Ternary conditional and assignment operator precedence

I'm confused about direct assignment and ternary conditional operators precedence:

#include<stdio.h>
int main(void)
{
    int j, k;

    j = k = 0;
    (1 ? j : k) = 1; // first
    printf("%d %d\n", j, k);

    j = k = 0;
    1 ? j : k = 1; // second
    printf("%d %d\n", j, k);
    return 0;
}

I would expect the output to be:

1 0
1开发者_JAVA百科 0

But it happens to be:

1 0
0 0

Plus I get this warning:

main.cpp:20: warning: statement has no effect

which is about the line I commented as second.

Since the direct assignment operator has less precedence than the ternary conditional operator, I was expecting lines commented as first and second to be equivalent. But alas it is not the case.

I tried this with g++ --version (Ubuntu 4.4.3-4ubuntu5) 4.4.3

The operator precedence in the C/C++ language in not defined by a table or numbers, but by a grammar. Here is the grammar for conditional operator from C++0x draft chapter 5.16 Conditional operator [expr.cond]:

conditional-expression:
    logical-or-expression
    logical-or-expression ? expression : assignment-expression

The precedence table like this one is therefore correct when you use assignment on the left side of the doublecolon, but not when used on the right side. What is the reason for this asymmetry I have no idea. It may be a historical reason: in C the conditional result was not lvalue, therefore assigning something to it had no sense, and allowing assignment to be accepted without parentheses might seem smart at that time.

The second line is equivalent to:

1 ? (j) : (k = 1);

That's the same as:

j;

That's the same as:

;

The key is that the two operands of the ternary conditional operator can be expressions, so operator precedence isn't relevant here. It's simply that the second operand is the assignment expression k = 1.

(1 ? j : k) = 1;

is equivalent to,

if(true) j = 1;
else k = 1;

And,

1 ? j : k = 1;

is equivalent to,

if(true) j;  // warning: statement has no effect
else k = 1;

In the second case,

1 ? j : k = 1;

is evaluated as:

(1) ? (j) : (k = 1);

and since one evaluates to true, the expression evaluates to j which does nothing.