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Declarations and Initializations Questions

NA
SHSTTON
263
Solv. Corr.
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Solv. In. Corr.
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21 / 54

Choose the correct option.

Which of the following operators in 'C' programming language takes only integer operands?


A+

B*

C/

D%

Answer: Option D

Explanation:

% is the answer cause it always takes both the inputs as integers in c language.

Workspace

NA
SHSTTON
211
Solv. Corr.
222
Solv. In. Corr.
433
Attempted
0 M:50 S
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22 / 54

Which of the following option is correct for the below program?
#include<stdio.h>
void main()
{
int a,b,c;
b=2;
a= 2*(b++);
c = 2*(++b);
}

Aa=4,c=6

Ba=3,c=8

Cb=3, c=6

Da=4, c=8

ENone of these

Answer: Option D

Explanation:

b=2;  

a= 2*(b );..........a=2*2=4  Here post increment of b will take place and b value after this line has executed will become 2 1= 3.

c = 2*( b);  Here pre increment will take place and before the excecution of this line b value will become 3 1=4 hence c= 2*4=8



so o/p= a=4 and c=8.

Workspace

NA
SHSTTON
318
Solv. Corr.
504
Solv. In. Corr.
822
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0 M:42 S
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23 / 54

Which of the following option is correct for the below 'C' program?
#include<stdio.h>
int main () {
int  ( *Commprintf) (const char*, ... ) = printf;
Commprintf (  "Hello World");
return 0;
}

ANo output

BUndefined symbol Commprintf

CCompile Error: Prototype mismatch

DHello World

Answer: Option D

Explanation:

Here is no explanation for this answer

Workspace

NA
SHSTTON
395
Solv. Corr.
473
Solv. In. Corr.
868
Attempted
1 M:13 S
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24 / 54

What is the problem in the following declarations?
int func(int);
double func(int);
int func(float);

AA function with same name connot have different signatures.

BA function with same name cannot have different return types.

CA function with same name cannot have different number of parameters

DAll of the mentioned.

Answer: Option D

Explanation:

A function with the same name cannot have different signatures.



 int func(int); double func(int);



A function with the same name cannot have different return types. 



double func(int);

int func(float);



  A function with the same name cannot have a different number of parameters



.int func(int);int func(float);



 

Workspace

NA
SHSTTON
585
Solv. Corr.
295
Solv. In. Corr.
880
Attempted
0 M:56 S
Avg. Time

25 / 54

What is the output of the following code?
#include <stdio.h>
void main()
{
int a=0, b=0;
a = (b =75)+9;
printf("\n%d, %d ",a,b);
}

A75, 9

B75, 84

C84, 75

DNone of these Options

Answer: Option C

Explanation:

a = (b =75) 9;......b value is assigned to 75 so a value will become 75+9= 84.



final value will become a=84 and b=75

Workspace

NA
SHSTTON
143
Solv. Corr.
258
Solv. In. Corr.
401
Attempted
0 M:8 S
Avg. Time

26 / 54

Choose the correct option.

Assignment operator targets to


AL-value

BH-value

CNone of the above

DBoth (a) and (b)

Answer: Option A

Explanation:

Assignment operators are used to assign the the values present on the right-hand side to whatever is present on the left-hand side of the operator. The left-hand side can be a variable or a pointer, and the right-hand side can be a variable, a constant, an expression or a function call.

For example-

a=5;    //assigns '5' to 'a'.

On the other hand, 5=a;    //error.

Very often in C, we use the terminology of l-value in error messages. It basically refers to the operands present on the left-hand side of the assignment operator. It can be a variable or a pointer, but should not be a constant.

Hence, from the above theory, we can conclude for sure that assignment operator targets to L-value.

Workspace

NA
SHSTTON
134
Solv. Corr.
259
Solv. In. Corr.
393
Attempted
0 M:9 S
Avg. Time

27 / 54

What will be the output of the following "C" code?
#include<stdio.h>
void main()
{
100;
printf("%d",100);
}

ACompilation Error

B100

C1

DNone of these

Answer: Option B

Explanation:

100;  ...........there was no use of this 100. Its only to confuse the students.

printf("%d",100);  ...... its as simple as to print hello world.  hence 100 is printed.

Workspace

NA
SHSTTON
118
Solv. Corr.
272
Solv. In. Corr.
390
Attempted
0 M:0 S
Avg. Time

28 / 54

What will be the output of the following "C" code?
#include<stdio.h>
void main()
{
int i=10;
i=!i>14;
printf("%d", i);
}

A0

B10

C1

DCompilation Error

Answer: Option A

Explanation:

Here is no explanation for this answer

Workspace

NA
SHSTTON
58
Solv. Corr.
325
Solv. In. Corr.
383
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0 M:0 S
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29 / 54

What will be the output of the following "C" code?
#include<stdio.h> 
void main ( )
{
int i;
for( i=0; i<10; i++,printf("%d", i));
}

A2345678910

B123456789

CCompilation Error

D12345678910

Answer: Option D

Explanation:

In a for loop, first the initialization part is checked, then the test condition. If the test condition is true then the body of the loop gets executed. Then, updation part is done, again the test condition is checked. If true then again the body gets executed, and this process goes on until the test condition becomes false.

Now, let's dry run this particular code-

Value of i                        Condition(i<10)                    Updation                        Status

i=0                                    True                                       i=0+1=1                  1 gets printed

i=1                                    True                                       i=1+1=2                  2 gets printed

i=2                                    True                                       i=2+1=3                  3 gets printed

i=3                                    True                                       i=3+1=4                  4 gets printed

i=4                                    True                                       i=4+1=5                  5 gets printed

i=5                                    True                                       i=5+1=6                  6 gets printed          

i=6                                    True                                       i=6+1=7                  7 gets printed

i=7                                    True                                       i=7+1=8                  8 gets printed

i=8                                    True                                       i=8+1=9                  9 gets printed

i=9                                    True                                       i=9+1=10                10 gets printed

i=10                                  False                                            --                               --

Hence, the output will be 12345678910




Workspace

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What will be output of the following "c" code?
#include<stdio.h> 
void main ( )
{
 int a=500, b=100,c;
if( !(a>=400))
b=200;
c=200;
printf( "%d %d", b,c);
}

A200 100

B100 200

C200 200

DCompilation Error

Answer: Option B

Explanation:

In line no 4, a and b are initialized with values 500 and 100 respectively.

In the next line, we have an if condition, let's try to analyze it-

a>=400, which is true as 500>400, so the result is 1, but !1 results to 0. So, the 'if' part now becomes something like if(0), which is false. So, the 'if' part is not executed, which means b is not updated to 200. b remains 100 as it was.

c gets initialzed to 200.

So, the output will be 100 200.

Workspace

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