Understanding functions and pointers in C_问答_开发者

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 canno开发者_开发问答t--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?