memory-layout

问题 Is there a generic way (not platform dependent) to get at compile time the size of a class object in the memory, without counting the vtable pointers? 回答1: As you are asking for a portable way: class MyClass { private: struct S { DataMemberType1 dataMember1; ... DataMemberTypeN dataMemberN; } m; public: static const size_t MemberSize = sizeof(S); }; 回答2: Use sizeof on this class , it doesn't include size of the vtable just the pointer. 来源： https://stackoverflow.com/questions/28433973

问题 Does Linux provide an inaccessible memory area below the lower stack end that has a guaranteed minimum size? And if such a guaranteed minimum size exists, what is it? Or in other words, when should I start to worry about alloca() or so giving me pointers into valid, non-stack memory? 回答1: As the alloca man page says: There is no error indication if the stack frame cannot be extended. (However, after a failed allocation, the program is likely to receive a SIGSEGV signal if it attempts to

问题 Short introduction : I have written an Augmented Reality Application with the Oculus Rift in C++ (DirectX). One of my fragment shaders computes the undistortion for a omnidirectional camera model. The only Problem I have now is to read out the rendered undistortion texture2D and convert it further for 3DPose Tracking/Mapping of the real world using OpenCV. The funny Thing is, I have already done the other way around, which means I have already created shader resource views with my distorted

问题 This question already has answers here : C / C++ MultiDimensional Array Internals (4 answers) Closed 3 years ago . I know this answer is in violation of the reinterpret_cast rules but it also presumes that sub-arrays will be allocated linearly. I believed this was not guaranteed, but as I search the standard, I find my confidence wavering. If I statically allocate a 2D array, like this: int foo[][4] = { { 5, 7, 8 }, { 6, 6 }, {}, { 5, 6, 8, 9 } }; Am I allowed to assume that all elements will

问题 The standard defines when two types are layout-compatible . But, I don't see anywhere in the standard what the consequences are when two types are layout-compatible . It seems that layout-compatible is a definition which is not used anywhere. What is the purpose of layout-compatible ? Note: Supposedly, it could mean that the types have the same layout ( offsetof is the same for each corresponding member), so for example, for trivially copyable types, underlying bytes can be copied between

问题 I'm trying to write some code in Fortran which requires the re-ordering of an n-dimensional array. I thought the reshape intrinsic combined with the order argument should allow this, however I'm running into difficulties. Consider the following minimal example program test implicit none real, dimension(:,:,:,:,:), allocatable :: matA, matB integer, parameter :: n1=3, n2=5, n3=7, n4=11, n5=13 integer :: i1, i2, i3, i4, i5 allocate(matA(n1,n2,n3,n4,n5)) !Source array allocate(matB(n3,n2,n4,n1

问题 I'm learning about the layout of executable binaries. My end goal is to analyze a specific executable for things that could be refactored (in its source) to reduce the compiled output size. I've been using https://www.embeddedrelated.com/showarticle/900.php and https://www.geeksforgeeks.org/memory-layout-of-c-program/ as references for this initial learning. From what I've learned, a linker script specifies the addresses where sections of compiled binaries are placed. E.g. > ld --verbose |

问题 This is the output of the given program: sizeof(Empty) 1 sizeof(Derived1) 1 sizeof(Derived2) 4 sizeof(Derived3) 1 sizeof(Derived4) 8 sizeof(Dummy) 1 This is the program: #include <iostream> using namespace std; class Empty {}; class Derived1 : public Empty {}; class Derived2 : virtual public Empty {}; class Derived3 : public Empty { char c; }; class Derived4 : virtual public Empty { char c; }; class Dummy { char c; }; int main() { cout << "sizeof(Empty) " << sizeof(Empty) << endl; cout <<

问题 Here, in this code, the size of ob1 is 16 which is fine(because of the virtual pointer) but I can't understand why the size of ob2 is 24. #include <iostream> using namespace std; class A { int x; }; class B { int y, z; }; class C : virtual public A { int a; }; class D : virtual public B { int b; }; int main() { C ob1; D ob2; cout << sizeof(ob1) << sizeof(ob2) << "\n"; } I expect the size of ob2 as 20, but the output is 24 回答1: One possible layout for objects of type D is: +----------+ | y |

问题 It appears to be smart enough to only use one byte for A, but not smart enough to use one byte for B, even though there are only 8*8=64 possibilities. Is there any way to coax Rust to figure this out or do I have to manually implement a more compact layout? Playground link. #![allow(dead_code)] enum A { L, UL, U, UR, R, DR, D, DL, } enum B { C(A, A), } fn main() { println!("{:?}", std::mem::size_of::<A>()); // prints 1 println!("{:?}", std::mem::size_of::<B>()); // prints 2 } 回答1: Both bytes