CSCI-0046 Systems Programming in Linux - MIDTERM
True or False: int data[10 ]; In this array, we can store different data types
False
True or False: Once an array is defined with a certain size, we can always decrease or increase the size depending upon how the program works.
False
all data cells in this array will have a value of 4 int data [ 6 ] = { 4 } ;
False
if ( 0 ) printf ( "Yes \n" ); else printf ( "No\n"); The above statement would print Yes. True or False
False
int data [ 3 ] ; data++ ; // Is this a valid statement
False
int data1 [10 ] ; int data2 [10 ]; data1 = data2; // is this possible ?
False
int x = 10; if ( x = 0 ) printf ( "Yes\n") else printf ( "No\n"); printf ( "x=%d \n", x ) ; The value of x printed is 10 True or False
False
statement 1: for ( i = 0 ; i < 20 ; i++ ) continue; printf ( "i = %d \n", i) ; statement 2: for ( i = 0 ; i < 10 ; ++i ) continue; printf ( "i = %d \n", i) ; both statements produce the same result. True or False
False
Which of the following expressions results in false value A. true && false B. true && true C. true || true D. false || true
A
The following statement will get compilation error int data[4]; data++ ;
True
if ( -10 ) printf ( "Yes \n" ); else printf ( "No\n"); The above statement would print Yes. True or False
True
int data [ ] = { 8, 6, 4, 2 } ; data++ ; you cannot do data++ on an array name.
True
int data [ ] = { 8, 6, 4, 2 } ; int * ptr = &data[0]; printf ( "%d", *ptr ) ; will print 8
True
int data [3 ] = { 3, 4, 5 }; int data2 [ 3 ] ; data2 = data ; // THIS IS ILLEGAL. You cannot assign an array to another array
True
int i = 0; int x = 1; for ( ; i < x ; break ) i+=10 ; printf ( "i=%d \n", i ); The above program clip will not compile. True
True
statement 1: for ( i = 0 ; i < 10 ; i++ ) printf ( "\n"); statement 2: for ( i = 0 ; i < 10 ; ++i ) printf ( "\n"); Both statements will produce the same result . True or False
True
statement 1: for ( i = 0 ; i < 20 ; i++ ) break; statement 2: for ( i = 0 ; i < 10 ; ++i ) break; both statements produce same result . True or False
True
int * ptr = 20 ; int y = *ptr ; The above statement WILL give you run-time error TRUE OR FALSE
True - Yes, ptr = 20 is illegal because you are assigning 20 (a address somewhere in the kernel ) to ptr. When you access the value *ptr, we don't have permission to access that space, resulting in segmentation fault.
steps to Dereference a pointer:
#include <stdio.h> int main() { int x=9; // First, we declare the integer variable to which the pointer points int *ptr; //Now, we declare the integer pointer variable. ptr=&x; // After the declaration of an integer pointer variable, we store the address of 'x' variable to the pointer variable 'ptr'. *ptr=8;// We can change the value of 'x' variable by dereferencing a pointer 'ptr' as given The above line changes the value of 'x' variable from 9 to 8 because 'ptr' points to the 'x' location and dereferencing of 'ptr', i.e., *ptr=8 will update the value of x. printf("value of x is : %d", x); return 0;}
int count = 10; int *extraPtr1 = ________ count ; Fill in the blank
&
logical operators
&& [and] || [or] ! [not]
int data [ ] = { 8, 6, 4, 2 } ; How would you print the address of the first cell ? printf ( "%p", __________ ) ;
&data [ 0 ]
When you write function, you have to ask :
--What kind of variables should be function take as input - pass by value --what kind of variables should this function take as output/input - pass by reference What kind of values should this function return: int, float, char, double , short, ... what kind of address should this function return : char *, int *, float *, ... Why do functions define const in front of the variables
Pay attention to = vs == operator ! Pay attention to || vs && operaot ! What is printed ? int x = 20 , y = 35 , z = 0; if ( ( x == 30 ) && ( y = 40 ) ) z = 10; printf ( "%d \n", z ) ;
0
Pay attention to = vs == operator ! Pay attention to || vs && operaot ! What is printed ? int x = 20 , y = 35 , z = 0; if ( ( x == 30 ) || ( y = 0 ) ) z = 10; printf ( "%d \n", z ) ;
0
int i = 0; int x = 0; for ( ; i < x ; i ++ ) i+=10 ; printf ( "i=%d \n", i ); what does it print
0
int data [ 4 ] = { 3, 4 } ; The first cells gets 3 The second cell gets 4 The third cell gets ____ and the fourth cells gets ___
0 and 0
int i = 0 ; do { i++ ; break; i = 20 ; } while ( i < 10 ) ; printf ( " %d \n", i ) ; The value of i printed here is :
1
char data [ 2 ] [ 3 ] = { { 2, 3 , 0} , { 1 , 4, 5 } } ; int i ; for ( i = 0 ; i < 3 ; i++ )printf ( "%d ", data [ 1 ] [ i ] ); The above code would print
1 4 5
Pay attention to = vs == operator ! Pay attention to || vs && operaot ! What is printed ? int x = 20 , y = 35 , z = 0; if ( ( x == 30 ) || ( y = 40 ) ) z = 10; printf ( "%d \n", z ) ;
10
What is the value of x printed in main assume header files are included and no syntax errors int x = 30 ; void changeX ( int x ) { x = 20 ; } main ( ) { int x = 10 ; changeX ( x ) ; printf ( " x %d \n", x ) ; }
10
int i = 0 ; do { i++ ; continue ; i = 20 ; } while ( i < 10 ) ; printf ( " %d \n ", i ) ; The value of i printed here is :
10
pay attention to conditions with just one variable. non-zero value of the condition is true, zero value is false. What is printed ? int x = 0 , y = 10; while ( x ) { y++ ; } printf ( "%d \n", y) ;
10
int i = 0; int x = 1; for ( ; i < x ; i ++ ) i+=10 ; printf ( "i=%d \n", i ); what does it print
11
int i = 0; do { i++ ; i = 12; break ; } while ( i < 10 ); printf ( "%d \n", i); The value of i printed here is :
12
int i = 0; do { i++ ; i = 12; continue; } while ( i < 10 ); printf ( "%d \n", i); The value of i printed here is :
12
what is the size (in terms of bytes) of this array char data [ 12 ] ;
12
Pay attention to variables declared inside the { }. They are not visible outside What is printed ? int x = 10, y = 15; for ( x = 10 ; x < 20 ; x++ ) { int y = 20; break; }
15
int data [ ] = { 8, 6, 4, 2 } ; int *ptr1 = data ; ptr1++ ; ptr1++ ; int *ptr2 ; ptr2 = ptr1 ; ptr2++ ; printf ( "%d\n", *ptr2) ; what is printed ?
2
What is the value of x printed in main assume header files are included. int x = 30 ; void changeX ( int *x ) { *x = 20 ; } main ( ) { int x = 10 ; changeX ( &x ) ; printf ( " x %d \n", x ) ; }
20
What is the value of x printed in main assume header files are included. int x = 30 ; void changeX ( int x ) { x = 20 ; } main ( ) { changeX ( x ) ; printf ( " x %d \n", x ) ; }
30
int data [ ] = { 8, 6, 4, 2 } ; int *ptr1 = data ; ptr1++ ; ptr1++ ; int *ptr2 ; ptr2 = ptr1 ; printf ( "%d\n", *ptr2) ; what is printed ?
4
What is the size of this array int data [ 4 ] = { 4, 4 } ; ?
4 * 4 = 16 bytes
int x = 30; void changeX(int *y) { *y = x + 20; } int main(){ // int x = 10; changeX(&x); printf(" x %d \n", x); return (0); } what is the value of x printed ?
50
int data [ ] = { 8, 6, 4, 2 } ; int * ptr ; ptr = data ; ptr++ ; printf ( "%d\n", *ptr); what is printed ?
6
int data [ ] = { 8, 6, 4, 2 } ; int * ptr ; ptr = data ; printf ( "%d\n", *ptr); What is printed ?
8
statement 1: for ( i = 0 ; i < 10 ; i++ ) printf ( "\n"); statement 2: for ( i = 0 ; i < 10 ; ) printf ( "\n"); i++ ; Both statements produce same result . True or False
False- the second statement would produce infinite loop
#include <stdio.h> int main ( ) { int x = 10; switch x { case 10 : printf ( "Yes" ); break ; } } // The above code compiles fine. True or False
False- Yes, there should be ( ) around x in the switch statement
Let us try this one more time.. int data [ 3 ] = { 2, 3, 4 }; int * ptr ; data = ptr ; The above statement is valid . True or False
False. Yes, you cannot assign a value to a pointer constant.
int i = 0 ; int x = 10 ; switch ( 0 ) { case 0: printf ( "Hello World"); break; case 1 : printf ( "Hi World"); break; case 10 : printf ( "Good Bye World"); break; } ; What does it print ?
Hello World
Cells in an array are contiguous
True
Pay attention to = vs == operator ! Pay attention to || vs && operaot ! What is printed ? int x = 10; if (( x == 10 ) || ( x = 0 ) ) printf ( " True \n"); else printf ( "False \n");
True
What is wrong with this program ? if there is no problem, what is printed ? int x = 11, y = 20 ; if ( x < 10 ) printf ( " %d \n", x) ; else ( x > 10 ) printf ( "%d \n", y );
compilation error, keyword "if" is missing after else
How to declare an array-1D?
dataType arrayName[arraySize];
int i = 0; int x = 1; for ( ; i < x ; break ) i+=10 ; printf ( "i=%d \n", i ); The above program clip will not compile. True
i=12
How to define a pointer variable?
int *ptr ; ptr is now a pointer variable
What is the proper way to print the address of a variable int x = 20 ;
printf ( " %p \n", &x ) ;
The printf function is defined as:
printf ( "format specifier", var1, var2, ... ) ;
int i = 0; do { i++ ; i = 12; break; } while ( 1 ) ; printf ( "i= \n", i ) ; What does this do ?
prints i = 12
&& (AND) Operator
returns TRUE when ALL the conditions under consideration are true and returns FALSE when any ONE or MORE than one condition is false.
The || (Or) operator
returns TRUE when ONE or more conditions are true, and FALSE when all conditions are false.
! operator
returns TRUE when condition is FALSE and returns FALSE when condition is TRUE. So the opposite of the condition.
int data [ 12 ] ; int x = _______ ( data ) ; which operator should we be using to get the number of bytes allocated to the array ? Fill in the blank
sizeof
// Jack wrote a function to increment the value of x. But he is using a pointer as the parameter // Which statement inside the function will do the job of incrementing . int func ( int *x ) { // which is statement is valid ? // statement 1 : *x++ ; // statement 2 * ( x++ ) ; // statement 3 ( *x ) ++ ; } main ( ) { int x = 10 ; func ( &x ) ; printf ( " x = %d \n " x ); }
statement 3
int i = 0 ; int x = 0 ; switch ( i ) { case 0: printf ( "Hello World"); break; case 1 : printf ( "Hi World"); break; case x : printf ( "Good Bye World"); break; } ; What does it print ?
syntax error, won't compile
What is wrong with this program ? int x = 20, y = 10 ; switch ( x ) { case x : printf ( "%d \n", x ) ; break; case y: printf ( "%d \n", y); break; // default : // break; }
you can't have expressions in case statements (marked red)
