C Programming :: Declarations and Initializations

Answer: Option B

statement 1 will only be printed when i value=0. But i value is initialized as i=4 so it doesn't go with the first condition hence the second condition is accepted and statement 2 is printed.

Answer: Option A

4th line of the program says j=10 . 5th line says j means post increment so after that line is executed j value will become 11 but during the execution of the 5th line j value is still 10 so i= 10-10=0

Now in the 5th line itself there has been two times increment of j so final value will be 12.

Answer: Option A

p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

Answer: Option D

#define max(a,b) ((a).(b)?(a):(b)) This is the complete and correct way to define the function max . In the code its incomplete hence it shows compiler error.

Answer: Option A

The value of x initialised is 10. Also i=x so i=10. Now when comparing condition  (i==x) it will give boolean result that is if the condition is true it will return 1 else it will return 0. So here the condition is true coz x=i hence output is 1.

Answer: Option B

Here integer (unsigned) is assumed as 2 bytes so that for loop is till 16 i.e (2byte = 16 bits.)

As unsigned integer so user is expected to enter positive number.

statement in printf (num << i & 1 << 15)? 1:0 is used to convert decimal to binary equivalent.

in the above statement below things has been used.

<< ----> left shift

& -----> And operator

? : ------> trinery operator

Ex: if the number is 10 and its binary equivalent 1010 or (00000000000001010 in 16 bit)

the printf statement will be

for i = 0

(10 << 0 & 1 << 15)>?1:0

for i = 1

(10 << 1 & 1 << 15)>?1:0

similarly for i = 2, 3, 4 , ---- 15

Answer: Option B

x = (y*=2) (z=a=y); Meaning of y*=2 means y=y*2 hence y value becomes 2*2=4. 

Current value of y =4 so z=a=y=4.

Hence x=4+4=8

Answer: Option D

The value of y%2 is 0. This value is assigned to x.
The condition will if (0) so z goes uninitialized.

Answer: Option A

Here is no explanation for this answer

Answer: Option C

Here char is signed so,the size of char is 1 bytes and its value range are -128 to 127,

and it will be something like below

-128 -127 -126 .......... 0 ....... 126 127 in the cycle. 

Now as per questions c = 125 it is under the above limit.

but when 10 added to it, it becomes 125 10 = 135

125 126 127 -128 -127 -126 -125 -124 -123 -122 -121 

So, count above 135 will fall at -121, so Answer will be -121

If in the question it would have given unsigned char then answer will be 135