memory-layout

问题 As far as I know, the Rust compiler is allowed to pack, reorder, and add padding to each field of a struct. How can I specify the precise memory layout if I need it? In C#, I have the StructLayout attribute, and in C/C++, I could use various compiler extensions. I could verify the memory layout by checking the byte offset of expected value locations. I'd like to write OpenGL code employing custom shaders, which needs precise memory layout. Is there a way to do this without sacrificing

问题 It is clear from the reference manual that the memory layout of structs is unspecified (when the repr attribute is not used). This rule gives the compiler the possibility to pack structures tighter by reordering the fields. What about the memory layout of tuples and tuple structs? How is it (un)specified and why? 回答1: The memory layout of tuples and tuple structs is undefined, just like the layout of normal structs, with one exception: The exception to this is the unit tuple ( () ) which is

问题 Consider the following: // Just a sequence of adjacent fields of same the type #[repr(C)] #[derive(Debug)] struct S<T> { a : T, b : T, c : T, d : T, } impl<T : Sized> S<T> { fn new(a : T, b : T, c : T, d : T) -> Self { Self { a, b, c, d, } } // reinterpret it as an array fn as_slice(&self) -> &[T] { unsafe { std::slice::from_raw_parts(self as *const Self as *const T, 4) } } } fn main() { let s = S::new(1, 2, 3, 4); let a = s.as_slice(); println!("s :: {:?}\n\ a :: {:?}", s, a); } Is this code

问题 Consider the following: // Just a sequence of adjacent fields of same the type #[repr(C)] #[derive(Debug)] struct S<T> { a : T, b : T, c : T, d : T, } impl<T : Sized> S<T> { fn new(a : T, b : T, c : T, d : T) -> Self { Self { a, b, c, d, } } // reinterpret it as an array fn as_slice(&self) -> &[T] { unsafe { std::slice::from_raw_parts(self as *const Self as *const T, 4) } } } fn main() { let s = S::new(1, 2, 3, 4); let a = s.as_slice(); println!("s :: {:?}\n\ a :: {:?}", s, a); } Is this code

来源： https://stackoverflow.com/questions/50300643/storage-order-for-multidimensional-arrays-in-c

问题 why when i debug asm source in gdb is 0x8048080 the address chosen for the starting entry point into code? this is just a relative offset, not an actual offset of into memory of an instruction, correct? 回答1: There is no special significance to address 0x8048080 , but there is one for address 0x08048000 . The latter address is the default address, on which ld starts the first PT_LOAD segment on Linux/x86. On Linux/x86_64, the default is 0x400000 , and you can change the default by using a

问题 why when i debug asm source in gdb is 0x8048080 the address chosen for the starting entry point into code? this is just a relative offset, not an actual offset of into memory of an instruction, correct? 回答1: There is no special significance to address 0x8048080 , but there is one for address 0x08048000 . The latter address is the default address, on which ld starts the first PT_LOAD segment on Linux/x86. On Linux/x86_64, the default is 0x400000 , and you can change the default by using a

问题 [Not a duplicate of Structure padding and packing. That question is about how and when padding occurs. This one is about how to deal with it.] I have just realized how much memory is wasted as a result of alignment in C++. Consider the following simple example: struct X { int a; double b; int c; }; int main() { cout << "sizeof(int) = " << sizeof(int) << '\n'; cout << "sizeof(double) = " << sizeof(double) << '\n'; cout << "2 * sizeof(int) + sizeof(double) = " << 2 * sizeof(int) + sizeof(double

问题 I have troubles understanding the exact meaning of a paragraph of C99 draft standard (N1256) about bit-fields (6.7.2.1:10): 6.7.2.1 Structure and union specifiers [...] Semantics [...] An implementation may allocate any addressable storage unit large enough to hold a bit-field. If enough space remains, a bit-field that immediately follows another bit-field in a structure shall be packed into adjacent bits of the same unit. If insufficient space remains, whether a bit-field that does not fit

问题 I have troubles understanding the exact meaning of a paragraph of C99 draft standard (N1256) about bit-fields (6.7.2.1:10): 6.7.2.1 Structure and union specifiers [...] Semantics [...] An implementation may allocate any addressable storage unit large enough to hold a bit-field. If enough space remains, a bit-field that immediately follows another bit-field in a structure shall be packed into adjacent bits of the same unit. If insufficient space remains, whether a bit-field that does not fit