expression evaluation

**lazy_hack** · 08-21-2009

Hi,

Although the expression *((a<20)?&b:&c)=30; works but

((a<20)?b:c)=30 gives

error: invalid lvalue in assignment.

what could be the reason for it? Please need sum urgent help.

Thanks in advance.

**hk_mp5kpdw** · 08-21-2009

The result of the left hand side of:

Code:

((a<20)?b:c)=30

is a value (the value stored in either b or c). Let's say for example that b is 20 and c is 30. If a is 10 then the result of the left-hand side is 20 and you are effectively trying to say:

Code:

20=30

which would of course give you that error.

**lazy_hack** · 08-21-2009

full program:

Code:

main()
{
  int a=10,b,c;
  ((a<20)?b:c)=30;
  printf("%d",b);
 }

this also gives error invalid lvalue in assignment.

but if i replace the line with *((a<20)?&b:&c)=30; it works.

please explain also why the second one works .

**laserlight** · 08-21-2009

Originally Posted by lazy_hack

but if i replace the line with *((a<20)?&b:&c)=30; it works.

please explain also why the second one works .

You are assigning to what the corresponding pointer points to, and that is not a problem. Where did you come across this code, anyway? It looks like a lazy hack, or rather, bad style.

**BEN10** · 08-21-2009

Originally Posted by lazy_hack

full program:

Code:

main()
{
  int a=10,b,c;
  ((a<20)?b:c)=30;
  printf("%d",b);
 }

this also gives error invalid lvalue in assignment.

but if i replace the line with *((a<20)?&b:&c)=30; it works.

please explain also why the second one works .

In the code above b and c are having garbage values, that's why as told to you in post#2, it'll give you error, while using & you are directly "attacking" in the address, thus no error.

**laserlight** · 08-21-2009

Originally Posted by BEN10

In the code above b and c are having garbage values, that's why as told to you in post#2, it'll give you error, while using & you are directly "attacking" in the address, thus no error.

That is off the mark. That b and c were not initialised does not matter, since the intention is to assign a value to them. What matters is that the result of the conditional operator is not an lvalue, hence the attempted assignment does not work.

**BEN10** · 08-21-2009

I think hk_mp5kpdw's post is also wrong then.

**laserlight** · 08-21-2009

Originally Posted by BEN10

I think hk_mp5kpdw's post is also wrong then.

What do you find wrong about it? hk_mp5kpdw made the point that the result of the conditional operator is not a variable that can be assigned to, but a value that cannot be assigned to.

**BEN10** · 08-21-2009

From hk_mp5kpdw's post, let b has a value of 12345 in the code given by the OP, then this will happen

Code:

12345=30;

That's why the error. This is what I said in my previous post.

**roaan** · 08-21-2009

Originally Posted by BEN10

In the code above b and c are having garbage values, that's why as told to you in post#2, it'll give you error, while using & you are directly "attacking" in the address, thus no error.

No you said this. But as laserlight pointed out that the uninitialized b and c are not a problem here. The problem is the assignment being checked

for example as in your case

12345 = 30 // this is wrong

**laserlight** · 08-21-2009

Originally Posted by BEN10

That's why the error. This is what I said in my previous post.

Fair enough: looking at your post #5 again, I see that I missed the phrase "as told to you in post#2". It would have been clearer if you had given an example there and then, or at least suggested the scenario where 20 and 30 were the "garbage" values of b and c respectively.

**lazy_hack** · 08-21-2009

thanks for the explanation...

"the result of the conditional operator is not an lvalue" line is the crux i believe...

the code was from test ur c skills book....just anther intrsting book..

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