Question

applied gate test series

What is the output of the following c-code[ Note: Assume integer as 2 bytes] ?

#include<stdio.h>

int main()

{  int x = 500;

char *p=&x;

*++p=2;

printf(“%d”,x);

return 0;

}

Answer 1

 

At the last x = 512 + 244 = 756

here 512 is upper nibble and 244 is lower nibble value

Value change when we use a big-endian processor in place of a little-endian. 

Comments

thanks bro just one doubt now where will p be pointing after this? at address 1000 or 1001

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1001

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What upper nibble lower nibble? Nibble means 4 bit

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@raja11sep lower nibble means the lower address

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ok,,thanks