lvalue

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 7 years ago . #include<stdio.h> //line 1 #include<conio.h> //line 2 void main() //line 3 { //line 4 int a=6,g=7,b=3; //line 5 clrscr(); //line 6 printf("%d",a>?g=a:g=b); //line 7 getch(); //line 8 } Case 1: before saving the

问题 What is the reasoning behind the naming of "lvalue" and "rvalue" in C/C++ (I know how they function)? 回答1: The standard mentions this: An lvalue (so called, historically, because lvalues could appear on the left-hand side of an assignment expression) [...] An rvalue (so called, historically, because rvalues could appear on the right-hand side of an assignment expression) [...] That is, an lvalue was something you could assign to and an rvalue was something you could assign from. However, this

问题 There is an error. Is it wrong to assign a value to a[i] in the following code? Or something is wrong with conditional operators? #include<stdio.h> #include<string.h> int main(){ char a[12]="sumit tyagi"; int i=0; while(a[i]!='\0'){ a[i]>90 ? a[i]=a[i]-32 : a[i]=a[i]+32; //error in this line i++; } printf("\n %s",a); 回答1: a[i]>90 ? a[i]=a[i]-32 : a[i]=a[i]+32; is not evaluated as a[i]>90 ? (a[i]=a[i]-32) : (a[i]=a[i]+32); since = has lower precedence than ?: . In standard C you can't write it

问题 I have the following code, which I cannot get to work: struct foo {}; foo foo1 = {}; template <foo& F> class FooClass {}; template <foo& F> void foobar(FooClass<F> arg) { } int main() { FooClass<foo1> f; foobar(f); } The error is: main.cpp:14:5: error: no matching function for call to 'foobar' note: candidate template ignored: substitution failure : deduced non-type template argument does not have the same type as the its corresponding template parameter ('foo' vs 'foo &') Is it at all

问题 I'm using N3936 as a reference here (please correct this question if any of the C++14 text differs). Under 3.10 Lvalues and rvalues we have: Every expression belongs to exactly one of the fundamental classifications in this taxonomy: lvalue, xvalue, or prvalue. However the definition of lvalue reads: An lvalue [...] designates a function or an object. In 4.1 Lvalue-to-rvalue conversion the text appears: [...] In all other cases, the result of the conversion is determined according to the

问题 One of the cool new features of the upcoming C++ standard, C++0x, are "rvalue references." An rvalue reference is similar to an lvalue (normal) reference, except that it can be bound to a temporary value (normally, a temporary can only be bound to a const reference): void FunctionWithLValueRef(int& a) {...} void FunctionWithRValueRef(int&& a) {...} int main() { FunctionWithLValueRef(5); // error, 5 is a temporary FunctionWithRValueRef(5); // okay } So, why did they invent a whole new type,

问题 It is very common to have struct-like types that are not expected to be modified by distant copyholders. A string is a basic example, but that's an easy case because it's excusably immutable -- Python is unusual in even allowing things like method calls on literal strings. The problem is that (in most languages) we frequently have things like, say an (x,y) Point class. We occasionally want to change x and y independently. I.e., from a usage perspective, a Point LVALUE should be mutable (even

问题 There are two kinds of expressions in Objective-C 1. RValue The term rvalue refers to a data value that is stored at some address in memory 2. LValue Expressions that refer to a memory location is called "lvalue" expression. An lvalue may appear as either the left-hand or right-hand side of an assignment I didn't get it . can someone explain it to me? 回答1: RValue is a value that is evaluated but does not have a designated memory address to be stored until assigned to such a memory location.

问题 Below is a triple nested indexing scheme. My pointer to an array of pointers is dereferenced on the commented line... in theory this should give me a pointer. I subtract one from it and then reference it and reassign that to my pointer to the array of pointers. But that line gives an lvalue error for the operand "&". Let me be perfectly clear. I want to both know why the error is occurring here AND get a method that works for assigning the address of the previous element in the array of

问题 #include <iostream> int foo() { return 0; } int main() { const int& a = foo(); std::cout << &a << std::endl; } In this code, a binds to a rvalue. Is it legal to take its address? (And by legal I mean: in the code ill-formed? Am I causing an undefined behaviour?) 回答1: This is fine. In C++11 you can even do this: int&& a = foo(); a = 123; You can kind of think about temporaries like this (conceptually and in general): x = func(); // translated as: auto __temporary = func(); x = __temporary; _