what will be output when you execute following c code include stdio..., Data types and variables, Adv C

Question:What will be output when you will execute following c code?#include<stdio.h>
int main(){
    char a=250;
    int expr;
    expr= a+ !a + ~a + ++a;
    printf("%d",expr);
    return 0;
}

A 249 
B 250 
C 0 
D -6 
E Compilation error

/69

+ AnswerD

+ Explanationchar a = 250;
250 is beyond the range of signed char. Its corresponding cyclic value is: -6
So, a = -6
Consider on the expression:
expr= a+ !a + ~a + ++a;
Operator! , ~ and ++ have equal precedence. And it associative is right to left.
So, First ++ operator will perform the operation. So value a will -5
Now,
Expr = -5 + !-5 + ~-5 + -5
= -5 + !-5 + 4 - 5
= -5 + 0 + 4 -5
= -6

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Question:What will be output when you will execute following c code?#include<stdio.h> int main(){ char a=250; int expr; expr= a+ !a + ~a + ++a; printf("%d",expr); return 0; }

A 249

B 250

C 0

D -6

E Compilation error

D

Question:What will be output when you will execute following c code?
#include<stdio.h> int main(){ char a=250; int expr; expr= a+ !a + ~a + ++a; printf("%d",expr); return 0; }