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See more: C++ C
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?
Posted 16-Jan-13 1:17am

1 solution

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Solution 1

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.
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.

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