constants

问题 Anyone care to elaborate on how val in scala is different from const in java? What are the technical differences? I believe I understand what "const" is in c++ and java. I get the feeling that "val" is somehow different and better in some sense but I just can't put my finger on it. Thanks 回答1: const in Java has no function—it's reserved but you can't actually use it for anything. Declaring a Java variable as final is roughly equivalent . Declaring a variable as a val in Scala has similar

问题 I have int, float, boolean and string from Properties file. Everything has loaded in Properties. Currently, I am parsing values as I know expected value for particular key. Boolean.parseBoolean("false"); Integer.parseInt("3") What is better way of setting these constants values, If I don't know what could be primitive value datatype for a key. public class Messages { Properties appProperties = null; FileInputStream file = null; public void initialization() throws Exception { appProperties =

问题 I got an assignment to implement a template array class. One of the requirement is to overload the [] operator. I made this two const and non-const version which seems to be working fine. const T& operator[](const unsigned int index)const and T& operator[](const unsigned int index) My question is how will the compiler know which one to run when i will do something like: int i=arr[1] On a non-const array? 回答1: The non-const function will always be called on a non-const array, and the const

问题 I got an assignment to implement a template array class. One of the requirement is to overload the [] operator. I made this two const and non-const version which seems to be working fine. const T& operator[](const unsigned int index)const and T& operator[](const unsigned int index) My question is how will the compiler know which one to run when i will do something like: int i=arr[1] On a non-const array? 回答1: The non-const function will always be called on a non-const array, and the const

问题 I got an assignment to implement a template array class. One of the requirement is to overload the [] operator. I made this two const and non-const version which seems to be working fine. const T& operator[](const unsigned int index)const and T& operator[](const unsigned int index) My question is how will the compiler know which one to run when i will do something like: int i=arr[1] On a non-const array? 回答1: The non-const function will always be called on a non-const array, and the const

问题 I got an assignment to implement a template array class. One of the requirement is to overload the [] operator. I made this two const and non-const version which seems to be working fine. const T& operator[](const unsigned int index)const and T& operator[](const unsigned int index) My question is how will the compiler know which one to run when i will do something like: int i=arr[1] On a non-const array? 回答1: The non-const function will always be called on a non-const array, and the const

问题 This question already has answers here : typedef pointer const weirdness (6 answers) What is the difference between char * const and const char *? (19 answers) Closed 3 years ago . Why aren't const B and const A* indistinguishable, when B is typedef'ed to A* ? When compiling this simple example: struct A {}; typedef A* B; void f1(const A* a1); void f2(const B a2); int main() { const A a; f1(&a); f2(&a); } I get the following compiler output (G++ 6.3.1): test.cpp: In function ‘int main()’:

问题 I am a beginner to C++ and have the following question regarding the purpose of pointers where neither the address nor the value that it is pointing to can be changed ( constEverything in the example ). Example: int main() { int i = 1; int j = 2; int* const constPoint = &i; *constPoint = 3; // legal constPoint = &j; // illegal const int* constVal = &i; *constVal = 3; // illegal constVal = &j; // legal const int* const constEverything = &i; *constEverything = 3; // illegal constEverything = &j

问题 I was looking at an example which showed that why typedef'ng a pointer is a bad practice. The part I didn't understand about the example is that why the compiler wasn't able to catch the problem. I elaborated the example into the following code: #include <stdio.h> typedef int *TYPE; void set_type(TYPE t) { *t = 12; } void foo(const TYPE mytype) { set_type(mytype); // Error expected, but in fact compiles } int main() { TYPE a; int b = 10; a = &b; printf("A is %d\n",*a); foo(a); printf("A is %d

问题 In the following code... #include <stdlib.h> #include <stdint.h> extern void get_buffer_from_HW_driver(volatile uint32_t **p); void getBuffer(volatile uint32_t **pp) { // Write an address into pp, that is obtained from a driver // The underlying HW will be DMA-ing into this address, // so the data pointed-to by the pointer returned by this // call are volatile. get_buffer_from_HW_driver(pp); } void work() { uint32_t *p = NULL; getBuffer((volatile uint32_t **)&p); } ...the compiler rightfully