I am trying to understand pointer to pointers and I can not understand why in first case I need (&) ampersand ( I receive a message [Error] cannot convert 'char*' to 'char**' in assignment )
and in the second case I don't need ampersand
first case :
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main ()
{
char *p ={"jack is a good boy"};
char**p1;
p1=p; //why I need & in this case
return0;
}
second case :
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main ()
{
char *p[5] ={"jack","is", "a","good","boy"};
int i=0;
char**p1;
p1=p;
//p1=&p[0];
for(p1=p; *p1; p1++)
{
printf("\n the words are %s",*p1);
}
return 0;
}
The unary & operator could be seen as a pointer-to operator.
From the first example, p is a pointer to char and p1 is a pointer to a pointer to a char. But when you have &p it becomes a pointer to a pointer to a char, of the type char **, which makes is the same type as p1.
In the second example, p is an array of pointers. And as all arrays it can decay to a pointer to its first element. So just using p alone is the same as &p[0]. And as p[0] is a pointer to char (i.e. char *) then &p[0] is a pointer to a pointer to char (i.e. char **).
So in the first example the pointer-to operator must be used to get a pointer to a pointer to char. This isn't needed in the second example because the pointer-to operator is used implicitly.
p1=p; //why I need & in this case
The variable p has the type char *
char *p ={"jack is a good boy"};
While the variable p1 has the type char **
char**p1;
So you are trying to use incompatible pointer types in the assignment statement.
You need to write
p1 = &p;
In this case the pointer p1 will point to the object p that has the type char *. That is a pointer to the object p will have the type char **.
In the second program you declared an array with the element type char *.
char *p[5] ={"jack","is", "a","good","boy"};
Array designator used in expressions with rare exception is implicitly converted to pointer to its first element
So in this assignment statement
p1=p;
that is equivalent to
p1 = &p[0];
the expression p used as an initializer is converted to the type char **. So the left and the right operands of the assignment have the same pointer type.
Pay attention to that this for loop
for(p1=p; *p1; p1++)
{
printf("\n the words are %s",*p1);
}
invokes undefined behavior because the array p does not have a null pointer among its elements. Thus the condition of the for loop *p1 that is the same as *p1 != NULL will not be evaluated for elements of the array p.
If the loop will be correct you need to include an element that is equal to null pointer, For example
char *p[6] ={"jack","is", "a","good","boy"};
^^^
or
char *p[] = {"jack","is", "a","good","boy", NULL };
Related
The compiler states "assignment from incompatible pointer type" when the row of the 2D array is not mentioned, I always thought an array without brackets means the address of the first element, in this case address of the element twodstring[0][0]
Compiler does not state an error when the row is mentioned, I was wondering why is this the case?
#include<stdio.h>
int main()
{
char onedstring[]={"1D Array"};
char twodstring[][5]={"2D","Array"};
char *p1,*p2;
p1=onedstring;
p2=twodstring;
p2=twodstring[1];
}
A two-dimensional array
char a[M][N];
can be declared using a typedef the following way
typedef char T[N];
T a[M];
So a pointer to the first element of the array a can be declared like
T *p = a;
where T is an alias for the type char[N]. Now making the reverse substitution we can write
char ( *p )[N] = a;
That is elements of a two-dimensional array are one-dimensional arrays.
This declaration
char ( *p )[N] = a;
is equivalent to
char ( *p )[N] = &a[0];
where a[0] has the type char[N]. So the pointer points to the first "row" of the array.
Dereferencing the pointer you will get an object of the type char[N].
Pay attention to that a two-dimensional array can be declared like
char ( a[M] )[N];
So substituting the one-dimensional array declarator a[M] for pointer you will get
char ( a[M] )[N];
char ( *p )[N] = a;
If you will declare a pointer like this
char *p1;
then you may write for example
p1 = a[1];
in this expression a[1] is a one-dimensional array of the type char[N]. Using the expression as an initializer the array is converted to pointer to its first element that has the type char *.
So this expression statement
p1 = a[1];
is equivalent to
p1 = &a[1][0];
Dereferencing this pointer you will get an object of the type char.
Let's look at these lines:
char twodstring[][5]={"2D","Array"};
char *p2;
p2=twodstring;
In the last line, you just write the identifier of an array. This evaluates to a pointer to the first array element. But this first array element doesn't have the type char -- it is an array itself with the type char [5].
If you want this assignment to work, you have to declare p2 with the correct type like this:
char twodstring[][5]={"2D","Array"};
char (*p2)[5];
p2=twodstring;
So I'm cruising through Learn C the Hard Way (just enrichment) - but one bonus problem I've come across is the following -
#include <stdio.h>
char * mystrcpy(char *, char *);
int main(void) {
char dest[100];
char src[] = "Test string\n";
char *p;
if (p = mystrcpy(dest,src)) {
printf("%s\n",p);
return(0);
}
else {
printf("null pointer received\n");
return(1);
}
}
/* mystrcpy: Copy a string from s to a buffer pointer to by d.
d = destination buffer
s = source string
return value = a pointer to the beginning of the string in the
destination buffer.
*/
char * mystrcpy(char *d, char *s) {
}
I know that *d and *s are both pointer variables, but I'm not exactly sure what that means, nor how to use them in the context of this problem. Any help or "pointers" (no pun intended) would be greatly appreciated.
Except when it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" will be adjusted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array.
In the call to mystrcpy, the expressions src and dest "decay" from type "array of char" to "pointer to char", so what the function actually receives are the pointer values corresponding to the addresses of the first element of each array.
You can access the contents of each array by using the [] subscript operator on the pointer expressions1:
d[i] = s[i];
or, you can dereference each pointer directly:
*d = *s;
and adjust the value of each pointer
d++;
s++;
Note that these operations may be combined:
*d++ = *s++;
From there it's just a matter of determining where the source string ends.
The expression a[i] is defined as *(a + i); given the pointer value a, offset i elements (not bytes!) from that address and dereference the result. This is why the array conversion rule mentioned above exists in the first place.
My XCode (default compiler should be clang?) shows me on this code a warning:
Incompatible pointer types passing 'char *(*)[2]' to parameter of type 'char ***' (when calling func)
void func (char ***arrayref, register size_t size) {
/// ...
}
int main()
{
char *array[2] = {"string", "value"};
func(&array, 2);
}
while this code is no problem (=no warning):
void func (char **arrayref, register size_t size) {
/// ...
}
int main()
{
char *array[2] = {"string", "value"};
func(array, 2);
}
While this removes the warning
func((char***)&array, 2);
I still don't know why the first emits a warning, while the latter doesn't.
Also, when the first is a problem, I'd also expect that the first emits a warning like:
Incompatible pointer types passing 'char *[2]' to parameter of type 'char **'
Time for a brief refresher on array semantics.
Except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element in the array.
The expression array in your code has type "2-element array of char *", or char *[2]. When you pass it as an argument to func, as in
func( array, 2 );
the expression is converted to an expression of type "pointer to char *", or char **, which is the type your function is expecting, and the value of the expression is the address of the first element: array == &array[0]. This is why you don't get the warning for the second code snippet.
In the first snippet, the array is an operand of the unary & operator, so the conversion to a pointer type doesn't happen; instead, the type of the expression is "pointer to 2-element array of char *", or char *(*)[2], which is not compatible with char **. The address value is the same (the address of the first element of the array is the same as the address of the array itself), but the types don't match, hence the warning.
So why does this matter? A pointer is just an address, and all address are the same, right? Well, no. A pointer is an abstraction of an address, with associated type semantics. Pointers to different types don't have to have the same size or representation, and pointer arithmetic depends on the type of the pointed-to type.
For example, if I declare a pointer as char **p;, then the expression p++ will advance the pointer to point to the next object of type char *, or sizeof (char *) bytes from the current address. If p is declared as char *(*p)[2], however, the expression p++ will advance p to point to the next two-element array of char *, which is 2 * sizeof (char *) bytes from the current address.
char *array[2] = {"string", "value"};
is an array with 2 elements of char *.
Using array as an address results to a pointer to the first element, i. e. of type char **.
Using &array results to a pointer to the same place, but of type char *(*)[2] (not sure if the spelling is right).
This is not the same as a char *** - the representation in memory is completely different.
To be more verbose,
+++++++++++++++++++++++
+ array[0] + array[1] +
+++++++++++++++++++++++
this is the array.
char ** p1 = array; // is a pointer to the first element, which in turn is a pointer.
char *(*p2)[2] = &array; // is a pointer to the whole array. Same address, but different type, i. e. sizeof(*p1) != sizeof(*p2) and other differences.
char ***p3 = &p1; // Now, p1 is a different pointer variable which has an address itself which has type `char ***`.
Here's an example of how to do what you want and change what array points to:
char *array2[] = {"string", "NewValue"};
void func0 (char **arrayref, register size_t size) {
puts(arrayref[1]);
}
void func1 (char ***arrayref, register size_t size) {
puts(arrayref[0][1]);
*arrayref= (char **) array2;
}
int main()
{
char *array[] = {"string", "value"};
char **foo = array;
func0(foo, 2);
func1(&foo,2);
func0(foo, 2);
}
You have an array of type char *[2] i.e. array of 2 pointers to char. It is an array of fixed size with automatic storage duration. The only thing that your function can do with this kind of array is to either use its elements or to change them (it can not resize it or deallocate it... therefore it makes no sense to try to make it possible to change the array itself ~> in other words: you don't really need a pointer to this kind of array).
Here's a simple example:
void func (char *arrayref[2]) {
printf("%s", arrayref[1]);
arrayref[1] = "new value";
}
int main() { {
char *array[2] = {"string", "value"};
func(array);
printf(" -> %s", array[1]);
return 0;
}
or alternatively changing func to take an array of unspecified size making it usable with char *[X] for any X, not just 2 (in that case it makes sense already to pass the array's size):
void func (char *arrayref[], size_t size) {
if (size > 1) {
printf("%s", arrayref[1]);
arrayref[1] = "new value";
}
}
with one way or other, this program would output value -> new value.
If you need your function to be able to resize this array or affect the array itself in some other way, you should consider using dynamically-allocated array and passing in form of char**.
I don't understand why in the url_split function I can use a++, but in the main function I can't use key_value++, they have the same type...
void url_split(char *src, char **host, char *a[])
{
char const *p1 = "?";
char const *p2 = "&";
*host = strtok(src, p1);
char *str;
while((str = strtok(NULL, p2)))
{
*a = str;
a++;
}
*a = str;
}
int main(int argc, char *argv[])
{
char *host;
char *key_value[100];
char url[] = "http://www.baidu.com/s?wd=linux&cl=3";
url_split(url, &host, key_value);
printf("host = %s\n", host);
while(*key_value)
{
printf("key-value : %s\n", *key_value);
key_value++;
}
return 0;
}
No, they are not actually the same thing: key_value in main is an array that you cannot change (you can change the contents but not the array variable itself).
When you pass it to a function, it becomes a pointer to the first element of that array, which can change (to point to other elements of that array, for example).
It's no different to:
int xyzzy[10]; // xyzzy cannot change
int *plugh = xyzzy; // but plugh can.
This "decay" of arrays to pointers actually happens in the vast majority of cases. From the C11 standard:
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.
They actually don't have the same type:
key_value is of type char *[100], an array of 100 char pointers.
a is actually of type char **, a pointer to a char*, not an array.
When you pass key_value to url_split it decays to a char**, which is why key_value is a valid argument to the function and why you use char** as the type of a function argument that is intended to be an array of char*.
The post-increment operator is incompatible with arrays, an array can't be assigned a new value, but it works perfectly fine for pointers, since they can be assigned a new value. That's why a++ is valid and key_value++ is not.
Here is what is written in Kernighan book about this:
"When an array name is passed to a function, what is passed is the location of the initial element. Within the called function, this argument is a local variable, and so an array name
parameter is a pointer, that is, a variable containing an address."
Which means in main function key_value is an array name, you cannot modify it, since it's just a synonim to the first element in the array. But when passed to function, another pointer is created which points to the same location as first element of array.
I'm trying to learn C now, I'm coming from Java and there is some stuff that is new to me.
I want to print a string and send an int and a string(char array) to another method. But I keep getting some errors that I don't know how to fix.
Would really appreciate if someone could take their time and explain to me what's wrong in my code. I'm quite disoriented at the moment with these pointers. When to use %s and %c when printing etc...
Code:
#include <stdio.h>
void main()
{
int k = 10;
char string;
char *sptr;
string = "hello!";
int *ptr;
sptr = &string;
ptr = &k;
printf("%s \n", &sptr);
printf("Sending pointer.\n");
sendptr(ptr, sptr);
}
And the errors.
test.c: In function ‘main’:
test.c:8:9: warning: assignment makes integer from pointer without a cast
test.c:15:2: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘char **’
tezt.c: In function ‘sendptr’:
tezt.c:8:8: error: incompatible types when assigning to type ‘char[6]’ from type ‘char’
Thanks for your time! :)
First functions solved.
Second function i get this..
tezt.c: In function ‘sendptr’:
tezt.c:5:2: error: invalid initializer
#include <stdio.h>
void sendptr(int *test, char *fname)
{
char fnamn[] = &fname;
int pt;
pt = *test;
printf("%p \n", test);
printf("%d \n", pt);
printf("%s \n", fnamn);
}
char string;
string = "hello!";
First problem: you're declaring string as a single char, not as an array. Also, you can only initialize the array to a string literal in a single statement.
char string[] = "hello!";
Second problem: sptr is a pointer-to-char, so it has to point to the first element of your string. Either of these will do:
char *sptr = string;
char *sptr = &string[0];
Then, when printing the string, just pass sptr directly.
printf("%s \n", sptr);
EDIT for your next question.
char fnamn[] = &fname;
You're trying to assign a char** (pointer to pointer to char) to an array. That just won't work. If you want to copy the string pointed to by fname into fnamn then you need to use a function such as strncpy.
char fnamn[MAX_STRING_SIZE];
strncpy(fnamn, fname, MAX_STRING_SIZE);
Having said that, if you just want to print the string, then print fname directly without copying it into your array first.
Here's a corrected version of the program with some annotation:
#include <stdio.h>
int main(void) // int and (void) for standard mains.
{
int k = 10;
char *string; // a C string is a char array, you need a pointer to point to it
char *sptr;
int *ptr;
string = "hello!";
sptr = string;
ptr = &k;
printf("%s \n", sptr); // no &. The %s format expects a char*.
printf("Sending pointer.\n");
// sendptr(ptr, sptr); // don't know what this function is, ignoring
return 0;
}
In C language, the & operator means you want to use the address of the variable (ie & = "the address of the variable").
int an_integer=2; // an_integer is a memory part where you want to store 2 ;)
printf("%d", &an_integer); // here you will print the address of the memory part where an_integer is stored (not 2, more something like 2510849).
The * operator in a declaration of variable means that you want to have a pointer to a memory part, when using it in the code, it means the "the value contained at the address of"
int an_integer=2;
int *ptr_integer; // you declare a pointer to an integer
ptr_integer = &an_integer; // here you set the pointer ptr_integer to the address of an_integer
printf("%d", *ptr_integer); // here you print the value contained at the memory address stored in the ptr_integer
The [] operator means you want to store an array of something. In C, an array can be seen as a pointer to a memory space.
int an_integer[2]; // you declare an array of 2 integers
int *ptr_integer; // you declare a pointer to an integer
ptr_integer = (int *)an_integer; // here you set the value of the pointer to the address of the array, you have to cast it into an (int *) to avoid compilation warnings.
For a start, I would suggest changing:
char string;
to:
char *string;
It's pretty clear that you want the string variable to be a string rather than a single character.
In addition, you probably want to change the two lines:
sptr = &string;
printf("%s \n", &sptr);
to:
sptr = string;
printf("%s \n", sptr);
but you could equally well just pass string itself to printf.
As for the sendptr(ptr, sptr);, we can't help that much without knowing more details about it.
To fix your second function (from your edit), change:
char fnamn[] = &fname;
to:
char *fnamn = fname;
or just use fname directly. You don't have to make a copy of the pointer and the former is for things like:
char fnamn[] = "I am a string literal";
I thought it might be helpful to adding something about the difference between a char array and a pointer to a string.
In function1 below, the local variable stringPtr is a pointer to memory which contains the string "hello!". The memory containing this string will be located in a read-only section of the program. The compiler decides where to place the string "hello!" and ensures that your local variable is initialised with this memory address.
You can modify the pointer stringPtr and change it to point somewhere else. But you cannot modify the memory it points at.
Also, it is perfectly valid to use the array access notation stringPtr[2] even though it is a pointer.
In function2 the compiler will set aside 9 bytes of space on the stack for the local variable stringArray and it will ensure that this array is initialised with the string "Goodbye!". As this memory is on the stack you can modify the contents of the array.
#include <stdio.h>
void function1(void)
{
char *stringPtr = "hello!";
printf("The first char is %c\n", stringPtr[0]);
printf("The next char is %c\n", *(stringPtr+1));
// This would cause a segmentation fault, stringPtr points to read-only memory
// stringPtr[0] = 'H';
}
void function2(void)
{
char stringArray[] = "Goodbye!";
printf("The first char is %c\n", stringArray[0]);
}
int main(void)
{
function1();
function2();
return 0;
}
First of all, the return type for main should be int, not void. void main() is only well-defined if your compiler documentation explicitly lists it as a legal signature. Otherwise you invoke undefined behavior. Use int main(void) instead.
Secondly, it's time for a quick crash course on strings, arrays, and pointers.
Unlike Java, C doesn't have a dedicated string datatype; rather, strings are represented as sequences of char values terminated by a 0. They are stored as arrays of char. The string literal "hello" is stored as a 6-element array of char (const char in C++). This array has static extent, meaning it is allocated at program startup and held until the program terminates. Attempting to modify the contents of a string literal invokes undefined behavior; it's best to act as though they're unwritable.
When an array expression appears in most contexts, the type of the expression is converted from "N-element array of T" to "pointer to T", and the value of the expression is the address of the first element of the array. That's one of the reasons the string = "hello"; statement doesn't work; in that context, the type of the expression "hello" is converted from "6-element array of char" to "pointer to char", which is incompatible with the target type (which, being char, isn't the correct type anyway). The only exceptions to this rule are when the array expression is an operand of either the sizeof or unary & operators, or if it is a string literal being used to initialize another array in a declaration.
For example, the declaration
char foo[] = "hello";
allocates foo as a 6-element array of char and copies the contents of the string literal to it, whereas
char *bar = "hello";
allocates bar as a pointer to char and copies the address of the string literal to it.
If you want to copy the contents of one array to another, you need to use a library function like strcpy or memcpy. For strings, you'd use strcpy like so:
char string[MAX_LENGTH];
strcpy(string, "hello");
You'll need to make sure that the target is large enough to store the contents of the source string, along with the terminating 0. Otherwise you'll get a buffer overflow. Arrays in C don't know how big they are, and running past the end of an array will not raise an exception like it does in Java.
If you want to guard against the possibility of a buffer overflow, you'd use strncpy, which takes a count as an additional parameter, so that no more than N characters are copied:
strncpy(string, "hello", MAX_LEN - 1);
The problem is that strncpy won't append the 0 terminator to the target if the source is longer than the destination; you'll have to do that yourself.
If you want to print the contents of a string, you'd use the %s conversion specifier and pass an expression that evaluates to the address of the first element of the string, like so:
char string[10] = "hello";
char *p = string;
printf("%s\n", "hello"); // "hello" is an array expression that decays to a pointer
printf("%s\n", string); // string is an array expression that decays to a pointer
printf("%s\n", p); // p is a pointer to the beginning of the string
Again, both "hello" and string have their types converted from "N-element array of char" to "pointer to char"; all printf sees is a pointer value.
Here's a handy table showing the types of various expressions involving arrays:
Declaration: T a[M];
Expression Type Decays to
---------- ---- ---------
a T [M] T *
&a T (*)[M]
*a T
a[i] T
&a[i] T *
Declaration: T a[M][N];
Expression Type Decays to
---------- ---- ---------
a T [M][N] T (*)[N]
&a T (*)[M][N]
*a T [N] T *
a[i] T [N] T *
&a[i] T (*)[N]
*a[i] T
a[i][j] T
&a[i][j] T *
Remember that the unary & operator will yield the address of its operand (provided the operand is an lvalue). That's why your char fnamn[] = &fname; declaration threw up the "invalid initializer" error; you're trying to initialize the contents of an array of char with a pointer value.
The unary * operator will yield the value of whatever its operand points to. If the operand isn't pointing anywhere meaningful (it's either NULL or doesn't correspond to a valid address), the behavior is undefined. If you're lucky, you'll get a segfault outright. If you're not lucky, you'll get weird runtime behavior.
Note that the expressions a and &a yield the same value (the address of the first element in the array), but their types are different. The first yields a simple pointer to T, where the second yields a pointer to an array of T. This matters when you're doing pointer arithmetic. For example, assume the following code:
int a[5] = {0,1,2,3,4};
int *p = a;
int (*pa)[5] = &a;
printf("p = %p, pa = %p\n", (void *) p, (void *) pa);
p++;
pa++;
printf("p = %p, pa = %p\n", (void *) p, (void *) pa);
For the first printf, the two pointer values are identical. Then we advance both pointers. p will be advanced by sizeof int bytes (i.e., it will point to the second element of the array). pa, OTOH, will be advanced by sizeof int [5] bytes, so that it will point to the first byte past the end of the array.
#include <stdio.h>
void main()
{
int k = 10;
char string;
char *sptr;
sptr = "hello!";
int *ptr;
ptr = &k;
printf("%s \n", sptr);
printf("Sending pointer.\n");
sendptr(ptr, sptr);
}