CH 14

Question 1

What headers?

Answer 1

stdio. h

ctype. h

Question 2

2 #include ctype.h

Answer 2

declares ,functions ,testing ,mapping ,characters.

Question 3

1 #include studio.h

Answer 3

defines three variable types,

macros,functions for input and output

Question 4

Functions for Later

Answer 4

//  forward declaration  
5 int can_print_it( char ch); 
6 void print_letters( char arg[]);

Question 5

Define can_print_it, which simply returns the truth value (0 or 1) of
isalpha(ch) || isblank(ch) back to its caller, print_letters.

Answer 5

32 int can_print_it( char ch)
33 {
34 return isalpha(ch) || isblank(ch);
35 }

Question 6

Finally, make the whole chain of functions go.

Answer 6

37 int main( int argc, char * argv[]) 
38 { 
39 print_arguments(argc, argv); 
40 return 0 ;
41 }

Question 7

void?

Answer 7

A pointer to void can be converted to any pointer type

Question 8

Every program is starting with what definition?

Answer 8

main(int argc, char *argv[])

Question 9

minimum possible value of argc

Answer 9

is 1 which is the execution command itself

Question 10

argv[0] holds

argv[1] holds

Answer 10

argv[0] the string ./mymainprogram

argv[1] is next argument

Question 11

For loop covering print_arguments
Define the print_arguments function, supply it with the command line info
Start the counter on argc
Redefine the print_letters function

Answer 11

8 void print_arguments( int argc, char * argv[]) 
9 { 
10 int i = 0 ; 
11
12 for (i = 0 ; i < argc; i ++ ) { 
13 print_letters(argv[i]); 
14 } 
15 }

Question 12

For loop covering print_letters

Define the print_arguments function, supply it with a array variable
Start a counter for that array variable
spit the array variable into a char variable
pass the variable through the can_print_it function
then print the result

Answer 12

17 void print_letters( char arg[]) 
18 { 
19 int i = 0 ; 
20
21 for (i = 0 ; arg[i] != '\0' ; i ++ ) { 
22 char ch = arg[i]; 
23
24 if (can_print_it(ch)) { 
25 printf( "'%c' == %d " , ch, ch); 
26 } 
27 } 
28
29 printf( "\n" ); 
30 }

Question 13

forward declarations means

Answer 13

TYPE VAR_NAME( TYPE OTHER_VAR);

Question 14

Are these two Variables
5 int can_print_it
6 void print_letters
or functions called from the header?

Answer 14

Variables

Question 15

isalpha(int c)

Answer 15

Explanation: Even though it accepts an int, it really deals with characters. Based on the ASCII value of the character it will determine if it is an alphabetical character or not.

Question 16

isblank()

Answer 16

isblank() considers blank characters the tab character (‘\t’) and the space character (‘ ‘).

Question 17

isblank()

Answer 17

int isblank(int ch)

Question 18

what are these?
5 int can_print_it( char ch);
6 void print_letters( char arg[]);

Answer 18

functions

Question 19

explain this program

Answer 19

To play with this program, you just feed it different command line arguments, which get passed
through your functions.

Question 20

a deeper explanation of forward declarations

Answer 20

defined functions the same
help C out by telling it
if you’re going to use functions
it hasn’t encountered yet in the file

Question 21

Why the new header file?

Answer 21

cstyle.h gives us

so we can gain access to isalpha and isblank.

Question 22

Ref loop covering print_arguments

8 void print_arguments( int argc, char * argv[])

Answer 22

argv pointing at Array returning a char
and
argc returns an Int
called by
print_arguments
returning
a void
This is a function

Question 23

Ref loop covering print_arguments
How this function works
9 { 
10 int i = 0 ; 
11
12 for (i = 0 ; i < argc; i ++ ) { 
13 print_letters(argv[i]); 
14 } 
15 }

Answer 23

Set integer to 0
Initialise function at 0
Test if it is less that argc
Run Code
Kicks off at argv 0 up to /0
delivers the print_letters function to the program
Go up in an incriment of 1

Question 24

Syntax of the if statement in

24 if (can_print_it(ch)) { 
25 printf( "'%c' == %d " , ch, ch); 
26 }

Answer 24

if ( statement is TRUE )

Execute this line of code

Question 25

when is can_print_it true?

Answer 25

when it is non zero and non null

Question 26

what is happening here
24 if (can_print_it(ch)) { 
25 printf( "'%c' == %d " , ch, ch); 
26 }

Answer 26

Variable ch is being tested if true then the decimal version(int) and the single letter it translates into (char)

Question 27

8 void print_arguments( int argc, char * argv[])
what is this bit called?
»»»»char * argv[]

Answer 27

pointer declaration

type *identifier;

Question 28

how to think about 1 dimensional arrays

Answer 28

Think about arrays like this:
[][][][][][]
Each of the bracket pairs is a slot(element) in the array, and you can put information into each one of them.
a group of variables side by side.

Question 29

a compiler needs to know about a function

Answer 29

compiler what the function will return, what the function will be called, as well as what arguments the function can be passed

Question 30

arguments in a function

Answer 30

there can be more than one argument passed to a function or none at all (where the parentheses are empty), and it does not have to return a value.

Question 31

How to think about function
int mult ( int x, int y );

Answer 31

the function mult will accept two arguments,
both integers
it will return an integer.
Do not forget the trailing semi-colon.

Question 32

How to think about function

Answer 32

What will it return? Void
Call it a functional name
(How is this Argument returned
What arguments it needs)

Question 33

what is it thinking here
17 void print_letters( char arg[]) 
18 { 
19 int i = 0 ; 
20
21 for (i = 0 ; arg[i] != '\0' ; i ++ ) { 
22 char ch = arg[i];

Answer 33

it is called print letters
arg[] is variable
char returns a character
( char arg[]) gets replaced with print_letters(argv[i]);
it gets replaced with whats in the brackets
just needs to be same types
is getting spat into char ch for later use

Question 34

Functions that don’t return values have a type of ?

Answer 34

void.