you either need to pass the size yourself, using another parameter, or you need to compute it yourself in the function, if doing so is possible (e.g., in your scenario with a C string, you can easily find the end of the string).
You can't use sizeof here. In C arrays are decayed to pointers when passed to functions, so sizeof gives you 4 or 8 - size of pointer depending on platform. Use strlen(3) as suggested, or pass size of the array as explicit second argument.
C strings are just arrays of char. Arrays are not passed by value in C; instead, a pointer to their first element is passed.
So these two are the same:
void foo(char blah[]) { ... }
void foo(char *blah) { ... }
and these two are the same:
char str[] = "Hello";
foo(str);
char *p = str;
foo(p);
You cannot pass an array as a function parameter, so you can't use the sizeof trick within the function. Array expressions are implicitly converted to pointer values in most contexts, including function calls. In the context of a function parameter declaration, T a[] and T a[N] are synonymous with T *.
You'll need to pass the array size as a separate parameter.
Recently, I was debugging one of my programs and found a mistake that I've constantly make, but it was not shown as a warning during compilation, so I've just assume everything was in place and was OK. I a bit confused on what's happening in the code below:
void foo(char b[2]);
char a[2] = {1, 2};
foo(a); // I always assumed that this would pass the entire array to be
// duplicate in stack, guess I was wrong all this while
// Instead the address of the array was passed
void foo(char b[2])
{
// Value of b[0], b[1]?
// Does this mean : 1) b[0] == &a[0]?
// or 2) b[0] == &a[0+2]?
// Compiler didn't complain, so I assume this is a valid syntax
}
When you pass an array as a parameter to a function it decays into a pointer, this is defined in the C standard in 6.7.1:
On entry to the function the value of each argument expression shall be converted to the type
of its corresponding parameter, as if by assignment to the parameter. Array expressions and
function designators as arguments are converted to pointers before the call. A declaration of a
parameter as “array of type” shall be adjusted to “pointer to type,”
This essentially means that in your function declaration it's equivalent to use
void foo(char b[2]); or
void foo(char b[]); or
void foo(char *b)
`
It is valid syntax, and yes, when passing an array the memory address of the first element is copied, but when you dereference the address, you are modifying the original array.
This is the same as the following:
// The array "char b[2]" "decays" to a pointer "char *b"
void foo(char *b)
{
// b[0] == a[0]
// also, b == a (two addresses, both the same)
}
You can read up on how arrays and pointers in C behave very similarly (but not exactly the same). Arrays decay to pointers if they're function arguments (but not anywhere else). The real gotcha here is that on a 64-bit system, sizeof(b) == 8 and sizeof(a) == 2, which is somewhat surprising unless you know about arrays decaying into pointers.
When you declare an array as a function parameter, it is treated as if it were a pointer. Your foo is completely identical to
void foo(char *b)
{
...
}
Arrays decay to pointers. In other words, in some uses (e.g. sizeof(a)) a is an array, but in others where a pointer is expected, the name a means the address of a[0].
b[0] = &a[0] if you do foo(a) as the argument.
If you pass foo((a+1)) then b[0] = &a[1] (you shouldn't do that though since b[1] would be undefined) and so on.
I should make a correction: the address of the array is not passed - it's the address of the array's first element. The simplest way of putting it is:
In C, the value of an array is a pointer to its first element.
If you work with arrays of arrays, you need to supply information about the sizes of arrays after the first dereference:
// either the following or: void foo(char param[][30][20])
void foo(char (*param)[30][20])
{
// ...
}
int main(void)
{
// bar is an array of 10 arrays of 30 arrays of 20 chars.
// its value's TYPE is 'pointer to an array of 30 arrays of 20 chars,
// which is what foo above requires.
char bar[10][30][20];
foo(bar);
}
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.
Confused with the problem here. New to C, as made obvious by the below example:
#include <stdlib.h>
#include <stdio.h>
void pass_char_ref(unsigned char*);
int main()
{
unsigned char bar[6];
pass_char_ref(&bar);
printf("str: %s", bar);
return 0;
}
void pass_char_ref(unsigned char *foo)
{
foo = "hello";
}
To my understanding, bar is an unsigned character array with an element size of 6 set away in static storage. I simply want to pass bar by reference to pass_char_ref() and set the character array in that function, then print it back in main().
You need to copy the string into the array:
void pass_char_ref(unsigned char *foo)
{
strcpy( foo, "hello" );
}
Then when you call the function, simply use the array's name:
pass_char_ref( bar );
Also, the array is not in "static storage"; it is an automatic object, created on the stack, with a lifetime of the containing function's call.
Two things:
You don't need to pass &bar; just pass bar.
When you pass an array like this, the address of its first (0th) element is passed to the function as a pointer. So, call pass_char_ref like this:
pass_char_ref(bar);
When you call pass_char_ref like this, the array name "decays" into a pointer to the array's first element. There's more on this in this tutorial, but the short story is that you can use an array's name in expressions as a synonym for &array_name[0].
Pointers are passed by value. You have:
void pass_char_ref(unsigned char *foo)
{
foo = "hello";
}
In some other languages, arguments are passed by reference, so formal parameters are essentially aliases for the arguments. In such a language, you could assign "hello" to foo and it would change the contents of bar.
Since this is C, foo is a copy of the pointer that's passed in. So, foo = "hello"; doesn't actually affect bar; it sets the local value (foo) to point to the const string "hello".
To get something like pass by reference in C, you have to pass pointers by value, then modify what they point to. e.g.:
#include <string.h>
void pass_char_ref(unsigned char *foo)
{
strcpy(foo, "hello");
}
This will copy the string "hello" to the memory location pointed to by foo. Since you passed in the address of bar, the strcpy will write to bar.
For more info on strcpy, you can look at its man page.
In C, arrays are accessed using similar mechanics to pointers, but they're very different in how the definitions work - an array definition actually causes the space for the array to be allocated. A pointer definition will cause enough storage to be allocated to refer (or "point") to some other part of memory.
unsigned char bar[6];
creates storage for 6 unsigned characters. The C array semantics say that, when you pass an array to another function, instead of creating a copy of the array on the stack, a pointer to the first element in the array is given as the parameter to the function instead. This means that
void pass_char_ref(unsigned char *foo)
is not taking an array as an argument, but a pointer to the array. Updating the pointer value (as in foo = "hello";, which overwrites the pointer's value with the address of the compiled-in string "hello") does not affect the original array. You modify the original array by dereferencing the pointer, and overwriting the memory location it points to. This is something that the strcpy routine does internally, and this is why people are suggesting you use
void pass_char_ref(unsigned char *foo)
{
strcpy(foo, "hello");
}
instead. You could also say (for sake of exposition):
void pass_char_ref(unsigned char *foo)
{
foo[0] = 'h';
foo[1] = 'e';
foo[2] = 'l';
foo[3] = 'l';
foo[4] = 'o';
foo[5] = 0;
}
and it would behave correctly, too. (this is similar to how strcpy will behave internally.)
HTH
Please see here to an explanation of pointers and pass by reference to a question by another SO poster. Also, here is another thorough explanation of the differences between character pointers and character arrays.
Your code is incorrect as in ANSI C standard, you cannot pass an array to a function and pass it by reference - other data-types other than char are capable of doing that. Furthermore, the code is incorrect,
void pass_char_ref(unsigned char *foo)
{
foo = "hello";
}
You cannot assign a pointer in this fashion to a string literal as pointers use the lvalue and rvalue assignment semantics (left value and right value respectively). A string literal is not an rvalue hence it will fail. Incidentally, in the second link that I have given which explains the differences between pointers and arrays, I mentioned an excellent book which will explain a lot about pointers on that second link.
This code will probably make more sense in what you are trying to achieve
void pass_char_ref(unsigned char *foo)
{
strcpy(foo, "hello");
}
In your main() it would be like this
int main()
{
unsigned char bar[6];
pass_char_ref(bar);
printf("str: %s", bar);
return 0;
}
Don't forget to add another line to the top of your code #include <string.h>.
Hope this helps,
Best regards,
Tom.
Since bar[] is an array, when you write bar, then you are using a pointer to the first element of this array. So, instead of:
pass_char_ref(&bar);
you should write:
pass_char_ref(bar);
Time again for the usual spiel --
When an expression of array type appears in most contexts, its type is implicitly converted from "N-element array of T" to "pointer to T" and its value is set to point to the first element of the array. The exceptions to this rule are when the array expression is the operand of either the sizeof or & operators, or when the array is a string litereal being used as an initializer in a declaration.
So what does all that mean in the context of your code?
The type of the expression bar is "6-element array of unsigned char" (unsigned char [6]); in most cases, the type would be implicitly converted to "pointer to unsigned char" (unsigned char *). However, when you call pass_char_ref, you call it as
pass_char_ref(&bar);
The & operator prevents the implicit conversion from taking place, and the type of the expression &bar is "pointer to 6-element array of unsigned char" (unsigned char (*)[6]), which obviously doesn't match the prototype
void pass_char_ref(unsigned char *foo) {...}
In this particular case, the right answer is to ditch the & in the function call and call it as
pass_char_ref(bar);
Now for the second issue. In C, you cannot assign string values using the = operator the way you can in C++ and other languages. In C, a string is an array of char with a terminating 0, and you cannot use = to assign the contents of one array to another. You must use a library function like strcpy, which expects parameters of type char *:
void pass_char_ref(unsigned char *foo)
{
strcpy((char *)foo, "hello");
}
Here's a table of array expressions, their corresponding types, and any implicit conversions, assuming a 1-d array of type T (T a[N]):
Expression Type Implicitly converted to
---------- ---- -----------------------
a T [N] T *
&a T (*)[N]
a[0] T
&a[0] T *
Note that the expressions a, &a, and &a[0] all give the same value (the address of the first element in the array), but the types are all different.
The use of the address of operator (&) on arrays is no longer allowed. I agree that it makes more sense to do &bar rather than bar, but since arrays are ALWAYS passed by reference, the use of & is redundant, and with the advent of C++ the standards committee made it illegal.
so just resist the urge to put & before bar and you will be fine.
Edit: after a conversation with Roger, I retract the word illegal. It's legal, just not useful.