I have two questions,
1
the output is 1 1Code:main(){ int i =257; int *iptr =&i; printf (“%d %d”,*((char*)iptr),* ((char *)iptr+1)); }
here, iptr at first should print the value stored in first 8 bits of variable i and in next it should print value of second 8 bit block.
would you please explain why the out put is so?
2--
why there is an linker error, how should we use the variable so that the error wont come?Code:main() { extern int iExtern; iExtern = 20; printf(“%d”,iExtern); }
What if you change it to:
Think about what 257 looks like in binary.Code:int i = 258;
"I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008
"the internet is a scary place to be thats why i dont use it much." - billet, 03/17/2010
1) you are casting your int into a char, and 257 as a char is 1.
2) Prove there is a linker error. I see no link command and no linker error message.
Mainframe assembler programmer by trade. C coder when I can.
There certainly is, Dino:
test.obj : error LNK2001: unresolved external symbol "int iExtern"
Try to be less sarcastic and more helpful.
In 1 write out the integer 257 in binary and you will see why the output of your program 1 1 is correct.
In 2 remove the extern qualifier so the scope of the variable iExtern becomes local instead of global and try not to use reserved keywords for variable names.
