Consider the following C code:

#include <stdio.h>

int main(void) {
    int a = 0;
    printf("%d\n",   a  = 1);
}

has precedence over = so it should be evaluated first. So we have a on the left and 1 on the right. Now based on my understanding a should result in a (only changes its value) which is lvalue so why this statement gives the following error:

lvalue required as left operand of assignment

CodePudding user response：

The prefix increment operator results in the incremented value of its operand, however it is not an lvalue. Section 6.5.3.1p2 of the C standard: describes the semantics as follows:

The value of the operand of the prefix operator is incremented. The result is the new value of the operand after incrementation. The expression E is equivalent to (E =1). See the discussions of additive operators and compound assignment for information on constraints, types, side effects, and conversions and the effects of operations on pointers.

Then section 6.5.16.2p3 regarding compound assignment operators states:

A compound assignment of the form E1 op = E2 is equivalent to the simple assignment expression E1 = E1 op (E2), except that the lvalue E1 is evaluated only once, and with respect to an indeterminately-sequenced function call, the operation of a compound assignment is a single evaluation.

And 6.5.16p3 regarding the assignment operator further states:

An assignment operator stores a value in the object designated by the left operand. An assignment expression has the value of the left operand after the assignment, but is not an lvalue.

So it is explicitly not allowed. Even if it was, an expression such as a = 1 would cause a to be modified more than once without an intervening sequence point which would trigger undefined behavior.

This is one of those place where C and C differ. C does in fact allow the result of the = operator, and by extension compound assingment and prefix /--, to be an lvalue.