Intermediate C Programming

Pointers 59

4.9 Exercises

4.9.1 Swap Function 1

Does this program have any syntax problems because of wrong types (such as assigning

an integer to a pointer’s value)? Will this function actually swap the values of u and t?

What is the program’s output? Please draw the call stack and explain.

// swap1 .c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

void swap1 ( i n t a , in t b )4

int k = a;6

a = b;7

b = k;8

int main ( i n t argc ,char * * argv )10

{11

int u;12

int t;13

u = 17;14

t = -96;15

printf (" before swap1 : u = %d , t = %d \ n " , u , t );16

swap1 ( u , t);17

printf (" after swap1 : u = %d , t = %d \n " , u , t );18

return EXIT _SUCCES S ;19

}20

4.9.2 Swap Function 2

How about this program?

// swap2 .c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

void swap2 ( i n t * a , i n t * b)4

int * k = a;6

a = b;7

b = k;8

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

{12

int u;13

int t;14

u = 17;15

t = -96;16

printf (" before swap2 : u = %d , t = %d \ n " , u , t );17

60 Intermediate C Programming

swap2 (& u , & t);18

printf (" after swap2 : u = %d , t = %d \n " , u , t );19

return EXIT _SUCCES S ;20

}21

4.9.3 Swap Function 3

How about this program?

// swap3 .c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

void swap3 ( i n t * a , i n t * b)5

int k = * a;7

a = b;8

* b = k ;9

}10

int main ( i n t argc ,char * * argv )13

{14

int u;15

int t;16

u = 17;17

t = -96;18

printf (" before swap3 : u = %d , t = %d \ n " , u , t );19

swap3 (& u , & t);20

printf (" after swap3 : u = %d , t = %d \n " , u , t );21

return EXIT _SUCCES S ;22

}23

4.9.4 Swap Function 4

How about this program?

// swap4 .c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

void swap4 ( i n t * a , i n t * b)5

int k = * a;7

* a = * b;8

* b = * k;9

}10

int main ( i n t argc ,char * * argv )13

{14

Pointers 61

int u;15

int t;16

u = 17;17

t = -96;18

printf (" before swap4 : u = %d , t = %d \ n " , u , t );19

swap4 (& u , & t);20

printf (" after swap4 : u = %d , t = %d \n " , u , t );21

return EXIT _SUCCES S ;22

}23

4.9.5 Swap Function 5

How about this program?

// swap5 .c1

#in clude < stdio .h >2

#in clude < stdlib .h >3

void swap5 ( i n t * a , i n t * b)5

int k = a;7

a = b;8

b = k;9

}10

int main ( i n t argc ,char * * argv )12

{13

int u;14

int t;15

u = 17;16

t = -96;17

printf (" before swap5 : u = %d , t = %d \ n " , u , t );18

swap5 (& u , * t);19

printf (" after swap5 : u = %d , t = %d \n " , u , t );20

return EXIT _SUCCES S ;21

}22

4.9.6 15,552 Variations

There are many variations of the swap function. To be speciﬁc, there are 15,552 variations

and only one of them is correct. Some variations have syntax errors (wrong types) and some

of them do not swap the values in the main function. Let me explain why there are so many

variations. First, this is the correct swap function and the correct way to call it:

void swap ( i nt * k , i nt * m )1

int s;3

s = * k ;4

* k = * m;5

* m = s ;6

62 Intermediate C Programming

void f( void )9

{10

int a = 83;11

int c = -74;12

swap (& a, & c ) ;13

}14

How do we get 15,552 variations? In the ﬁrst line, there are two options for k:

1. int k

2. int * k

int & k is illegal so it is not considered.

Similarly, there are two options for m and two options for s. So far, there are 8 variations

of the function up to the third line. Next, consider the number of options for s = k at the

fourth line; there are six options:

1. s = * k;

2. s = & k;

3. s = k;

4. * s = * k;

5. * s = & k;

6. * s = k;

& s = is illegal so it is not considered.

Similarly, there are also six options for k = m and another six options for m = s. So far

there are 8 × 6 × 6 × 6 = 1,728 variations for swap function.

From the main function, calling swap has three options for using a in the thirteenth line:

1. a

2. & a

3. * a

Similarly, there are another three options in using c. Thus, in total, there are 1,728 × 3

× 3 = 15,552 variations.

Among all these variations, if the swap function is called without using addresses, the

changes are lost when the swap function ﬁnishes. In other words, regardless what happens

inside swap, calling swap in this way,

swap (a , c) ;1

is always wrong.

4.10 Answers

4.10.1 Swap Function 1

There are no syntax errors or warnings but this function does not swap u and k. This is

the output of the program:

before swap1: u = 17 , t = -96

after swap1: u = 17 , t = -96

Pointers 63

4.10.2 Swap Function 2

There are no syntax errors or warnings but this function does not swap u and k. This is

the call stack after ﬁnishing line 7, before swap2 ﬁnishes. The values of a and b are swapped

but the values of u and t remain unchanged.

Frame Symbol Address Value

swap

k 105 100

b 104 100

a 103 101

Return Location 102 line 18

main

t 101 −96

u 100 17

This is the output of the program:

before swap1: u = 17 , t = -96

after swap1: u = 17 , t = -96

4.10.3 Swap Function 3

There are no syntax errors or warnings. The problem is the seventh line. This line assigns

b’s value to a’s value. This is the call stack after ﬁnishing the seventh line:

Frame Symbol Address Value

swap

k 105 17

b 104 101

a 103 101

Return Location 102 line 18

main

t 101 −96

u 100 17

This is the output of the program. Both u and t are 17, and the value −96 has been

discarded.

before swap3: u = 17 , t = -96

after swap3: u = 17 , t = 17

4.10.4 Swap Function 4

The eighth line has a problem: k is an integer and adding * in front of k is invalid. This

program will not compile.

4.10.5 Swap Function 5

The sixth line has a problem: k is an integer but a is a pointer. It is invalid to assign

a pointer’s value to an integer. The eighteenth line also has a problem: t is an integer and

adding * in front of t is invalid.

Table of Contents for Intermediate C Programming

Table of Contents for
Intermediate C Programming