LEC 2

Question 1

*

Answer 1

– evaluate expression to integer

– use integer as address to get value from memory

Question 2

*(&name)

Answer 2

get address associated with name

get value from memory at that address

Question 3

assignments are

Answer 3

expressions

Question 4

space allocated to declarations on ?

Answer 4

on stack

Question 5

while(expression

while(expression)
statement ->

Answer 5

1. evaluate expression
2. if non‐zero then
   i. execute statement
   ii. repeat from 1.
3. if zero then end iteration
   • break in statement ends enclosing iteration

Question 6

Iteration: for

for(exp1;exp2;exp3)
statement ->
exp1;
while(exp2)
{ statement
exp3;
}

Answer 6

1. execute statement exp1
2. repeatedly test exp2
3. each time exp2 is true
4. execute statement
5. execute statement exp3
   • all exps and statement
   are optional

Question 7

Condition: if

if(expression)
statement1
else
statement2 ->

Answer 7

1. evaluate expression
2. if non‐zero then execute statement1
3. if zero then execute statement2
   • else statement2 is optional

Question 8

Condition: switch

switch(expression)
{ case constant1: statements1
case constant2: statements2
...
default: statementsN
}

Answer 8

1. evaluate expression to a value
2. for first constanti with same value, execute statementsi
3. if no constants match expression, evaluate default
   statementsN

Question 9

Function declaration

type name(type1 name1,...,typeN nameN)
{ declarations
statements
}

Answer 9

1. evaluate expression to a value
2. for first constanti with same value, execute statements
3. if no constants match expression, evaluate default
   statementsN

Question 10

Function call

name(exp1,…,expN)

Answer 10

1. evaluate expifrom left to right
2. push expi values onto stack
3. execute body, allocating stack space to any
   declarations
4. reclaim stack space for parameters & any
   declarations
5. return result if any
   • NB end statement function call with ;

Question 11

Changing parameters

Answer 11

• parameters passed by value
• value of actual parameter copied to space for
formal parameter
• final value is not copied back from formal to
actual
• so changing the formal does not change the
actual

Question 12

Pointers

type*name;

Answer 12

• variable name holds address for byte
sequence for type
• allocates stack space for address but does not
create instance of type
• must allocate space explicitly
NULL
• empty pointer

Question 13

How to actually change parameters?

Answer 13

• to change actual parameter
– pass address (&) of actual to formal
– change indirect (*) on formal

Question 14

What is FILE?

Answer 14

• type for system representation of file
• all files are treated as text
– i.e. character sequences

Question 15

FILE * name;

Answer 15

• name is the address of a FILE value

Question 16

FILE * fopen(path,mode)

Answer 16

• open file with path string
• return
– FILE address
– NULL if no file

• mode
– “r” – read
– “w” – write – create new empty file – delete old version!
– “a” – append i.e. write at end of existing file

Question 17

fclose(FILE *)

Answer 17

• close file
• system may not do this for you when program stops
• for I/O, C handles characters as ints
• EOF == system end of file constant

Question 18

int getc(FILE *)

Answer 18

• return next byte from file address
– as int
• return EOF if at end of file

Question 19

int putc(int,FILE *)

Answer 19

• puts int to file address
– as byte
• return EOF if failure

Question 20

main(int argc; char ** argv)

Answer 20

• argc == number of command line arguments
– including call to executable

• argv == pointer to pointer to char
– i.e. pointer to sequence of char
• i.e. array of strings
– argv[0] == name of executable
– argv[1] == 1st argument
– argv[2] == 2nd argument

Question 21

exit(int)

Answer 21

end program
• return value of int to outer system
• in stdlib library

Question 22

file copy

$ copy path1 path

Answer 22

1. check correct number of arguments
2. open file1
   – check file1 exists
3. open file2
   – check file2 exists
4. copy characters from file1 to file2 until EOF
5. close files

Question 23

formatted input

Answer 23

int fscanf(FILE *,”format”,exp1,exp2...)
– expi == lvalue

Question 24

formatted output

Answer 24

int fprintf(FILE *,”format”,exp1,exp2...)
– expi == rvalue