C Operator Questions

Here C comma ( , ) works like a operator instead of a separator. So fst is first complied and thus called followed by snd .

Different compiler will give different output according to their precedence .

Copy the code and run in for language C++14 nd C++17 on the online compiler at the link given below to observe.

http://www.onlinegdb.com/

Consider octal number (26)₈ = (11010)₂ in binary.

Command skip = skip >>4; shifts the binary value (11010)₂ 4 place to right.

(11010)₂  ---- (00001)₂ 
Now after this is executed command skip = skip <<6; shifts the binary value (11010)₂ 6 place to left.

(00001)₂  ---- (1000000)₂  = (100)₈  in octadecimal
and (64)₁₀₀  in decimal.

Here the statements are executed step by step therefore

first a is post incremented ++a

then b is post decremented --b

then c added the current values of a b and d

then d is post incremented d++

resulting into c=c=(++a) + (--b) + (d++)

d\'s value is incremented only after c is calculated for the current values of and at the time c was executed.

Loop unrolling is a loop transformation technique that helps to optimize the execution time of a program. We basically remove or reduce iterations

instead of a for(i=0;i<3;i++) printf("Hello");

we use printf("Hello"); printf("Hello"); printf("Hello");

We have 8 operators =     ++(pre)    ==     ++(post)    ()    *    +    -

Following the precedence able for operators these will be solved in order :

()

++(post/pre)

*

+/-

=

therefore by order

g=5==(2)*1+4-1

g=5==2+4-1

g=5==5

g=1

In C a comma (,) acts as a separator. Note that here compiler initializes int i as 11 but then can not initialize 12 because it will not be a valid identifier, therefore it shows error.

The () operator has the highest precedence in the precedence table and it is evaluated from left to right therefore the rightmost value given will be assigned to i.

= has a higher precedence than  , operator therefore i=11 is first assigned to i and rest doesn\\\'t make any further changes to the value of i

The || (or) logic checks if the first statement is true if yes then it exits, otherwise evaluats the second statement and so on.

here  i++ gives l=0 (i becomes 1 after the execution of the command post increment ) therefore the logic gives FALSE value so compiler evaluates next statement fun(&j, &k) which swaps j and k and returns j which is now  6 i.e TRUE.

hence 0 || 6 gives TRUE therefore l=1.

therefore i=1 (post increment) , j=6 , k=9 , l=1.