指针算术和动态记忆？

Question

What does the following piece of code mean?

int* pointer = malloc (sizeof(int) + 3);
pointer++;

The allocated piece of memory can't be broken down into chunks of sizeof(int) . So what happens when pointer is asked to jump to the next "block"? Is it defined?

Answer 1

The code is valid but maybe unusual without more context.

Line 1: The malloc allocates 3 bytes larger than the size of an int . This is valid.

Line 2: The pointer++ is valid. It's just an address.

Further references to pointer (eg addition or subtraction or comparison) are valid. Dereferences (ie *pointer ) will result in undefined behaviour.

Not that those 3 "extra" bytes are valid storage space and can be addressed with a char * , for example.

Answer 2

pointer can be used for pointer comparison (C standard allows pointers to be one element past the last one). Read or write access is undefined.

Answer 3

*pointer is 3 bytes of 0 and sizeof(int) - 3 bytes of undefined. Which byte[s] (as related to significance in your int ) are undefined is platform dependent (on the system bytesex) so in terms of your C program, the whole thing might as well be undefined.