One common place where pointers are helpful is when you are writing functions. Functions take their arguments 'by value', which means that they get a copy of what is passed in and if a function assigns a new value to one of its arguments that will not affect the caller. This means that you couldn't write a "doubling" function like this:
void doubling(int x)
{
x = x * 2;
}
This makes sense because otherwise what would the program do if you called doubling like this:
doubling(5);
Pointers provide a tool for solving this problem because they let you write functions that take the address of a variable, for example:
void doubling2(int *x)
{
(*x) = (*x) * 2;
}
The function above takes the address of an integer as its argument. The one line in the function body dereferences that address twice: on the left-hand side of the equal sign we are storing into that address and on the right-hand side we are getting the integer value from that address and then multiply it by 2. The end result is that the value found at that address is now doubled.
As an aside, when we want to call this new function we can't pass in a literal value (e.g. doubling2(5)) as it won't compile because we are not properly giving the function an address. One way to give it an address would look like this:
int a = 5;
doubling2(&a);
The end result of this would be that our variable a would contain 10.
Videos
What are the types of pointers in C?
What is the void pointer in C?
One common place where pointers are helpful is when you are writing functions. Functions take their arguments 'by value', which means that they get a copy of what is passed in and if a function assigns a new value to one of its arguments that will not affect the caller. This means that you couldn't write a "doubling" function like this:
void doubling(int x)
{
x = x * 2;
}
This makes sense because otherwise what would the program do if you called doubling like this:
doubling(5);
Pointers provide a tool for solving this problem because they let you write functions that take the address of a variable, for example:
void doubling2(int *x)
{
(*x) = (*x) * 2;
}
The function above takes the address of an integer as its argument. The one line in the function body dereferences that address twice: on the left-hand side of the equal sign we are storing into that address and on the right-hand side we are getting the integer value from that address and then multiply it by 2. The end result is that the value found at that address is now doubled.
As an aside, when we want to call this new function we can't pass in a literal value (e.g. doubling2(5)) as it won't compile because we are not properly giving the function an address. One way to give it an address would look like this:
int a = 5;
doubling2(&a);
The end result of this would be that our variable a would contain 10.
A variable itself is a pointer to data
No, it is not. A variable represents an object, an lvalue. The concept of lvalue is fundamentally different from the concept of a pointer. You seem to be mixing the two.
In C it is not possible to "rebind" an lvalue to make it "point" to a different location in memory. The binding between lvalues and their memory locations is determined and fixed at compile time. It is not always 100% specific (e.g. absolute location of a local variable is not known at compile time), but it is sufficiently specific to make it non-user-adjustable at run time.
The whole idea of a pointer is that its value is generally determined at run time and can be made to point to different memory locations at run time.
Newbie here:
I know how pointers work, what do they do and how to deference them, but I don’t understand why would someone use pointers instead of the variable name. Can someone tell me what’s up?
- Why use pointers over normal variables?
Short answer is: Don't. ;-) Pointers are to be used where you can't use anything else. It is either because the lack of appropriate functionality, missing data types or for pure perfomance. More below...
- When and where should I use pointers?
Short answer here is: Where you cannot use anything else. In C you don't have any support for complex datatypes such as a string. There are also no way of passing a variable "by reference" to a function. That's where you have to use pointers. Also you can have them to point at virtually anything, linked lists, members of structs and so on. But let's not go into that here.
- How do you use pointers with arrays?
With little effort and much confusion. ;-) If we talk about simple data types such as int and char there is little difference between an array and a pointer.
These declarations are very similar (but not the same - e.g., sizeof will return different values):
char* a = "Hello";
char a[] = "Hello";
You can reach any element in the array like this
printf("Second char is: %c", a[1]);
Index 1 since the array starts with element 0. :-)
Or you could equally do this
printf("Second char is: %c", *(a+1));
The pointer operator (the *) is needed since we are telling printf that we want to print a character. Without the *, the character representation of the memory address itself would be printed. Now we are using the character itself instead. If we had used %s instead of %c, we would have asked printf to print the content of the memory address pointed to by 'a' plus one (in this example above), and we wouldn't have had to put the * in front:
printf("Second char is: %s", (a+1)); /* WRONG */
But this would not have just printed the second character, but instead all characters in the next memory addresses, until a null character (\0) were found. And this is where things start to get dangerous. What if you accidentally try and print a variable of the type integer instead of a char pointer with the %s formatter?
char* a = "Hello";
int b = 120;
printf("Second char is: %s", b);
This would print whatever is found on memory address 120 and go on printing until a null character was found. It is wrong and illegal to perform this printf statement, but it would probably work anyway, since a pointer actually is of the type int in many environments. Imagine the problems you might cause if you were to use sprintf() instead and assign this way too long "char array" to another variable, that only got a certain limited space allocated. You would most likely end up writing over something else in the memory and cause your program to crash (if you are lucky).
Oh, and if you don't assign a string value to the char array / pointer when you declare it, you MUST allocate sufficient amount of memory to it before giving it a value. Using malloc, calloc or similar. This since you only declared one element in your array / one single memory address to point at. So here's a few examples:
char* x;
/* Allocate 6 bytes of memory for me and point x to the first of them. */
x = (char*) malloc(6);
x[0] = 'H';
x[1] = 'e';
x[2] = 'l';
x[3] = 'l';
x[4] = 'o';
x[5] = '\0';
printf("String \"%s\" at address: %d\n", x, x);
/* Delete the allocation (reservation) of the memory. */
/* The char pointer x is still pointing to this address in memory though! */
free(x);
/* Same as malloc but here the allocated space is filled with null characters!*/
x = (char *) calloc(6, sizeof(x));
x[0] = 'H';
x[1] = 'e';
x[2] = 'l';
x[3] = 'l';
x[4] = 'o';
x[5] = '\0';
printf("String \"%s\" at address: %d\n", x, x);
/* And delete the allocation again... */
free(x);
/* We can set the size at declaration time as well */
char xx[6];
xx[0] = 'H';
xx[1] = 'e';
xx[2] = 'l';
xx[3] = 'l';
xx[4] = 'o';
xx[5] = '\0';
printf("String \"%s\" at address: %d\n", xx, xx);
Do note that you can still use the variable x after you have performed a free() of the allocated memory, but you do not know what is in there. Also do notice that the two printf() might give you different addresses, since there is no guarantee that the second allocation of memory is performed in the same space as the first one.
One reason to use pointers is so that a variable or an object can be modified in a called function.
In C++ it is a better practice to use references than pointers. Though references are essentially pointers, C++ to some extent hides the fact and makes it seem as if you are passing by value. This makes it easy to change the way the calling function receives the value without having to modify the semantics of passing it.
Consider the following examples:
Using references:
public void doSomething()
{
int i = 10;
doSomethingElse(i); // passes i by references since doSomethingElse() receives it
// by reference, but the syntax makes it appear as if i is passed
// by value
}
public void doSomethingElse(int& i) // receives i as a reference
{
cout << i << endl;
}
Using pointers:
public void doSomething()
{
int i = 10;
doSomethingElse(&i);
}
public void doSomethingElse(int* i)
{
cout << *i << endl;
}