Valid use of the conditional operator?

Question

Consider the following piece of code:

int i, k, m;
k = 12;
m = 34;
for (i = 0; i < 2; i++) ((i & 1) ? k : m) = 99 - i;
printf("k: %ld   m: %ld\n\n", k, m);

In this silly example, the conditional operator expression is a shortcut for:

if (i & 1) k = 99 - i; else m = 99 - i;

My compiler does not complain and executing this piece of code gives the expected output

k: 98   m: 99

My question, though, is if this is valid code according to the C standard? I have never seen anything like it used before.

Answer 1

Footnote 110 in the C11 standard:

A conditional expression does not yield an lvalue.

And 6.5.16 paragraph 2:

An assignment operator shall have a modifiable lvalue as its left operand.

So no, that code does not conform to the C standard.

In C++11, it is valid:

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

So this is another one of those dusty corners where C and C++ differ significantly. If your compiler doesn't produce an error, then I'm guessing you're using a C++ compiler with a "C mode", rather than a proper C compiler; MSVC?

Answer 2

This is not legal C, and a compiler that accepts it is wrong. However, you can do the same thing with:

*((i & 1) ? &k : &m) = 99 - i;

and it becomes legal C.

Answer 3

It is valid to use conditional as an lval in C++, but not in C.

In C++ (ISO?IEC 14882:1998(E) 5.16.4)

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

If you would like to use similar trick in C, you should use pointers:

ISO/IEC 9899:TC2, 6.5.14.6

If both the second and third operands are pointers or one is a null pointer constant and the other is a pointer, the result type is a pointer to a type qualified with all the type qualifiers of the types pointed-to by both operands.

*((i & 1) ? &k : &m) = 99 - i;