lvalue

int main () { int a = 5,b = 2; printf("%d",a+++++b); return 0; } This code gives the following error: error: lvalue required as increment operand But if I put spaces throughout a++ + and ++b , then it works fine. int main () { int a = 5,b = 2; printf("%d",a++ + ++b); return 0; } What does the error mean in the first example? printf("%d",a+++++b); is interpreted as (a++)++ + b according to the Maximal Munch Rule ! . ++ (postfix) doesn't evaluate to an lvalue but it requires its operand to be an lvalue . ! 6.4/4 says the next preprocessing token is the longest sequence of characters that could

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 following rules: [...] Otherwise, the value contained in the object indicated by the glvalue is the prvalue

I'm having trouble printing a member of a struct that is returned from a function: #include <stdio.h> struct hex_string { char a[9]; }; struct hex_string to_hex_string_(unsigned x) { static const char hex_digits[] = "0123456789ABCDEF"; struct hex_string result; char * p = result.a; int i; for (i = 28; i >= 0; i -= 4) { *p++ = hex_digits[(x >> i) & 15]; } *p = 0; printf("%s\n", result.a); /* works */ return result; } void test_hex(void) { printf("%s\n", to_hex_string_(12345).a); /* crashes */ } The printf call inside to_hex_string_ prints the correct result, but the printf call inside test_hex

Assuming the definition: int i = 10; int *p = &i; Why is *p a valid lvalue here: *p+=10; Shouldn't *p evaluate to the value of the int stored at &i, ie. 10, and hence generate a "Not an lvalue" error? SiegeX An lvalue is an expression that refers to a region of storage that can be manipulated. *p is such an expression that refers to a region of storage. This is different than say 10+=10; because 10 doesn't refer to a region of storage like a variable would. I believe you are confused with the definition of p . p is, in fact, a variable of type pointer to int , and its value is initialized to

This question already has an answer here: Literal initialization for const references 3 answers We cannot write int& ref = 40 because we need lvalue on right side. But we can write const int& ref = 40 . Why is this possible? 40 is rvalue instead lvalue I know that this is an exception but why? As Stroustrup says: The initializer for a const T& need not be an lvalue or even of type T. In such cases: [1] First, implicit type conversion to T is applied if necessary. [2] Then, the resulting value is placed in a temporary variable of type T. [3] Finally, this temporary variable is used as the value

I have here char text[60]; Then I do in an if : if(number == 2) text = "awesome"; else text = "you fail"; and it always said expression must be a modifiable L-value. You cannot change the value of text since it is an array, not a pointer. Either declare it as char pointer (in this case it's better to declare it as const char* ): const char *text; if(number == 2) text = "awesome"; else text = "you fail"; Or use strcpy: char text[60]; if(number == 2) strcpy(text, "awesome"); else strcpy(text, "you fail"); 来源： https://stackoverflow.com/questions/6008733/expression-must-be-a-modifiable-l-value

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, instead of just removing the restrictions on normal references to allow them to be bound to temporaries?

In lisp you can say: (setf (aref a 1) 5) In perl you can say: substr( $string, $start, $stop ) =~ s/a/b/g Is it possible something like this in python? I mean is it possible to use function result as a lvalue (as a target for assignment operation)? No. Assigning to the result of a function call is specifically prohibited at the compiler level: >>> foo() = 3 File "<stdin>", line 1 SyntaxError: can't assign to function call There are however two special cases in the Python syntax: # Slice assignment a = [1,2,3,4] a[0:2] = 98, 99 # (a will become [98, 99, 3, 4]) # Tuple assignment (x, y, z) = (10