3 ms·

I'd like to see a C equivalent of this, just for comparisons sake to zig

tiehuis 9y ago
#include <stdint.h> #include <string.h> typedef struct { int32_t a; int32_t b; } Foo; int main(void) { uint8_t array[1024]; memset(array, 1, sizeof(array)); Foo *foo = (Foo*)(&array[0]); foo->a += 1; } Using clang 3.8.0-2. Compiling examples with `clang -S llvm-ir`. It appears that the array is aligned with the minimum ABI requirement 16 by default? May be a note of this in the standard, can't recall of the top of my head. %array = alloca [1024 x i8], align 16 ... %6 = load i32, i32* %5, align 4 ... store i32 %7, i32* %5, align 4 We can also explicitly specify the alignment required in C11. #include <stdalign.h> #include <stdint.h> #include <string.h> typedef struct { int32_t a; int32_t b; } Foo; int main(void) { uint8_t alignas(alignof(Foo)) array[1024]; memset(array, 1, sizeof(array)); Foo *foo = (Foo*)(&array[0]); foo->a += 1; } Results in the following IR. %array = alloca [1024 x i8], align 4 ... %6 = load i32, i32* %5, align 4 ... store i32 %7, i32* %5, align 4
- bjourne 9y ago
  On 64bit Linux, stack frames are always aligned at 16 byte boundaries. The first 8 bytes of the frame contains the return address then there are 8 bytes of padding and then comes the stack allocations. I think the example is poorly constructed, because it is inconceivable that the address to the start of an array would not be aligned sizeof(int*) bytes.
  - dbaupp 9y ago
    The example is illustrative enough: all the array needs to be misaligned in practice is a small value on the stack near it, e.g. if the Rust code has `let x: u8 = 1;` inserted after the array (or, I imagine, `uint8_t x = 1;` in the C, etc.), then the array's address is odd.
    - bjourne 9y ago
      Why would it be? I tried it with msvc and it always manages to put arrays in stack frames at 8 byte boundaries. I can't see why a compiler would not do that.
- viperscape 9y ago
  Also curious, would one ever do this in C? I understand the need in Rust and Zig, but in C is this usual?
deleted 9y ago
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#include <stdint.h> #include <string.h> typedef struct { int32_t a; int32_t b; } Foo; int main(void) { uint8_t array[1024]; memset(array, 1, sizeof(array)); Foo *foo = (Foo*)(&array[0]); foo->a += 1; } Using clang 3.8.0-2. Compiling examples with `clang -S llvm-ir`. It appears that the array is aligned with the minimum ABI requirement 16 by default? May be a note of this in the standard, can't recall of the top of my head. %array = alloca [1024 x i8], align 16 ... %6 = load i32, i32* %5, align 4 ... store i32 %7, i32* %5, align 4 We can also explicitly specify the alignment required in C11. #include <stdalign.h> #include <stdint.h> #include <string.h> typedef struct { int32_t a; int32_t b; } Foo; int main(void) { uint8_t alignas(alignof(Foo)) array[1024]; memset(array, 1, sizeof(array)); Foo *foo = (Foo*)(&array[0]); foo->a += 1; } Results in the following IR. %array = alloca [1024 x i8], align 4 ... %6 = load i32, i32* %5, align 4 ... store i32 %7, i32* %5, align 4

bjourne 9y ago
On 64bit Linux, stack frames are always aligned at 16 byte boundaries. The first 8 bytes of the frame contains the return address then there are 8 bytes of padding and then comes the stack allocations. I think the example is poorly constructed, because it is inconceivable that the address to the start of an array would not be aligned sizeof(int*) bytes.
- dbaupp 9y ago
  The example is illustrative enough: all the array needs to be misaligned in practice is a small value on the stack near it, e.g. if the Rust code has `let x: u8 = 1;` inserted after the array (or, I imagine, `uint8_t x = 1;` in the C, etc.), then the array's address is odd.
  - bjourne 9y ago
    Why would it be? I tried it with msvc and it always manages to put arrays in stack frames at 8 byte boundaries. I can't see why a compiler would not do that.
viperscape 9y ago
Also curious, would one ever do this in C? I understand the need in Rust and Zig, but in C is this usual?

deleted 9y ago

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