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> gcc will compile this code and advance the pointer address by 1 byte. Yes, it's not standard conformant. https://godbolt.org/z/cjoYvjxGq https://godbolt.org/

> gcc will compile this code and advance the pointer address by 1 byte. Yes, it's not standard conformant. https://godbolt.org/z/cjoYvjxGq https://godbolt.org/z/cjoYvjxGq <source>: In function 'void foo(int)': <source>:4:28: warning: pointer of type 'void *' used in arithmetic [-Wpointer-arith] 4 | int * p = malloc(size) + 1; | ~~~~~~~~~~~~~^~~ <source>:4:28: error: invalid conversion from 'void*' to 'int*' [-fpermissive] 4 | int \* p = malloc(size) + 1; | ~~~~~~~~~~~~~^~~ | | | void\* Compiler returned: 1

jstimpfle 2y ago
first, yes it did accept the pointer arithmetic even though it issued a warning. Second, the error about implicit conversion from void-pointer to typed-pointer is in C++ only -- not in C, where you won't even get a warning. The error happens because you didn't cast the void pointer to the target type and doesn't have to do anything with the fact that gcc _does_ let you (as a GCC extension) do arithmetic on void pointers. If you remove the "+ 1" you'll still see the same error in C++. In C++, change the target pointer to a void pointer: void *ptr = malloc(42) + 1; or, alternatively, cast the RHS expression properly (doesn't matter with "+ 1" or without) int *ptr = (int *) (malloc(42) + 1); and the error will go away.
- menaerus 2y ago
  I don't think there's a need to explain to me how pointers work, or what to do to make this error/warning go away, since it should be pretty obvious from my response that this is something I know very well. I was only disputing the claim that ending up with the unaligned ptr is not trivial for given line of code by providing explanation, and counter-example, why this is not possible. At least not in C++ because that's the language I had in mind. What you're saying is that it's possible to do void pointer arithmetic in C by using the GNU extension - fine, I can't disagree with that.
  - jstimpfle 2y ago
    You corrected me and I corrected you. No need to take personal offense. I don't care what you know -- I obviously took the compiler output that you presented above as a refutal to what I said earlier, and I pointed out why it's not a valid refutal.

jstimpfle 2y ago

first, yes it did accept the pointer arithmetic even though it issued a warning. Second, the error about implicit conversion from void-pointer to typed-pointer is in C++ only -- not in C, where you won't even get a warning. The error happens because you didn't cast the void pointer to the target type and doesn't have to do anything with the fact that gcc _does_ let you (as a GCC extension) do arithmetic on void pointers. If you remove the "+ 1" you'll still see the same error in C++. In C++, change the target pointer to a void pointer: void *ptr = malloc(42) + 1; or, alternatively, cast the RHS expression properly (doesn't matter with "+ 1" or without) int *ptr = (int *) (malloc(42) + 1); and the error will go away.

menaerus 2y ago
I don't think there's a need to explain to me how pointers work, or what to do to make this error/warning go away, since it should be pretty obvious from my response that this is something I know very well. I was only disputing the claim that ending up with the unaligned ptr is not trivial for given line of code by providing explanation, and counter-example, why this is not possible. At least not in C++ because that's the language I had in mind. What you're saying is that it's possible to do void pointer arithmetic in C by using the GNU extension - fine, I can't disagree with that.
- jstimpfle 2y ago
  You corrected me and I corrected you. No need to take personal offense. I don't care what you know -- I obviously took the compiler output that you presented above as a refutal to what I said earlier, and I pointed out why it's not a valid refutal.