Can you have a incrementor and a decrementor on the same variable in the same statement in c

问题

Is

--foo++;

a valid statement in C? (Will it compile/run) And is there any practical application for this?

Sorry for changing the question in an edit but I found something out.

According to my C++ compiler (Visual Studio 2010):

--++foo;

is a valid command but

foo--++; 

is not. Is there any reason for this?

回答1:

No, it is not valid because the result of the increment / decrement operators is not a lvalue.

EDIT: the OP edited his question by adding two more examples . So here we go, for the same reason:

--++foo;
--foo++;
foo--++;

are all invalid expression statements because the result of increment / decrement operators is not a lvalue. A compiler could extend the language and accepts these expressions, but a strictly conforming program cannot contain any of these expressions.

回答2:

As this is a C expression it should follow some cascading evaluation steps to find out result.

your expressions are

--++foo;

ok lets evaluate this first step:-

++foo;

must be evaluated to proceed

it is some how pre increment so foo become foo+1 .

then assignment happen

that is this evaluate to foo=foo+1;

and value of foo return for assignment to a variable(if nothing present then ignored).

so now our expression is like below

--(++foo); evaluated to

--(a constant value);//result of foo+1

again the statement evaluated like

a constant value=constant_valu+1;//right part is ok but left part is not a variable

as left side of assignment operator is not a variablle and raise an error.

so now no valid operand for -- /decrement operator So lvalue required error

来源：https://stackoverflow.com/questions/11679258/can-you-have-a-incrementor-and-a-decrementor-on-the-same-variable-in-the-same-st