unions

问题 Case 1. File: test1.c : unsigned long val = (unsigned long)&"test"; int main() { return 0; } Compiler invocation: cl test1.c /c Results: Microsoft (R) C/C++ Optimizing Compiler Version 19.25.28611 for x64 Copyright (C) Microsoft Corporation. All rights reserved. test1.c test1.c(1): warning C4311: 'type cast': pointer truncation from 'char (*)[5]' to 'unsigned long' test1.c(1): error C2099: initializer is not a constant Case 2. File: test2.c : union { unsigned long val; } val = { (unsigned

问题 Case 1. File: test1.c : unsigned long val = (unsigned long)&"test"; int main() { return 0; } Compiler invocation: cl test1.c /c Results: Microsoft (R) C/C++ Optimizing Compiler Version 19.25.28611 for x64 Copyright (C) Microsoft Corporation. All rights reserved. test1.c test1.c(1): warning C4311: 'type cast': pointer truncation from 'char (*)[5]' to 'unsigned long' test1.c(1): error C2099: initializer is not a constant Case 2. File: test2.c : union { unsigned long val; } val = { (unsigned

问题 Since my question here couldn't be confidently answered, I ask here again in hope that someone knows for sure: Is there any difference (besides syntactical) between a pointer to a union and a union that contains pointers to its elements? The generated assembly in this example is identical. As long as I'm never accessing the other members, is it allowed to allocate memory for only one of the members (which isn't the largest)? Regarding the 2nd question, 6.5.2.1 of the C89 draft says: The size

问题 My story starts off the same as this person's here: Unions in C++11: default constructor seems to be deleted The resolution here (now about three years old) is a bit unsatisfactory, because the "Digging into the standard" that the author did ended up with concluding that the behavior was as described in the standard, but unfortunately the quote is from a Note , and those are supposed to be non-normative (I've been told). Anyways, there's a link to an old bug report on gcc that is claimed to

问题 C++'s unions are more restrictive than those of C, because they introduce the concept of an "active member" (the one last assigned to) as the only one safe to access. The way I see it, this behavior of unions is a net negative. Can someone please explain what is gained by having this restriction? 回答1: Short answer In C, the union is only a question of how to interpret the data that is stored at a given location. The data is passive. In C++, unions can have members of different classes. And

问题 C++'s unions are more restrictive than those of C, because they introduce the concept of an "active member" (the one last assigned to) as the only one safe to access. The way I see it, this behavior of unions is a net negative. Can someone please explain what is gained by having this restriction? 回答1: Short answer In C, the union is only a question of how to interpret the data that is stored at a given location. The data is passive. In C++, unions can have members of different classes. And

问题 When optional and required property are merged via intersection, required wins type A = { who: string } type B = { who?: string } // $ExpectType {who:string} type R = A & B This may lead to runtime errors, when for instance, dealing with default params pattern within a function type Params = { who: string greeting: string } const defaults: Params = { greeting: 'Hello', who: 'Johny 5', } function greeting(params: Partial<Params>){ // $ExpectType Params const merged = {...defaults, ...params}

问题 When optional and required property are merged via intersection, required wins type A = { who: string } type B = { who?: string } // $ExpectType {who:string} type R = A & B This may lead to runtime errors, when for instance, dealing with default params pattern within a function type Params = { who: string greeting: string } const defaults: Params = { greeting: 'Hello', who: 'Johny 5', } function greeting(params: Partial<Params>){ // $ExpectType Params const merged = {...defaults, ...params}

问题 When optional and required property are merged via intersection, required wins type A = { who: string } type B = { who?: string } // $ExpectType {who:string} type R = A & B This may lead to runtime errors, when for instance, dealing with default params pattern within a function type Params = { who: string greeting: string } const defaults: Params = { greeting: 'Hello', who: 'Johny 5', } function greeting(params: Partial<Params>){ // $ExpectType Params const merged = {...defaults, ...params}

问题 When optional and required property are merged via intersection, required wins type A = { who: string } type B = { who?: string } // $ExpectType {who:string} type R = A & B This may lead to runtime errors, when for instance, dealing with default params pattern within a function type Params = { who: string greeting: string } const defaults: Params = { greeting: 'Hello', who: 'Johny 5', } function greeting(params: Partial<Params>){ // $ExpectType Params const merged = {...defaults, ...params}