What is the output of the program.docx

7/27/2019 What is the output of the program.docx

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What is the output of the program?

#include

int main()

{

union a{

int i;

char ch[2];

};

union a u;

u.ch[0] = 3;

u.ch[1] = 2;

printf("%d, %d, %d\n", u.ch[0], u.ch[1], u.i);

return 0;

}

A.

3, 2, 515 B.515, 2, 3C.3, 2, 5 D.None of these

Answer & Explanation

Answer: Option A

Explanation:

printf (" %d, %d, %d\n" , u.ch[0], u.ch[1], u.i);It prints the value ofu.ch[0] = 3, u.ch[1] = 2

and it prints the value ofu.imeans the value of entire union size.

So the output is 3, 2, 515.

#include

int main()

{


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unsigned int i = 65535; /* Assume 2 byte integer*/

while(i++ != 0)

printf("%d",++i);

printf("\n");

return 0;

}A.Infinite loop

B. 0 1 2 ... 65535

C.0 1 2 ... 32767 - 32766 -32765 -1 0

D.No output


Answer: Option A

Explanation:

Here unsigned in tsize is 2 bytes. It varies from 0,1,2,3, ... to 65535.

Step 1:unsigned int i = 65535;

Step 2:

Loop 1: while(i++ != 0)this statement becomes whi le(65535 != 0). Hence the while(TRUE)

condition is satisfied. Then the printf(" %d" , ++i);prints '1'(variable ' i 'is already

increemented by '1' in while statement and now increemented by '1' in printf statement)

Loop 2: while(i++ != 0)this statement becomes while(1 != 0). Hence the while(TRUE)

condition is satisfied. Then the printf(" %d" , ++i);prints '3'(variable ' i 'is already

increemented by '1' in while statement and now increemented by '1' in printf statement)

........

The while loop will never stops executing, because variable iwill never become '0'(zero).

Hence it is an 'Infinite loop'.

What will be the output of the program?

#include

int main()

{

float a = 0.7;

if(0.7 > a)printf("Hi\n");

else

printf("Hello\n");

return 0;

}

A.Hi B.Hello

C.Hi Hello D.None of above


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Answer: Option A

Explanation:

if (0.7 > a)here ais a float variable and 0.7is a double constant. The double constant 0.7is

greater than the float variable a. Hence the i fcondition is satisfied and it prints 'H i '

Example:

#include

int main()

{

float a=0.7;

printf("%.10f %.10f\n",0.7, a);

return 0;

}

Output:

0.7000000000 0.6999999881

#include

int main()

{

int i=3;

switch(i)

{

case 1:

printf("Hello\n");

case 2:

printf("Hi\n");

case 3:

continue;

default:

printf("Bye\n");

}

return 0;

}

A.Error: Misplaced continue B.Bye

C.No output D.Hello Hi


Answer: Option A


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Explanation:

The keyword continuecannot be used in swi tch case. It must be used in foror whileor do

whileloop. If there is any looping statement in switch case then we can use continue.


#include

int main()

{

int i = 1;

switch(i)

{

printf("Hello\n");

case 1:

printf("Hi\n");

break;

case 2:printf("\nBye\n");

break;

}

return 0;

}

A.Hello

HiB.

Hello

Bye

C.Hi D.Bye


Answer: Option C

Explanation:

switch(i)has the variable iit has the value '1'(one).

Then case 1:statements got executed. so, it prints "Hi". The break;statement make the

program to be exited from switch-case statement.

switch-casedo not execute any statements outside these blocks caseand default

Hence the output is "Hi".Which of the following statements are correct about the below program?

#include

int main()

{

int i = 10, j = 20;

if(i = 5) && if(j = 10)


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printf("Have a nice day");

return 0;

}

A.Output: Have a nice day

B. o output

C. rror: Expression syntaxD. rror: Undeclared identifier i f


Answer: Option C

Explanation:

"Expression syntax" error occur in this line i f(i = 5) & & if(j = 10).

It should be like if((i == 5) && (j == 10)).

Assunming, integer is 2 byte, What will be the output of the program?#include

int main()

{

printf("%x\n", -2


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Answer: Option A

Explanation:

The range oflong doubleis 3.4E-4932

to 1.1E+4932

#include

int main()

{

float f=43.20;

printf("%e, ", f);

printf("%f, ", f);

printf("%g", f);

return 0;

}

A.

4.320000e

+

, 43.200001, 43.2 B.4.3, 43.22, 43.21C.4.3e, 43.20f, 43.00 D.Error


Answer: Option A

Explanation:

printf(" %e, " , f);Here '%e' specifies the "Scientific Notation" format. So, it prints the

43.20 as 4.320000e+01

.

printf(" %f, " , f);Here '%f' specifies the "Decimal Floating Point" format. So, it prints the43.20 as 43.200001.

printf(" %g, " , f);Here '%g' "Use the shorter of %e or %f". So, it prints the 43.20 as 43.2.


#include

int reverse(int);

int main()

{

int no=5;reverse(no);

return 0;

}

int reverse(int no)

{

if(no == 0)

return 0;


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else

printf("%d,", no);

reverse (no--);

}

A.Print 5, 4, 3, 2, 1 B.Print 1, 2, 3, 4, 5

C.

Print 5, 4, 3, 2, 1, 0 D.Infinite loopAnswer & Explanation

Answer: Option D

Explanation:

Step 1: int no=5;The variable nois declared as integer type and initialized to 5.

Step 2: reverse(no);becomes reverse(5);It calls the function reverse()with '5' as parameter.

The function reverse accept an integer number 5 and it returns '0'(zero) if(5 == 0) if thegiven number is '0'(zero) or else printf(" %d," , no);it prints that number 5 and calls the

function reverse(5);.

The function runs infinetely because the there is a post-decrement operator is used. It will

not decrease the value of 'n' before calling the reverse() function. So, it calls reverse(5)

infinitely.

Note: If we use pre-decrement operator like reverse(--n), then the output will be 5, 4, 3, 2, 1.

Because before calling the function, it decrements the value of 'n'.

5. What will be the output of the program?

#include

void fun(int);

typedef int (*pf) (int, int);

int proc(pf, int, int);

int main()

{

int a=3;

fun(a);

return 0;

}void fun(int n)

{

if(n > 0)

{

fun(--n);

printf("%d,", n);

fun(--n);


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}

}

A. , 2, 1, 0 B. 1, 1, 2, 0

C. , 1, 0, 2 D.0, 1, 2, 0


Answer: Option D


#include

int fun(int(*)());

int main()

{

fun(main);

printf("Hi\n");

return 0;

}int fun(int (*p)())

{

printf("Hello ");

return 0;

}

A. nfinite loop B. Hi

C. ello Hi D.Error


Answer: Option C

#include

int main()

{

int a=10;

void f();

a = f();

printf("%d\n", a);

return 0;

}

void f()

{

printf("Hi");

}

A.Error: Not allowed assignment

B. Error: Doesn't print anything

C.No error

D.None of above



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Answer: Option A

Explanation:

The function void f ()is not visible to the compiler while going through main() function. Soe have to declare this prototype void f ();before to main() function. This kind of error will

not occur in modern compilers.

#include

int main()

{

printf("%p\n", main());

return 0;

}

A.It prints garbage values infinitely

B.

Runs infinitely without printing anythingC.Error: main()cannot be called inside printf()

D.No Error and print nothing


Answer: Option B

Explanation:

In printf(" %p\n" , main());it calls the main() function and then it repeats infinetly, untill

stack overflow.

If return type for a function is not specified, it defaults to intA. rue B.False


Answer: Option A

Explanation:

True, The default returntype for a function is int.

#include

#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);

int main(){

char *str1="India";

char *str2="BIX";

JOIN(str1, str2);

return 0;

}

A.str1=IndiaBIX str2=BIX B.str1=India str2=BIX


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C.str1=India str2=IndiaBIX D.Error: in macro substitution


Answer: Option B

#include

#define CUBE(x) (x*x*x)

int main()

{

int a, b=3;

a = CUBE(b++);

printf("%d, %d\n", a, b);

return 0;

}

A.9, 4 B.27, 4

C.27, 6 D.Error


Answer: Option C

Explanation:

The macro function CUBE(x) (x*x* x)calculates the cubic value of given number(Eg: 103.)

Step 1: in t a, b=3;The variable aand bare declared as an integer type and varaible bid

initialized to 3.

Step 2: a = CUBE(b++);becomes

=> a = b++ * b++ * b++;

=> a = 3 * 3 * 3;Here we are using post-increement operator, so the 3 is not incremented in

this statement.

=> a = 27;Here, 27 is store in the variable a. By the way, the value of variable bis

incremented by 3. (ie: b=6)

Step 3: printf(" %d, %d\n" , a, b);It prints the value of variable aand b.

Hence the output of the program is 27, 6.

#include

int main()

{

static char *s[] = {"black", "white", "pink", "violet"};


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char **ptr[] = {s+3, s+2, s+1, s}, ***p;

p = ptr;

++p;

printf("%s", **p+1);

return 0;

}A.ink B.ack

C.ite D.let


Answer: Option A


#include

int main()

{

int arr[3] = {2, 3, 4};

char *p;

p = arr;

p = (char*)((int*)(p));

printf("%d, ", *p);

p = (int*)(p+1);

printf("%d", *p);

return 0;

}

A.2, 3 B. 2, 0

C.2, Garbage value D.0, 0


Answer: Option B

15. What will be the output of the program ?

#include

int main()

{

printf("%c\n", 7["IndiaBIX"]);

return 0;

}

A.Error: in printf B. Nothing will print


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C.print "X" of IndiaBIX D.print "7"


Answer: Option C


#include

int main()

{

char str1[] = "India";

char str2[] = "BIX";

char *s1 = str1, *s2=str2;

while(*s1++ = *s2++)

printf("%s", str1);

printf("\n");

return 0;

}

A.IndiaBIX B. BndiaBIdiaBIXia

C.India D.(null)


Answer: Option B


#include

#include

int main()

{

int i, n;

char *x="Alice";

n = strlen(x);

*x = x[n];

for(i=0; i


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A.Alice B. ecilA

C.Alice lice ice ce e D.lice ice ce e


Answer: Option D

Explanation:

If you compile and execute this program in windows platform with Turbo C, it will give

"lice ice ce e".

It may give different output in other platforms (depends upon compiler and machine).

The online C compiler given in this site will give the Option C as output (it runs on

Linux platform).


#include

int main()

{

int i, a[] = {2, 4, 6, 8, 10};

change(a, 5);

for(i=0; i


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float a=3.14;

char *j;

j = (char*)&a;

printf("%d\n", *j);

return 0;

}A.

It prints ASCII value of the binary number present in the first byte of a float variable

a.

B.It prints character equivalent of the binary number present in the first byte of a float

variable a.

C.It will print 3

D.It will print a garbage value


Answer: Option A

In the following program add a statement in the function fact()such that the factorial

gets stored inj.

#include

void fact(int*);

int main()

{

int i=5;

fact(&i);

printf("%d\n", i);

return 0;

}

void fact(int *j)

{

static int s=1;

if(*j!=0)

{

s = s**j;

*j = *j-1;

fact(j);

/* Add a statement here */

}

}

A. =s; B. *j=s;

C.*j=&s; D.&j=s;


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Answer: Option B


#include

int main()

{

int a[5] = {5, 1, 15, 20, 25};

int i, j, m;

i = ++a[1];

j = a[1]++;

m = a[i++];

printf("%d, %d, %d", i, j, m);

return 0;

}A.2, 1, 15 B. 1, 2, 5

C.3, 2, 15 D.2, 3, 20


Answer: Option C

Explanation:

Step 1: in t a[5] = {5, 1, 15, 20, 25};The variable arr is declared as an integer array witha size of 5 and it is initialized to

a[0] = 5, a[1] = 1, a[2] = 15, a[3] = 20, a[4] = 25.

Step 2: int i , j, m;The variable i,j,m are declared as an integer type.

Step 3: i = ++a[1];becomes i = ++1;Hence i = 2and a[1] = 2

Step 4:j = a[1]++;becomesj = 2++;Hencej = 2and a[1] = 3.

Step 5: m = a[i++];becomes m = a[2];Hence m = 15and iis incremented by 1(i++means 2++ so i=3)

Step 6: printf (" %d, %d, %d" , i, j, m);It prints the value of the variables i, j, m

Hence the output of the program is 3, 2, 15

. What will be the output of the program If characters 'a', 'b' and 'c' enter are supplied


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as input?

#include

int main()

{ void fun();

fun();

printf("\n");

return 0;

}

void fun()

{

char c;

if((c = getchar())!= '\n')

fun();

printf("%c", c);}

A.abc abc B. bca

C.Infinite loop D.cba


Answer: Option D

Explanation:

Step 1: void fun ();This is the prototype for the function fun().

Step 2: fun();The function fun()is called here.

The function fun()gets a character input and the input is terminated by an enter

key(New line character). It prints the given character in the reverse order.

The given input characters are "abc"

Output: cba


#include

int main()

{

printf("India", "BIX\n");

return 0;


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}

A.Error B. India BIX

C.India D.BIX


Answer: Option C

Explanation:

printf (" India" , " BIX\n" );It prints "India". Because ,(comma) operator has Left to

Right associativity. After printing "India", the statement got terminated.


#include

#include

int main()

{

static char s[] = "Hello!";

printf("%d\n", *(s+strlen(s)));

return 0;

}

A.8 B. 0

C.16 D.Error


Answer: Option B


#include

int main()

{

static char s[25] = "The cocaine man";

int i=0;

char ch;

ch = s[++i];

printf("%c", ch);

ch = s[i++];

printf("%c", ch);

ch = i++[s];

printf("%c", ch);

ch = ++i[s];


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printf("%c", ch);

return 0;

}

A.hhe! B. he c

C.The c D.

Hhec


Answer: Option A


#include

int main()

{

char str[] = "India\0BIX\0";

printf("%d\n", sizeof(str));

return 0;

}

A.10 B. 6

C.5 D.11


Answer: Option D

Explanation:

The following examples may help you understand this problem:

1. sizeof(" " )returns 1 (1*).

2. sizeof(" I ndia" )returns 6 (5 + 1*).

3. sizeof(" BI X" )returns 4 (3 + 1*).

4. sizeof(" I ndia\0BI X" )returns 10 (5 + 1 + 3 + 1*).

Here '\0' is considered as 1 char by sizeof() function.

5. sizeof(" I ndia\0BI X\0" )returns 11 (5 + 1 + 3 + 1 + 1*).

Here '\0' is considered as 1 char by sizeof() function.


#include


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int main()

{

char str[25] = "IndiaBIX";

printf("%s\n", &str+2);

return 0;

}A.Garbage value B. Error

C.No output D.diaBIX


Answer: Option A

Explanation:

Step 1: char str[25] = " I ndiaBIX" ;The variable stris declared as an array of

characteres and initialized with a string "IndiaBIX".

Step 2: printf(" %s\n" , & str+2);

=> In the printf statement %sis string format specifier tells the compiler to print the

string in the memory of& str+2

=> & stris a location of string "IndiaBIX". Therefore & str+2is another memory

location.

Hence it prints the Garbage value.

#include

void swap(char *, char *);

int main()

{

char *pstr[2] = {"Hello", "IndiaBIX"};

swap(pstr[0], pstr[1]);

printf("%s\n%s", pstr[0], pstr[1]);

return 0;

}

void swap(char *t1, char *t2)

{

char *t;

t=t1;

t1=t2;

t2=t;

}

A.IndiaBIX B.Address of "Hello" and "IndiaBIX"


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Hello

C.Hello

IndiaBIXD.

Iello

HndiaBIX


Answer: Option C

Explanation:

Step 1: void swap(char * , char * );This prototype tells the compiler that the function swap

accept two strings as arguments and it does not return anything.

Step 2: char * pstr[2] = {" Hel lo" , " IndiaBIX" };The variable pstris declared as an pointer to

the array of strings. It is initialized to

pstr[0] = " Hello", pstr[1] = " I ndiaBIX"

Step 3: swap(pstr [0], pstr [1] );The swapfunction is called by "call by value". Hence it does

not affect the output of the program.

If the swapfunction is "called by reference" it will affect the variable pstr.

Step 4: printf (" %s\n%s" , pstr[0], pstr[1]);It prints the value ofpstr[0]and pstr[1].

Hence the output of the program is

Hello

IndiaBIX

27. What will be the output of the program (Turbo C in 16 bit platform DOS) ?

#include

#include

int main()

{

char *str1 = "India";

char *str2 = "BIX";char *str3;

str3 = strcat(str1, str2);

printf("%s %s\n", str3, str1);

return 0;

}

A.IndiaBIX India B. IndiaBIX IndiaBIX


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C.India India D.Error


Answer: Option B

Explanation:

It prints 'IndiaBIX IndiaBIX' in TurboC (in 16 bit platform).

It may cause a 'segmentation fault error' in GCC (32 bit platform).

2. Which of the following statements are correct about the below declarations?

char *p = " Sanjay" ;

char a[] = " Sanjay" ;

1: There is no difference in the declarations and both serve the same purpose.

2:pis a non-const pointer pointing to a non-const string, whereas ais a const pointer

pointing to a non-const pointer.

3:The pointer pcan be modified to point to another string, whereas the individual

characters within array acan be changed.

4: In both cases the '\0'will be added at the end of the string "Sanjay".

A.1, 2 B. 2, 3, 4

C.3, 4 D.2, 3


Answer: Option B

What will be the output of the program ?

#include

int main()

{

struct value{

int bit1:1;

int bit3:4;

int bit4:4;

}bit={1, 2, 13};

printf("%d, %d, %d\n", bit.bit1, bit.bit3, bit.bit4);


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return 0;

}

A.1, 2, 13 B. 1, 4, 4

C.-1, 2, -3 D.-1, -2, -13


Answer: Option C

Explanation:

Note the below statement inside the struct:

int bit1:1;--> 'int' indicates that it is a SIGNED integer.

For signed integers the leftmost bit will be taken for +/- sign.

If you store 1 in 1-bit field:

The left most bit is 1, so the system will treat the value as negative number.

The 2's complement method is used by the system to handle the negative values.

Therefore, the data stored is 1. The 2's complement of 1 is also 1 (negative).

Therefore -1 is printed.

If you store 2 in 4-bits field:

Binary 2: 0010 (left most bit is 0, so system will treat it as positive value)

0010 is 2

Therefore 2 is printed.

If you store 13 in 4-bits field:

Binary 13: 1101 (left most bit is 1, so system will treat it as negative value)

Find 2's complement of 1101:

1's complement of 1101 : 0010

2's complement of 1101 : 0011 (Add 1 to the result of 1's complement)


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0011 is 3 (but negative value)

Therefore -3 is printed.


#include

int main()

{

struct byte

{

int one:1;

};

struct byte var = {1};

printf("%d\n", var.one);

return 0;

}A.1 B. -1

C.0 D.Error


Answer: Option B

12. If a char is 1 byte wide, an integer is 2 bytes wide and a long integer is 4 bytes wide then

ill the following structure always occupy 7 bytes?

struct ex

{

char ch;

int i;

long int a;

};

A.Yes B.No


Answer: Option B

Explanation:

A compiler may leave holes in structures by padding the first char in the structure with

another byte just to ensures that the integer that follows is stored at an location. Also,

there might be 2extra bytes after the integer to ensure that the long integer is stored at

an address, which is multiple of 4. Such alignment is done by machines to improve the

efficiency of accessing values.


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What will be the output of the program (myprog.c) given below if it is executed from

the command line?

cmd> myprog one two three

/* myprog.c */

#include

int main(int argc, char **argv)

{

printf("%c\n", **++argv);

return 0;

}

A.myprog one two three B. myprog one

C.o D.two


Answer: Option C

8. What will be the output of the program (sample.c) given below if it is executed from the

command line?

cmd> sample fr iday tuesday sunday

/* sample.c */

#include

int main(int argc, char *argv[])

{

printf("%c", **++argv);

return 0;

}

A.s B. f

C.sample D.friday


Answer: Option B

What will be the output of the program (myprog.c) given below if it is executed from the

command line?

cmd> myprog f r iday tuesday sunday

/* myprog.c */

#include


{


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printf("%c", *++argv[1]);

return 0;

}

A.r B.f

C.

m D.y


Answer: Option A

10. What will be the output of the program (sample.c) given below if it is executed from the

command line?

cmd> sample one two thr ee

/* sample.c */

#include


{

int i=0;

i+=strlen(argv[1]);

while(i>0)

{

printf("%c", argv[1][--i]);

}

return 0;

}

A.three two one B. owt

C.eno D.eerht


Answer: Option C

15. What will be the output of the program (myprog.c) given below if it is executed from

the command line?


/* myprog.c */

#include


{

int i;

for(i=1; i


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}

A.oot B. ott

C.nwh D.eoe


Answer: Option B

21. What will be the output of the program (myprog.c) given below if it is executed from

the command line?


/* myprog.c */

#include

#include

int main(int argc, char **argv)

{

int i;

for(i=1; i


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Answer: Option C

Explanation:

Both ceil()and floor()return the integer found as a double.

floor(2.5)returns the largest integral value(round down) that is not greater than 2.5. So

output is 2.000000.

ceil(2.5)returns 3, while converting the doubleto intit returns '0'.

So, the output is '2.000000, 0'.

If the binary eauivalent of 5.375 in normalised form is 0100 0000 1010 1100 0000 0000

0000 0000, what will be the output of the program (on intel machine)?

#include

#includeint main()

{

float a=5.375;

char *p;

int i;

p = (char*)&a;

for(i=0; i 4

0000 -> 0

1010 -> A

1100 -> C

0000 -> 0

0000 -> 0

0000 -> 0

0000 -> 0


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Since the PC's (intel processors) use " LITTLE ENDIAN" byte order, the higher order

byte of number is stored in lowest address. The result is written from bottom to top.

#include

int main()

{

float a=0.7;

if(a < 0.7)

printf("C\n");

else

printf("C++\n");

return 0;

}

A.

C B.C++C.Compiler error D.Non of above

if (a < 0.7)here ais a float variable and 0.7is a double constant. The float variable ais less

than double constant 0.7. Hence the i fcondition is satisfied and it prints 'C'

#include

#define CUBE(x) (x*x*x)

int main()

{

int a, b=3;

a = CUBE(b++);printf("%d, %d\n", a, b);

return 0;

}

A.9, 4 B.27, 4

C.27, 6 D.Error


Answer: Option C

Explanation:

The macro function CUBE(x) (x*x* x)calculates the cubic value of given number(Eg: 103.)

Step 1: in t a, b=3;The variable aand bare declared as an integer type and varaible bid

initialized to 3.

Step 2: a = CUBE(b++);becomes


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=> a = b++ * b++ * b++;

=> a = 3 * 3 * 3;Here we are using post-increement operator, so the 3 is not incremented in

this statement.

=> a = 27;Here, 27 is store in the variable a. By the way, the value of variable bisincremented by 3. (ie: b=6)

Step 3: printf(" %d, %d\n" , a, b);It prints the value of variable aand b.

Hence the output of the program is 27, 6.


#include

#define FUN(i, j) i##j

int main(){

int va1=10;

int va12=20;

printf("%d\n", FUN(va1, 2));

return 0;

}

A.10 B.20

C.1020 D.12


Answer: Option B

Explanation:

The following program will make you understand about ## (macro concatenation) operator

clearly.

#include

#define FUN(i, j) i##j

int main()

{ int First = 10;

int Second = 20;

char FirstSecond[] = "IndiaBIX";

printf("%s\n", Fun(First, Second) );


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return 0;

}

Output:

-------

IndiaBIX

The preprocessor will replace Fun(F ir st, Second)as FirstSecond.

Therefore, the printf (" %s\n" , Fun(F ir st, Second) );statement will become as printf(" %s\n" ,

F ir stSecond );

Hence it prints IndiaBIXas output.

Like the same, the line printf(" %d\n" , FUN(va1, 2));given in the above question will

become as printf(" %d\n" , va12 );.

Therefore, it prints 20as output.

We can prevent the same file from getting included again by using a preprocessor directive

called #ifndef(short for "if not defined") to determine whether we've already defined a

preprocessor symbol called XSTRING_H. If we have already defined this symbol, the

compiler will ignore the rest of the file until it sees a #endif(which in this case is at the end

of the file).

Macro calls and function calls work exactly similarly.

A.True B. False


Answer: Option B

Explanation:

False, A macro just replaces each occurrence with the code assigned to it. e.g. SQUARE(3)

ith ((3)*(3))in the program.

A function is compiled once and can be called from anywhere that has visibility to the

funciton.


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A macro must always be defined in capital letters.

A.True B. False

Answer: Option B

Explanation:

FALSE, The macro is case insensitive.

Macros have a local scope.

A.True B.False


Answer: Option B

The scope of macros is globals and functions. Also the scope of macros is only from the

point of definition to the end of the file.

After preprocessing all the macro in the program are removed.

It is necessary that a header files should have a .h extension?

A.Yes B.No


Answer: Option B

Explanation:

No, the header files have any kind of extension.

Will the program compile successfully?

#include

int main()

{

#ifdef NOTE

int a;

a=10;

#else

int a;

a=20;

#endif

printf("%d\n", a);

return 0;


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}

A.Yes B. o


Answer: Option A

Explanation:

Yes, this program will compile and run successfully and prints 20.The macro #ifdef NOTE evaluates the given expression to 1. If satisfied it executes the

#ifdef block statements. Here #ifdef condition fails because the Macro NOTE is nowhere

declared.

Hence the #else block gets executed, the variable a is declared and assigned a value of 20.

printf("%d\n", a); It prints the value of variable a 20.

Will it result in to an error if a header file is included twice?

A.Yes

B. No

C.It is compiler dependent.

Answer & ExplanationAnswer: Option C

Explanation:

Unless the header file has taken care to ensure that if already included it doesn't get

included again.

Turbo C, GCC compilers would take care of these problems, generate no error.

What will be the output of the program if the array begins at 65472 and each integer

occupies 2 bytes?

#include

int main()

{

int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};

printf("%u, %u\n", a+1, &a+1);

return 0;

}

A.65474, 65476 B.65480, 65496

C.65480, 65488 D.65474, 65488


Answer: Option B

Explanation:

Step 1: in t a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};The array a[3][4] is declared as an

integer array having the 3 rows and 4 colums dimensions.


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Step 2: printf(" %u, %u\n" , a+1, & a+1);

The base address(also the address of the first element) of array is 65472.

For a two-dimensional array like areference to array has type "pointer to array of 4 ints".

Therefore, a+1is pointing to the memory location of first element of the second row inarray a. Hence 65472 + (4 ints * 2 bytes) = 65480

Then, & ahas type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1

denotes "12 ints * 2 bytes * 1 = 24 bytes".

Hence, begining address 65472 + 24 = 65496. So, &a+1= 65496

Hence the output of the program is 65480, 65496

3. What will be the output of the program (sample.c) given below if it is executed from thecommand line (Turbo C in DOS)?

cmd> sample 1 2 3

/* sample.c */

#include


{

int j;

j = argv[1] + argv[2] + argv[3];

printf("%d", j);

return 0;

}

A. B. sample 6

C. rror D.Garbage value


Answer: Option C

Explanation:

Here argv[1], argv[2]and argv[3]are string type. We have to convert the string to integer

type before perform arithmetic operation.

Example:j = atoi (ar gv[1]) + atoi (ar gv[2]) + atoi (ar gv[3]);

http://www.eskimo.com/~scs/cclass/int/sx4ab.html
http://www.eskimo.com/~scs/cclass/int/sx4ab.htmlhttp://www.eskimo.com/~scs/cclass/int/sx4ab.htmlhttp://www.eskimo.com/~scs/cclass/int/sx4ab.html


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Because of the nature of base-2 arithmetic, it turns out that shifting left and shifting right

are equivalent to multiplying and dividing by two. These operations are equivalent for the

same reason that tacking zeroes on to the right of a number in base 10 is the same as

multiplying by 10, and deleting digits from the right is the same as dividing by 10. You can

convince yourself that 0x56 > 1 is the same as

0x56 / 2. It's also the case that masking off all but the low-order bits is the same as taking aremainder; for example, 0x56 & 0x07 is the same as 0x56 % 8. Some programmers

therefore use , and & in preference to *, /, and % when powers of two are involved,

on the grounds that the bitwise operators are ``more efficient.'' Usually it isn't worth

worrying about this, though, because most compilers are smart enough to perform these

optimizations anyway (that is, if you write x * 4, the compiler might generate a left shift

instruction all by itself), they're not always as readable, and they're not always correct for

negative numbers.

The issue of negative numbers, by the way, explains why the right-shift operator >> is not

precisely defined when the high-order bit of the value being shifted is 1. For signed values,

if the high-order bit is a 1, the number is negative. (This is true for 1's complement, 2'scomplement, and sign-magnitude representations.) If you were using a right shift to

implement division, you'd want a negative number to stay negative, so on some computers,

under some compilers, when you shift a signed value right and the high-order bit is 1, new

1 bits are shifted in at the left instead of 0s. However, you can't depend on this, because not

all computers and compilers implement right shift this way. In any case, shifting negative

numbers to the right (even if the high-order 1 bit propagates) gives you an incorrect

answer if there's a remainder involved: in 2's complement, 16-bit arithmetic, -15 is 0xfff1,

so -15 >> 1 might give you 0xfff8shifted which is -8. But integer division is supposed to

discard the remainder, so -15 / 2 would have given you -7. (If you're having trouble seeing

the way the shift worked, 0xfff1 is 11111111111100012 and 0xfff8 is

11111111111110002. The low-order 1 bit got shifted off to the right, but

because the high-order bit was 1, a 1 got shifted in at the left.)

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