Thread: ponter doubt abc = *(int*)test;

Originally Posted by abhijit_mohanta

Above is a part of c program
Can anybody predict the contents of abc

Yeah, since "test" is a pointer to an array abc will contain a memory address. Or part of one on a 64-bit system.

Whatever the 4-bytes of "aaaa" points to; effectively the output of [97979797] in low-byte, high-byte order which is in this case 1633771873. Not that it matters b/c that code has an error in it, trying to shove 29 bytes into a 20-byte sequence so the OP should have wrote:

Code:

const char * foo = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaa";

Trick question. As-written the code will never compile (unless your compiler is an idiot) so what it will output is undetermined.

Originally Posted by tabstop

Eh, I think you missed the * in front. It would contain the contents of that memory address, interepreted as an int.

Oh yeah. Well it will definitely go out of bounds then. But as Mr. Cobb points out, hopefully the compiler will save the OP from his/her own silliness.

Originally Posted by Elkvis

well, test is an array of char, which converts implicitly to a char pointer, and in this code, it's being explicitly converted to an int pointer, and then dereferenced. this means that abc should contain the value 0x61616161.

"test is an array of char" .. that's too short. GCC sez:

Code:

error: initializer-string for array of chars is too long

And since code with errors will not compile, test will never be anything. And yes 0x61 is indeed 97 dec.