Question

The following program fragments unsigned i=1; int j =4; print("%u", i+j; prints​

Answer 1

Answer:

aninteger that changes from machine to machine

Explanation:

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.