3 ms·

I mean, just like with 1 dimensional arrays, it depends on the context. Array memory is on the stack. The size of that array is actually not known at run time,

I mean, just like with 1 dimensional arrays, it depends on the context. Array memory is on the stack. The size of that array is actually not known at run time, its only known at compile time, where any reference to that length gets resolved by the compiled. If your 2d array sits on the stack, then inferring memory layout is pretty easy. If you are dealing with pointer that was passed to a function, then you can't assume anything about data size or limits, which is why many functions that take pointers take a size parameter as well.

alexey-salmin 5mo ago
> If your 2d array sits on the stack, then inferring memory layout is pretty easy. If you are dealing with pointer that was passed to a function, then you can't assume anything about data size or limits, which is why many functions that take pointers take a size parameter as well. Right, but 2d arrays come into this picture with their own quirks again. You're not just passing the size as the parameter, you can pass it as a "special" parameter that influences how the compiler will interpret other parameters. E.g. in C99 you can do this: void do(size_t x, size_t y, int a[][y]); Here "y" plays the critical role because it will be used to compute offsets in the a[i][j] expression. For 1d arrays this doesn't happen. Of course it's still generalizable as "all but the outermost dimensions should be known" and for 1d array the outermost dimension is the only dimension. Still, this whole thing always felt a bit odd to me.
- uecker 5mo ago
  Well, you give the explanation yourself: The size for the outermost array is not always needed, and then C allows it to be omitted. But my recommendation is to always give the size and then everything is regular and the compiler can use the information for warnings.
simiones 5mo ago
> Array memory is on the stack. Array memory can sit on either the stack or the heap. > The size of that array is actually not known at run time, its only known at compile time, where any reference to that length gets resolved by the compiled. This is also a bit misleading, in two ways. First, it's not clear what you mean by "size" here - the size of the memory block(s), or the shape of the array? Second, many people think that the C runtime doesn't know the amount of memory allocated to an array, but this is actually false. It's just the C abstract model that for some reason chose to not expose this information - but the size is actually always stored and accessible, and this is virtually mandated by the standard: otherwise, `free(arr)` couldn't realistically work, it would have to be `free(arr, size)`. This is one of the weirdest inefficiencies of C, in fact - it requires you to store the size of arrays twice - once in user code, and another time in the internal logic of the allocator. Edit: and as a fun extra, C++ not only inherited this mistake from C, but reproduced it again, meaning that a C++ array allocated with new[] actually stores the size twice, at least with typical implementations - once in the C++ runtime and again in the allocator - and still requires the user-space code to store it a third time. This is because `delete[]` needs to call the destructors of all of the elements of the array, regardless of where and how the array was allocated, so the number of array elements needs to be stored alongside the object itself.
- ActorNightly 5mo ago
  >Array memory can sit on either the stack or the heap. No, if we are using the definition of an array that is like int c[] = ..., that is always going to be on the stack. Heap continuous memory =/= array. You can use the [] operator to access it like an array, but fundamentally, as far as structures in C language are concerned, those 2 are different, because they get treated by compiler differently. >but the size is actually always stored and accessible, and this is virtually mandated by the standard: otherwise, `free(arr)` couldn't realistically work, That would only be true if each element in the array was a char. The dynamic data structure stores total amount of memory allocated by address, it has no info about the size of the element, so it can't infer the actual number of items at runtime. You could write your own malloc that does this, but generally, that is left to the user for flexibility. For example, a really good practice in C coding that basically solves any double free is a mempool that allocates all the memory up front. That way, you never really even have to call free, and the memory you allocate can be partitioned any way you chose dynamically.
  - Joker_vD 5mo ago
    > that is always going to be on the stack. Unless your C implementation doesn't use a stack for data storage. Which existed, you know: IIRC the C compilers for Cray machines used linked lists to hold activation frames. And of course, there are PIC microcontrollers where you can't really use the hardware stack for anything except return addresses.
  - simiones 5mo ago
    > No, if we are using the definition of an array that is like int c[] = ..., that is always going to be on the stack. Heap continuous memory =/= array. You can use the [] operator to access it like an array, but fundamentally, as far as structures in C language are concerned, those 2 are different, because they get treated by compiler differently. Well, not necessarily. For one thing, if we have a function foo(int c[]), it's debatable if c is an array variable or a pointer variable. However, what's not debatable is that you can allocate a struct on the heap, and that struct can have an array member - e.g. `struct foo { int a[10]; }; [...] struct foo *x = malloc(sizeof(struct foo));` would allocate an array on the heap as part of the struct. > That would only be true if each element in the array was a char. That's why I said that it depends on what exactly you mean by the size of the array. It's also true that in today's world at least, malloc() will often allocate more memory than you actually ask for, to optimize against fragmentation - and then the internally stored size is the size of the actual allocation, not the logical size that you requested - which may not even fit into a whole number of array elements. So, I was being a little overly simplistic (lying) for dramatic effect. > For example, a really good practice in C coding that basically solves any double free is a mempool that allocates all the memory up front. While this is a very valid technique for certain purposes, especially when dynamic allocation is needed in very high performance code, it's very much not a valid solution for memory safety - quite the contrary, it's a terrible practice for that. In particular, this is almost exactly the issue that caused the infamous HeartBleed vulnerability in OpenSSL to stay hidden for so long: the use of a memory pool for the buffers used to store TLS packets was hiding the buffer overflow from UBSan and valgrind and similar tools, since the reads were perfectly valid from a language perspective (they were never reading from free()d/unallocated memory, only from memory that had been released to the memory pool).

alexey-salmin 5mo ago

> If your 2d array sits on the stack, then inferring memory layout is pretty easy. If you are dealing with pointer that was passed to a function, then you can't assume anything about data size or limits, which is why many functions that take pointers take a size parameter as well. Right, but 2d arrays come into this picture with their own quirks again. You're not just passing the size as the parameter, you can pass it as a "special" parameter that influences how the compiler will interpret other parameters. E.g. in C99 you can do this: void do(size_t x, size_t y, int a[][y]); Here "y" plays the critical role because it will be used to compute offsets in the a[i][j] expression. For 1d arrays this doesn't happen. Of course it's still generalizable as "all but the outermost dimensions should be known" and for 1d array the outermost dimension is the only dimension. Still, this whole thing always felt a bit odd to me.

uecker 5mo ago
Well, you give the explanation yourself: The size for the outermost array is not always needed, and then C allows it to be omitted. But my recommendation is to always give the size and then everything is regular and the compiler can use the information for warnings.

simiones 5mo ago

> Array memory is on the stack. Array memory can sit on either the stack or the heap. > The size of that array is actually not known at run time, its only known at compile time, where any reference to that length gets resolved by the compiled. This is also a bit misleading, in two ways. First, it's not clear what you mean by "size" here - the size of the memory block(s), or the shape of the array? Second, many people think that the C runtime doesn't know the amount of memory allocated to an array, but this is actually false. It's just the C abstract model that for some reason chose to not expose this information - but the size is actually always stored and accessible, and this is virtually mandated by the standard: otherwise, `free(arr)` couldn't realistically work, it would have to be `free(arr, size)`. This is one of the weirdest inefficiencies of C, in fact - it requires you to store the size of arrays twice - once in user code, and another time in the internal logic of the allocator. Edit: and as a fun extra, C++ not only inherited this mistake from C, but reproduced it again, meaning that a C++ array allocated with new[] actually stores the size twice, at least with typical implementations - once in the C++ runtime and again in the allocator - and still requires the user-space code to store it a third time. This is because `delete[]` needs to call the destructors of all of the elements of the array, regardless of where and how the array was allocated, so the number of array elements needs to be stored alongside the object itself.

ActorNightly 5mo ago
>Array memory can sit on either the stack or the heap. No, if we are using the definition of an array that is like int c[] = ..., that is always going to be on the stack. Heap continuous memory =/= array. You can use the [] operator to access it like an array, but fundamentally, as far as structures in C language are concerned, those 2 are different, because they get treated by compiler differently. >but the size is actually always stored and accessible, and this is virtually mandated by the standard: otherwise, `free(arr)` couldn't realistically work, That would only be true if each element in the array was a char. The dynamic data structure stores total amount of memory allocated by address, it has no info about the size of the element, so it can't infer the actual number of items at runtime. You could write your own malloc that does this, but generally, that is left to the user for flexibility. For example, a really good practice in C coding that basically solves any double free is a mempool that allocates all the memory up front. That way, you never really even have to call free, and the memory you allocate can be partitioned any way you chose dynamically.
- Joker_vD 5mo ago
  > that is always going to be on the stack. Unless your C implementation doesn't use a stack for data storage. Which existed, you know: IIRC the C compilers for Cray machines used linked lists to hold activation frames. And of course, there are PIC microcontrollers where you can't really use the hardware stack for anything except return addresses.
- simiones 5mo ago
  > No, if we are using the definition of an array that is like int c[] = ..., that is always going to be on the stack. Heap continuous memory =/= array. You can use the [] operator to access it like an array, but fundamentally, as far as structures in C language are concerned, those 2 are different, because they get treated by compiler differently. Well, not necessarily. For one thing, if we have a function foo(int c[]), it's debatable if c is an array variable or a pointer variable. However, what's not debatable is that you can allocate a struct on the heap, and that struct can have an array member - e.g. `struct foo { int a[10]; }; [...] struct foo *x = malloc(sizeof(struct foo));` would allocate an array on the heap as part of the struct. > That would only be true if each element in the array was a char. That's why I said that it depends on what exactly you mean by the size of the array. It's also true that in today's world at least, malloc() will often allocate more memory than you actually ask for, to optimize against fragmentation - and then the internally stored size is the size of the actual allocation, not the logical size that you requested - which may not even fit into a whole number of array elements. So, I was being a little overly simplistic (lying) for dramatic effect. > For example, a really good practice in C coding that basically solves any double free is a mempool that allocates all the memory up front. While this is a very valid technique for certain purposes, especially when dynamic allocation is needed in very high performance code, it's very much not a valid solution for memory safety - quite the contrary, it's a terrible practice for that. In particular, this is almost exactly the issue that caused the infamous HeartBleed vulnerability in OpenSSL to stay hidden for so long: the use of a memory pool for the buffers used to store TLS packets was hiding the buffer overflow from UBSan and valgrind and similar tools, since the reads were perfectly valid from a language perspective (they were never reading from free()d/unallocated memory, only from memory that had been released to the memory pool).