Testbook | Output of the following program

Question

main() {

int i = 511;

char *ptr = (char *)&i;

printf("%d", *ptr); }

Answer 1

i=

511

address 1000

 511 = 00000001 11111111

11111111	00000001
Lower byte (1000-1001)	upper byte (1002-1003)

ptr =1000 {char pointer}

11111111 which is in 2's complement o/p: -1

Comments

@srestha

{ char *ptr = (char *)&i; } type casting is done here

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we can also anlyse it other way ...just think char can range from (-128 to 127)..if there are any value like 

char c=128..then (printf("%d",c);  it will print -128....bocz  ...it will behave like circular queue...(-128,-127,....-1,0,1....,127)again (-128...)...

..solution ((511-128)-256)-128) no we would be at ... -1

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@hs_yadav

solution ((511-128)-256)-128) no we would be at ... -1

Why you are subtracting with 256 and again 128... Although I understood why you are subtracting first time with 128.

Please suggest your approach