Intermediate C Programming

254 Intermediate C Programming

as when passing other types of arguments, such as int and double: A copy of the argument

is passed. The following example shows a separate function for printing Vector objects:

// vector3 .c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

#in clude < string .h >4

#in clude " vector .h "5

void printVe ctor ( Vector v)6

printf (" The vector is (% d, %d , % d) .\ n ", v .x, v .y, v .z) ;8

int main ( i n t argc , char * argv [])11

{12

Vector v1 ;13

v1 .x = 3;14

v1 .y = 6;15

v1 .z = -2;16

printV ector ( v1 );17

return EXIT _SUCCES S ;18

}19

Frame Symbol Address Value

printVector

v.z 124 −2

v.y 120 6

v.x 116 3

return location 112 line 18

main

v1.z 108 −2

v1.y 104 6

v1.x 100 3

How do we know that the attributes are copied? In the following program changeVector

changes the Vector object passed to it. However, inside main, the attributes of v1 are

unchanged.

// vector4 .c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

#in clude < string .h >4

#in clude " vector .h "5

void printVe ctor ( Vector v)6

printf (" The vector is (% d, %d , % d) .\ n ", v .x, v .y, v .z) ;8

void change Vector ( Vector v )11

{12

v. x = 5;13

v. y = -3;14

v. z = 7;15

printV ector ( v) ;16

Programmer-Deﬁned Data Types 255

}17

int main ( i n t argc , char * argv [])19

{20

Vector v1 ;21

v1 .x = 3;22

v1 .y = 6;23

v1 .z = -2;24

printV ector ( v1 );25

chang eVector ( v1 );26

printV ector ( v1 );27

return EXIT _SUCCES S ;28

}29

The output of this program is:

The vector is (3, 6, -2).

The vector is (5, -3, 7).

The vector is (3, 6, -2).

What really happens when a function’s argument is an object? This can be explained

by showing the call stack before calling changeVector:

Frame Symbol Address Value

main

v1.z 108 −2

v1.y 104 6

v1.x 100 3

Calling changeVector pushes a new frame to the call stack. The argument is an object

that has three attributes. The values are copied from the calling function into the new

frame.

Frame Symbol Address Value

changeVector

v.z 124 −2

v.y 120 6

v.x 116 3

return location 112 line 27

main

v1.z 108 −2

v1.y 104 6

v1.x 100 3

The function changeVector changes the attributes of the object in its own frame.

Frame Symbol Address Value

changeVector

v.z 124 7

v.y 120 −3

v.x 116 5

return location 112 line 27

main

v1.z 108 −2

v1.y 104 6

v1.x 100 3

When changeVector ﬁnishes, the frame is popped and the program resumes at main.

The call stack is shown below:

256 Intermediate C Programming

Frame Symbol Address Value

main

v1.z 108 −2

v1.y 104 6

v1.x 100 3

Note that the attributes of v1 are unchanged.

16.3 Objects and Pointers

Is it possible to change an object’s attributes inside a function and keep the changes

even after the function returns? The answer is yes. To do this, we need to use pointers.

// vect orptr . c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

#in clude < string .h >4

#in clude " vector .h "5

void printVe ctor ( Vector v)6

printf (" The vector is (% d, %d , % d) .\ n ", v .x, v .y, v .z) ;8

void change Vector ( Vector * p )11

{12

p -> x = 5;13

p -> y = -3;14

p -> z = 7;15

printV ector (* p) ;16

}17

int main ( i n t argc , char * argv [])19

{20

Vector v1 ;21

v1 .x = 3;22

v1 .y = 6;23

v1 .z = -2;24

printV ector ( v1 );25

chang eVector (& v1 );26

printV ector ( v1 );27

return EXIT _SUCCES S ;28

}29

At line 11 changeVector’s argument is a pointer:

void change Vector ( Vector * p )11

This means that p is a pointer in the frame of the function changeVector. If you refer

back to Table 4.1, this is the ﬁrst way of using *. When calling changeVector at line 26,

main must provide the address of a Vector object, i.e., & v1. This is best understood by

showing the call stack:

Programmer-Deﬁned Data Types 257

Frame Symbol Address Value

changeVector

p 116 100

return location 112 line 27

main

v1.z 108 −2

v1.y 104 6

v1.x 100 3

Instead of copying the whole object, attribute by attribute, the argument p stores only

the address of the object v1. This is the address of the ﬁrst attribute.

What is the -> symbol inside changeVector? The -> symbol takes the value at the

address, and then gets the attribute as if applying a . to a structure. Pointers are used with

structures often and C has this special syntax. It is equivalent to saying:

p -> x13

is the same as

(* p) .x13

This dereferences p ﬁrst, and then applies . for x. Dereferencing is the second way of using

* as explained in Table 4.1. Note that this means -> can only be used on a pointer to a

structure. It is illegal to use it in any other circumstance. If -> is at the left hand side (LHS)

of an assignment, then the attribute is modiﬁed (i.e., written). If -> is at the right hand

side (RHS) of an assignment, then the attribute is read. The statement,

p -> x = 5;13

changes the value at address 100.

Frame Symbol Address Value

changeVector

p 116 100

return location 112 line 27

main

v1.z 108 −2

v1.y 104 6

v1.x 100 3 → 5

p -> y = -3;14

changes the value at address 104.

Frame Symbol Address Value

changeVector

p 116 100

return location 112 line 27

main

v1.z 108 −2

v1.y 104 6 → -3

v1.x 100 5

Why do we need to add * in front of p when changeVector calls printVector? In

changeVector, p is a pointer. However, printVector expects an object because there is

no * in the line:

void printVe ctor ( Vector v)6

In changeVector, adding * in front of p dereferences the pointer, as explained in

Table 4.1. Thus, the object stored at addresses 100–108 is copied to the argument of

printVector. How do we know that the object is copied? In C, arguments are always copied

when passed to functions. If the argument were Vector *, then a copy of the pointer would

258 Intermediate C Programming

be passed. There is no * after Vector in printVector and the argument is an object. As

a result, the object is copied. If v’s attributes are changed inside printVector, then the

changes will be lost when the function ﬁnishes. The syntax for using objects is:

• If p’s value is an address of an object, use p -> x. It is allowed to put a space before

or after -> but no space can be added between - and >.

• If v is an object (not an address), use v.x.

16.3.1 Returning an Object

Can a function return a Vector object? Yes. The following example shows a constructor

function that creates and initializes a new object:

// vector5 .c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

#in clude < string .h >4

#in clude " vector .h "5

Vector Ve ctor_ constr uct ( i n t a , i n t b , in t c)6

Vector v;8

v. x = a;9

v. y = b;10

v. z = c;11

return v ;12

}13

void Vector _print ( Vector v )15

{16

printf (" The vector is (% d, %d , % d) .\ n ", v .x, v .y, v .z) ;17

}18

int main ( i n t argc , char * argv [])20

{21

Vector v1 = Vecto r_con struct (3 , 6 , -2) ;22

Vecto r_print ( v1 );23

return EXIT _SUCCES S ;24

}25

What is the advantage of creating constructor functions? One good reason is that they

make programs easier to read. The three arguments remind programmers that a Vector

object has three attributes. Constructors should guarantee that all attributes are always

initialized. Uninitialized variables can make your programs behave in surprising ways. Before

calling the constructor, v1 is already on the call stack in the frame of the main function.

When the constructor returns the object, the attributes of v are copied to v1’s attributes

one by one. Then, the constructor’s frame is popped and v does not exist any more.

16.3.2 Objects and malloc

Is it possible to create an object that exists in heap memory, instead of stack memory?

Yes. Here is how to do it:

// ve ctormal l oc . c1

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