error: expression is not assignable ternary operator

Question

I have the following code and MPLABX XC8 compiler gives this error:

error: expression is not assignable

U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : uart1.oerr = 0;
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ^

Here is the relevant code section:

typedef union {
    struct {
        bool ferr  : 1;  // FERIF Framing Error
        bool aerr  : 1;  // ABDOVF Error
        bool oerr  : 1;  // RXFOIF Error
        bool ready : 1;  // Data Ready to be read
        uint8_t reserved : 4;
    };
    uint8_t status;
}uart1_status_t;

static volatile uart1_status_t uart1;

U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : uart1.oerr = 0;

Same thing does not give error when I use

if (U1ERRIRbits.RXFOIF) 
    uart1.oerr = 1;
else 
    uart1.oerr = 0;

Do not understand why?

Try instead `U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : (uart1.oerr = 0)`. Condition binds stronger than assignment. [C Operator Precedence](https://en.cppreference.com/w/c/language/operator_precedence). — Scheff's Cat, Mar 27 '19 at 12:53
@Blaze thank you! It looks way better @ Scheff thank you for the answer! :) — Arda30, Mar 27 '19 at 12:55
What does this mean? "Same thing does not give error when I use" This is not the same thing at all! A ternary operator is (as any operator) used as part of an expression. Expressions do not create lvalues. In your alternate version you have separate statements where you assign a value to an lvalue. — Gerhardh, Mar 27 '19 at 13:07
General rule: A ternary operator is not the same as an `if` clause — Gerhardh, Mar 27 '19 at 13:08
Another general rule: avoid using `?:` until it is actually needed. Which isn't very often at all. — Lundin, Mar 27 '19 at 14:06

score 4 · Answer 1 · answered Mar 27 '19 at 12:58

U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : uart1.oerr = 0;

is interpreted as:

(U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : uart1.oerr) = 0;

Which tries to assign 0 to ...? If you want to use such construct, you need braces:

U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : (uart1.oerr = 0);

Or better:

uart1.oerr = U1ERRIRbits.RXFOIF ? 1 : 0;

or:

uart1.oerr = !!U1ERRIRbits.RXFOIF;

or:

uart1.oerr = (bool)U1ERRIRbits.RXFOIF;

or really just:

uart1.oerr = U1ERRIRbits.RXFOIF;

as typeof(uart1.oerr) == bool, the value will be implicitly converted to 1 for nonzero values or 0 for zero.

score 2 · Answer 2 · answered Mar 27 '19 at 12:58

The answer is simple that, due to C operator precedence,

U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : uart1.oerr = 0;

is compiled as

(U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : uart1.oerr) = 0;

In C, a condition doesn't provide an LValue. Hence, this is a compiler error.

One solution would be to use parentheses:

U1ERRIRbits.RXFOIF ? uart1.oerr = 1 : (uart1.oerr = 0);

Please note, that uart1.oerr = 1 doesn't need parentheses as ? and : act like parentheses (a specialty of the ternary operator).

The even simpler solutions are already mentioned in Kamils answer...

error: expression is not assignable ternary operator

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