Click here to Skip to main content
15,507,388 members
Please Sign up or sign in to vote.
2.00/5 (1 vote)
See more:
int main( ) {
 char str[] = { 48, 49, 50, 51, 52, 53, 54};
 int *p = (int *) (str + 1);
 char *pStr = (char *)p;
 printf("%c\n", *pStr);

what is the o/p ?
in *p what gets stored and *pStr what gets stored?

1 solution

Try it!
Compile it, run it and see.

Output is "1"
p contains a pointer to the second byte in the array of characters
So does pStr.

It might be different if you try to actually use p as a pointer to an int and retrieve an integer value from it - some systems will fail, because the int pointer is (probably) not word aligned. But char values don't have to have any particular alignment, so that works fine.
Share this answer
Andreas Gieriet 16-Jan-13 8:11am    
My 5!
Key is, as you say, as long as you do not de-reference the non-aligned pointer, all is fine.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

CodeProject, 20 Bay Street, 11th Floor Toronto, Ontario, Canada M5J 2N8 +1 (416) 849-8900