error C2106: '=' : left operand must be l-value

[SunnyPriya]

Hi,
I get error (error C2106: '=' : left operand must be l-value) for

a>10?b=100:b=200;

but, not for

a>10?b=100b=200);

Please explain.

Regards
Sunnypriya

[SunnyPriya]

I could see a funny character
I get error for

a>10?b=100:b=200;

but not for

a>10?b=100:(b=200);

[exterminator]

I did not get any errors with VC++ 6.0. What compiler are you using? how are a and b defined (although this doesnt make any difference because you say using the brackets doesnt cause this error). May be there is error somewhere else and you are getting the error for that statement in this one.

[SunnyPriya]

Thats true,
I am also using VC++ 6.0.

If I write a C program (save it as .c file), then I will get error.
If I use C++ (save it as .cpp file,), I dont get any error.


By the way, the portion of code is.

	int a,b;
	a=5;
	a>10?b=100:b=200;

.

But still, why it gives l value error in C.

[SuperKoko]

I suppose that operator precedences are different in C, and operator= must have a lower precedence than operator?:
So:

a>10?b=100:b=200;

Is interpreted as:

(a>10?b=100:b)=200; // (a>10?b=100:b) is a r-value of type integer.

I think that C++ should also emit this error.

[SuperKoko]

I tried with borland C++ 5.5, and the result is the same:
operator= seems to have a higher precedence than operator?: in C++, and not in C.

[NMTop40]

the expression you probably want though is

b = a >10 ? 100 : 200;

which has the correct operator precedence (>, ?:, = )

[exterminator]

may be yes it is the operator precendence that is causing this error but in C++ even its the same but its working there. I am not sure but this is what I got at this link. I would quote it as:

The conditional operator does not produce an lvalue. Therefore, a statement such as a ? x : y = 10 is not valid.

I will try finding some more information over it though and let you know if I find something but this ensures that l-value error but the reason for the one with braces working is still mystery for me. It seems quite logical that the operator precedence is the cause (as pointed out by SuperKoko).

[Soni_Sharad]

Hi,
I get error (error C2106: '=' : left operand must be l-value) for

a>10?b=100:b=200;

but, not for

a>10?b=100b=200);

Please explain.

Regards
Sunnypriya

Hi,

I am using Visual Studio.NET 2003 for VC++ and I am not getting any error!

Rgs,
SoniSharad

[SuperKoko]

I will try finding some more information over it though and let you know if I find something but this ensures that l-value error but the reason for the one with braces working is still mystery for me. It seems quite logical that the operator precedence is the cause (as pointed out by SuperKoko).

This one

a>10?b=100:(b=200);

works well.
operator= can be used almost everywhere, as a normal operator: it assigns a value to its left operand which must be a l-value (that is b in this example), and returns the assigned value.
So, a>10?b=100b=200), evaluates to 100 or 200 depending on the value of a, after having assigned this value to b.
It is preferable, and clearer to use:

b=(a>10?100:200);

The fact that operator:? returns a r-value, means that:

int x=40,y=40;
int a=8;
(a>10?x:y)=42; // is invalid (if i refer to that rule).

but this code is accepted by bcc32 and gcc, because they are smart enough to understand that since, both x and y are l-value, the result of (a>10?x:y) is a l-value.

[SuperKoko]

Moreover:

(a>10?x:y)=42;

Is valid for the ISO standard.
http://library.n0i.net/programming/c/cp-iso/expr.html

-4- If the second and third operands are lvalues and have the same type, the result is of that type and is an lvalue.

[SuperKoko]

I finally understood the problem.
In fact, for both, C++ and C, the operator precedence is identical, and:

a>10?b=100:b=200; // is equivalent to (a>10?b=100:b)=200;

But with the ISO C++, (a>10?b=100:b) evaluates to a l-value, because the second and third operand are both l-values.
With ANSI C, (a>10?b=100:b) evaluates to a r-value, that is why, the assignment emit a compilation error!

[exterminator]

Bang!!! You got it SuperKoko. It surely somehow pointed that there was something to do with the standards. Great work.