I am not great on pointers but I have to learn in the field. If my understanding serves me correct these should all be valid statements below.
int* a;
int b = 10;
a = &b;
(*a) = 20;
(*a) == b; //this should be true
if you have a function like this:
void copy(int* out, int in) {
*out = in;
}
int m_out, m_in;
copy(&m_out, m_in);
m_out == m_in; // this should also be true
but I saw a function like this
create(float& tp, void* form, char* title);
I understand the void pointer, it can be cast to anything, I understand the character pointer which is basically a c style string.
I do not understand the first argument, which is the address of some type, let's say a float but it could be anything, a struct, a int, etc.
What is going on there?
First this
int m_out, m_in;
copy(&m_out, m_in);
is undefined behaviour - you passed uninitialized vaiable m_in to function - and hence trying to make copy of an uninitialized variable.
This:
create(float& tp, void* form, char* title);
doesn't make sense in C. Looks like reference from C++.
The first argument is a reference, it just means that if you modify this field in your function create, the field will still remain modified (even in the function where you called create()) because it points to an address and not a value.
Related
So, I am just trying to wrap my head around "pointer function that returns a pointer to an array"... but to start off slowly, I had to understand this:
void Print(const char c){
printf("\nPrint: %c\n", c);
}
int main () {
void (*FunctionPointer)(const char);
FunctionPointer = &Print;
FunctionPointer('a');
}
Which I do - pretty easy to guess what is going on... FunctionPointer just points to the location where the Print function "resides". Instead of jumping to a specific memory address (stored on a register) of a specific function, I can now be more flexible and point to any function that I want to access.
But I am stuck with the following...
int main () {
int (*FunctionPointer())[];
}
Now it seems that the function that is pointed by FunctionPointer, can in fact return a pointer to an array of type int. The compiler accepts the second line - so far so good - and I also understand the concept... but I am getting stuck regarding the implementation.
FunctionPointer needs - once again, to point to a function. That function can indeed return a pointer that points to an array of type int... soooooo:
int *Array(){
int ar[2] = {5,6};
return ar;
}
int main () {
int (*FunctionPointer())[];
FunctionPointer = &Array;
}
However, the last piece of code is just not accepted by the compiler.... So, what gives?
With
int (*FunctionPointer())[];
you've declared FunctionPointer as a function returning a pointer to an array of int -- not a function pointer. You want
int *(*FunctionPointer)();
If you use [] here, you'll get an error, as functions can't return arrays -- arrays are not first class types -- and unlike with function parameters, arrays will not be silently converted to pointers when used as the return value of a function type. With that, you'll still get the warning
t.c:3:12: warning: function returns address of local variable [-Wreturn-local-addr]
return ar;
^~
which is pretty self-explanatory
You have declared the array of function pointers. Arrays can't be assignable. Functions can't return arrays. You might wish
int* (*FunctionPointer)();
FunctionPointer = &Array;
Function pointers are much easier when you use typedefs. You can simply use the same notation as "normal" data pointers.
// func is a function type. It has one parater and returns pointer to int
typedef int *func(const char);
// funcptr is a pointer to func
func *funcptr;
#include<stdio.h>
typedef struct data
{
int a;
int b;
}dd;
dd *changed(dd **d);
dd changep(dd *d);
int main()
{
dd *d=(dd *)malloc(sizeof(*d));
d->a=5;
d->b=6;
changep(d);
printf("after entering into the functin %d\n",d->a);
changed(&d);
printf("%d\n",d->a);
}
dd changep(dd *d)
{
//d=(dd *)malloc(sizeof(*d));
d->a=14;
printf("%d\n",d->a);
}
dd *changed( dd **d)
{
*d=(dd *)malloc(sizeof(*d));
(*d)->a=3;
(*d)->b=4;
}
here changed and changep are changing the values in the structure why 2 use a double pointer then ??
and if i create new memory in changep then it is not changing its value why??
C uses pass by value in function parameter passing.
void changep(dd *pd) //notice the change
{
//d=malloc(sizeof(*d)); //do not cast
pd->a=14;
printf("%d\n",pd->a);
}
and it is called as
changep(d);
Here, pd is local to the function changep, i.e., a local copy of d. Any chnages made to pd will not be reflected to the caller function.
To make the changes relect to the caller [main()], you need a double pointer. That is why
void changed( dd ** pd)
{
if (pd)
{
if (! *pd)
*pd=malloc(sizeof(*d)); //do not cast
(*pd)->a=3;
(*pd)->b=4;
}
}
and the calling
changed(&d);
reflects the changes made to *pd to d in main().
this is difference of function using value parameters and function using reference parameters.
If you want to change a var pointed by a pointer by calling function, you must use a pointer as parameter. So if you want to change a pointer by calling function, you must use a pointer which point to that pointer!
Both of your functions change the values of the struct members. You confusion is over why single or double pointers? and when to use one vs the other? They are effectively the same, but there are a couple of subtle consideration. In both you are passing a pointer (a reference), so the function will receive an address as an argument, and can modify the value stored at that memory address such that the change is visible outside of the function in say main.
First, when passing a single pointer to a struct (or anything else), the intent is generally, not always, but generally to operate on that pointer and memory location such that the caller receives the new values without needing a return. In this general case, there is no general need to return anything, and the function can properly be of type void and return no value.
The second situation is where you need the function to be able to change the address for the data struct in some way, like when deleting a first or last node in a linked-list. In this situation, the function needs to have access to the address of the pointer (not just the memory address the pointer points to) in order to make the node changes.
Here, you generally think of passing the pointer as a double pointer so if a new list node takes its place, not only are the values pointed to by the pointer subject to change, but the address of the pointer itself may undergo a change. This is one of the few areas when passing a double pointer is required -- when you need to change the address of the pointer itself. Since you need to return this pointer to the caller, these functions are generally of the struct pointer type so that after the list address have all gone through their change, you can return the new address to the start of the list (or whatever) to the caller. So these type function are declared as struct name* to accommodate the requirement.
NOTE However, you can still operate on the values of the pointers provided to the function without providing a return, but if you do need to change the address of the struct provided, then the double pointer is required.
Your code acted appropriately with both, and for your double-pointer function, there was no need for a return. but the values in main reflected the new values:
#include <stdio.h>
#include <stdlib.h>
typedef struct data {
int a;
int b;
} dd;
dd *changed (dd ** d);
void changep (dd * d);
int main () {
dd *d = malloc (sizeof (*d));
d->a = 5;
d->b = 6;
changep (d);
printf ("after entering into the functin %d\n", d->a);
changed (&d);
printf ("%d\n", d->a);
return 0;
}
void changep (dd * d)
{
//d=(dd *)malloc(sizeof(*d));
d->a = 14;
printf ("%d\n", d->a);
}
dd *changed (dd ** d)
{
// *d = (dd *) malloc (sizeof (*d));
(*d)->a = 3;
(*d)->b = 4;
return *d;
}
output:
$./bin/spassvr
14
after entering into the functin 14
3
if I have a structure that has a pointer to a function like this
struct str{
int some_element;
char other_element;
int (*my_pointer_to_a_function)(int);
};
struct str my_struct;
int my_function(int);
and I asign values to it
my_struct.some_element = 1;
my_struct.other_element = 'a';
my_struct.my_pointer_to_a_function = my_function;
how do I call the function that the pointer is pointing to (using the pointer)?
My initial guess is this:
my_struct.(*my_pointer_to_a_function)(value);
or should it be
*my_struct.my_pointer_to_a_function(value);
?
Thank you.
Pointers to functions can be used as-is, without any dereference:
my_struct.my_pointer_to_a_function(value)
But if you insist in dereferencing it you have to use parenthesis this way:
(*my_struct.my_pointer_to_a_function)(value)
They both are totally equivalent, so I recommend the first one, that is simpler.
About you first try:
my_struct.(*my_pointer_to_a_function)(value); //Error!
That won't work because the expression in parenthersis has to be evaluated first: *my_pointer_to_a_function, but that alone means nothing.
And your second:
*my_struct.my_pointer_to_a_function(value); //Error!
The operator precedence rules evaluates first the ., then the function call, and lastly the *:
*(my_struct.my_pointer_to_a_function(value)); //Error!
So the function would be called, but the result of the call, an int, would be dereferenced, hence the error.
Suppose you have pointer to function as you struct member like:
struct newtype{
int a;
char c;
int (*f)(struct newtype*);
} var;
int fun(struct newtype* v){
return v->a;
}
You can call it as follows:
int main(){
var.f=fun;
var.f(&var);
// ^.....^..... have to pass `var` as an argument to f() :( :(
}
//Comment: here in var.f(&var); I miss this pointer and C++,
So for your case it should be just my_struct.my_pointer_to_a_function(value);
Additionally points:
Important to note in my example even you wants to access members of same structure variable you have to pass that. (its quite dissimilar than c++ object!)
virtual functions in C++ classes. They are implemented in a similar fashion under the hood.
Here is a project that will help you to use: Function pointers inside structures
Use this:
#define function mystruct.my_pointer_to_a_function
Then you can call the function :
int i = function(value);
I've just started to work with C, and never had to deal with pointers in previous languages I used, so I was wondering what method is better if just modifying a string.
pointerstring vs normal.
Also if you want to provide more information about when to use pointers that would be great. I was shocked when I found out that the function "normal" would even modify the string passed, and update in the main function without a return value.
#include <stdio.h>
void pointerstring(char *s);
void normal(char s[]);
int main() {
char string[20];
pointerstring(string);
printf("\nPointer: %s\n",string);
normal(string);
printf("Normal: %s\n",string);
}
void pointerstring(char *s) {
sprintf(s,"Hello");
}
void normal(char s[]) {
sprintf(s,"World");
}
Output:
Pointer: Hello
Normal: World
In a function declaration, char [] and char * are equivalent. Function parameters with outer-level array type are transformed to the equivalent pointer type; this affects calling code and the function body itself.
Because of this, it's better to use the char * syntax as otherwise you could be confused and attempt e.g. to take the sizeof of an outer-level fixed-length array type parameter:
void foo(char s[10]) {
printf("%z\n", sizeof(s)); // prints 4 (or 8), not 10
}
When you pass a parameter declared as a pointer to a function (and the pointer parameter is not declared const), you are explicitly giving the function permission to modify the object or array the pointer points to.
One of the problems in C is that arrays are second-class citizens. In almost all useful circumstances, among them when passing them to a function, arrays decay to pointers (thereby losing their size information).
Therefore, it makes no difference whether you take an array as T* arg or T arg[] — the latter is a mere synonym for the former. Both are pointers to the first character of the string variable defined in main(), so both have access to the original data and can modify it.
Note: C always passes arguments per copy. This is also true in this case. However, when you pass a pointer (or an array decaying to a pointer), what is copied is the address, so that the object referred to is accessible through two different copies of its address.
With pointer Vs Without pointer
1) We can directly pass a local variable reference(address) to the new function to process and update the values, instead of sending the values to the function and returning the values from the function.
With pointers
...
int a = 10;
func(&a);
...
void func(int *x);
{
//do something with the value *x(10)
*x = 5;
}
Without pointers
...
int a = 10;
a = func(a);
...
int func(int x);
{
//do something with the value x(10)
x = 5;
return x;
}
2) Global or static variable has life time scope and local variable has scope only to a function. If we want to create a user defined scope variable means pointer is requried. That means if we want to create a variable which should have scope in some n number of functions means, create a dynamic memory for that variable in first function and pass it to all the function, finally free the memory in nth function.
3) If we want to keep member function also in sturucture along with member variables then we can go for function pointers.
struct data;
struct data
{
int no1, no2, ans;
void (*pfAdd)(struct data*);
void (*pfSub)(struct data*);
void (*pfMul)(struct data*);
void (*pfDiv)(struct data*);
};
void add(struct data* x)
{
x.ans = x.no1, x.no2;
}
...
struct data a;
a.no1 = 10;
a.no1 = 5;
a.pfAdd = add;
...
a.pfAdd(&a);
printf("Addition is %d\n", a.ans);
...
4) Consider a structure data which size s is very big. If we want to send a variable of this structure to another function better to send as reference. Because this will reduce the activation record(in stack) size created for the new function.
With Pointers - It will requires only 4bytes (in 32 bit m/c) or 8 bytes (in 64 bit m/c) in activation record(in stack) of function func
...
struct data a;
func(&a);
...
Without Pointers - It will requires s bytes in activation record(in stack) of function func. Conside the s is sizeof(struct data) which is very big value.
...
struct data a;
func(a);
...
5) We can change a value of a constant variable with pointers.
...
const int a = 10;
int *p = NULL;
p = (int *)&a;
*p = 5;
printf("%d", a); //This will print 5
...
in addition to the other answers, my comment about "string"-manipulating functions (string = zero terminated char array): always return the string parameter as a return value.
So you can use the function procedural or functional, like in printf("Dear %s, ", normal(buf));
This is a very simple question but what does the following function prototype mean?
int square( int y, size_t* x )
what dose the size_t* mean? I know size_t is a data type (int >=0). But how do I read the * attached to it? Is it a pointer to the memory location for x? In general I'm having trouble with this stuff, and if anybody could provide a handy reference, I'd appreciate it.
Thanks everybody. I understand what a pointer is, but I guess I have a hard hard time understanding the relationship between pointers and functions. When I see a function prototype defined as int sq(int x, int y), then it is perfectly clear to me what is going on. However, when I see something like int sq( int x, int* y), then I cannot--for the life of me--understand what the second parameter really means. On some level I understand it means "passing a pointer" but I don't understand things well enough to manipulate it on my own.
How about a tutorial on understanding pointers?
In this case however, the pointer is probably used to modify/return the value. In C, there are two basic mechanisms in which a function can return a value (please forgive the dumb example):
It can return the value directly:
float square_root( float x )
{
if ( x >= 0 )
return sqrt( x );
return 0;
}
Or it can return by a pointer:
int square_root( float x, float* result )
{
if ( x >= 0 )
{
*result = sqrt( result );
return 1;
}
return 0;
}
The first one is called:
float a = square_root( 12.0 );
... while the latter:
float b;
square_root( 12.00, &b );
Note that the latter example will also allow you to check whether the value returned was real -- this mechanism is widely used in C libraries, where the return value of a function usually denotes success (or the lack of it) while the values themselves are returned via parameters.
Hence with the latter you could write:
float sqresult;
if ( !square_root( myvar, &sqresult ) )
{
// signal error
}
else
{
// value is good, continue using sqresult!
}
*x means that x is a pointer to a memory location of type size_t.
You can set the location with x = &y;
or set the value were x points to with: *x = 0;
If you need further information take a look at: Pointers
The prototype means that the function takes one integer arg and one arg which is a pointer to a size_t type. size_t is a type defined in a header file, usually to be an unsigned int, but the reason for not just using "unsigned int* x" is to give compiler writers flexibility to use something else.
A pointer is a value that holds a memory address. If I write
int x = 42;
then the compiler will allocate 4 bytes in memory and remember the location any time I use x. If I want to pass that location explicitly, I can create a pointer and assign to it the address of x:
int* ptr = &x;
Now I can pass around ptr to functions that expect a int* for an argument, and I can use ptr by dereferencing:
cout << *ptr + 1;
will print out 43.
There are a number of reasons you might want to use pointers instead of values. 1) you avoid copy-constructing structs and classes when you pass to a function 2) you can have more than one handle to a variable 3) it is the only way to manipulate variables on the heap 4) you can use them to pass results out of a function by writing to the location pointed to by an arg
Pointer Basics
Pointers And Memory
In response to your last comment, I'll try and explain.
You know that variables hold a value, and the type of the variable tells you what kind of values it can hold. So an int type variable can hold an integer number that falls within a certain range. If I declare a function like:
int sq(int x);
...then that means that the sq function needs you to supply a value which is an integer number, and it will return a value that is also an integer number.
If a variable is declared with a pointer type, it means that the value of that variable itself is "the location of another variable". So an int * type variable can hold as its value, "the location of another variable, and that other variable has int type". Then we can extend that to functions:
int sqp(int * x);
That means that the sqp function needs to you to supply a value which is itself the location of an int type variable. That means I could call it like so:
int p;
int q;
p = sqp(&q);
(&q just means "give me the location of q, not its value"). Within sqp, I could use that pointer like this:
int sqp(int * x)
{
*x = 10;
return 20;
}
(*x means "act on the variable at the location given by x, not x itself").
size_t *x means you are passing a pointer to a size_t 'instance'.
There are a couple of reasons you want to pass a pointer.
So that the function can modify the caller's variable. C uses pass-by-value so that modifying a parameter inside a function does not modify the original variable.
For performance reasons. If a parameter is a structure, pass-by-value means you have to copy the struct. If the struct is big enough this could cause a performance hit.
There's a further interpretation given this is a parameter to a function.
When you use pointers (something*) in a function's argument and you pass a variable you are not passing a value, you are passing a reference (a "pointer") to a value. Any changes made to the variable inside the function are done to the variable to which it refers, i.e. the variable outside the function.
You still have to pass the correct type - there are two ways to do this; either use a pointer in the calling routine or use the & (addressof) operator.
I've just written this quickly to demonstrate:
#include <stdio.h>
void add(int one, int* two)
{
*two += one;
}
int main()
{
int x = 5;
int y = 7;
add(x,&y);
printf("%d %d\n", x, y);
return 0;
}
This is how things like scanf work.
int square( int y, size_t* x );
This declares a function that takes two arguments - an integer, and a pointer to unsigned (probably large) integer, and returns an integer.
size_t is unsigned integer type (usually a typedef) returned by sizeof() operator.
* (star) signals pointer type (e.g. int* ptr; makes ptr to be pointer to integer) when used in declarations (and casts), or dereference of a pointer when used at lvalue or rvalue (*ptr = 10; assigns ten to memory pointed to by ptr). It's just our luck that the same symbol is used for multiplication (Pascal, for example, uses ^ for pointers).
At the point of function declaration the names of the parameters (x and y here) don't really matter. You can define your function with different parameter names in the .c file. The caller of the function is only interested in the types and number of function parameters, and the return type.
When you define the function, the parameters now name local variables, whose values are assigned by the caller.
Pointer function parameters are used when passing objects by reference or as output parameters where you pass in a pointer to location where the function stores output value.
C is beautiful and simple language :)
U said that u know what int sq(int x, int y) is.It means we are passing two variables x,y as aguements to the function sq.Say sq function is called from main() function as in
main()
{
/*some code*/
x=sr(a,b);
/*some other code*/
}
int sq(int x,int y)
{
/*code*/
}
any operations done on x,y in sq function does not effect the values a,b
while in
main()
{
/*some code*/
x=sq(a,&b);
/*some other code*/
}
int sq(int x,int* y)
{
/*code*/
}
the operations done on y will modify the value of b,because we are referring to b
so, if you want to modify original values, use pointers.
If you want to use those values, then no need of using pointers.
most of the explanation above is quite well explained. I would like to add the application point of view of this kind of argument passing.
1) when a function has to return more than one value it cannot be done by using more than one return type(trivial, and we all know that).In order to achieve that passing pointers to the function as arguments will provide a way to reflect the changes made inside the function being called(eg:sqrt) in the calling function(eg:main)
Eg: silly but gives you a scenario
//a function is used to get two random numbers into x,y in the main function
int main()
{
int x,y;
generate_rand(&x,&y);
//now x,y contain random values generated by the function
}
void generate_rand(int *x,int *y)
{
*x=rand()%100;
*y=rand()%100;
}
2)when passing an object(a class' object or a structure etc) is a costly process (i.e if the size is too huge then memory n other constraints etc)
eg: instead of passing a structure to a function as an argument, the pointer could be handy as the pointer can be used to access the structure but also saves memory as you are not storing the structure in the temporary location(or stack)
just a couple of examples.. hope it helps..
2 years on and still no answer accepted? Alright, I'll try and explain it...
Let's take the two functions you've mentioned in your question:
int sq_A(int x, int y)
You know this - it's a function called sq_A which takes two int parameters. Easy.
int sq_B(int x, int* y)
This is a function called sq_B which takes two parameters:
Parameter 1 is an int
Parameter 2 is a pointer. This is a pointer that points to an int
So, when we call sq_B(), we need to pass a pointer as the second
parameter. We can't just pass any pointer though - it must be a pointer to an int type.
For example:
int sq_B(int x, int* y) {
/* do something with x and y and return a value */
}
int main() {
int t = 6;
int u = 24;
int result;
result = sq_B(t, &u);
return 0;
}
In main(), variable u is an int. To obtain a pointer to u, we
use the & operator - &u. This means "address of u", and is a
pointer.
Because u is an int, &u is a pointer to an int (or int *), which is the type specified by parameter 2 of sq_B().
Any questions?