output of program

Question

If the following fragment (assume negative numbers are stored in 2's complement form)

unsigned i=1;

int j=-4;

printf("%u",i+j);

prints x then

printf("%d", 8*sizeof(int));

outputs an integer that is same as (log in the options are of  base 2)

a)8*log(x+3)     b)log(x+3)     c)unpredictable value    d)none of the above

Comments

sir,in first printf statement,format specifier is unsigned and -4 + 1 = -3 but -3 cant be prited as it is signed,hence 3 will be printed which is x.

now second printf statment prints 8* size of int which is 8*4 =32 //if size of int is 4 OR 8* 2 =16 if size of int is 2.

option A - 8 * log (3 + 3) =8 *log 6

option B - log 6

i guess answer depend upon size of INT which is system dependent.

pls correct me

Answer 1

Is answer a here

Comments

nopes

Answer 2

Option (d) None of the above.

The correct expression would be log(x + 2) (Not present in the options).

The first printf would produce (UINT_MAX – 2) == (4294967293) == ((2**32 – 1) – 2). The second printf basically prints the number of bits in an int. We can take x, add 2, (which would set all the bits), and then take a log2 to get the number of bits present in an int.