循环条件中的三元运算符：评估顺序/操作。 优先顺序不清楚

Question

Edit:

What is the real groups of the expression “3<8 ? (9<6 ? 7 : 5) : 2>0 ? 4 : 1” and the meaning of non-associative in PHP?

has been offered as a duplicate, but that concerns PHP, not C.

While building some test cases for a small program, I introduced a bug into the conditional part of a for loop, like this:

for(int row = 0; row < (mode == TEST) ? NO_OF_TESTS : ROWS; row++){}

(I know one should pull that out of the for loop, but that does not change the problem.)

This lead to a segmentation fault by overrunning the end of an array as the above results in an infinite loop.

Of course, the fix was easy enough:

for(row = 0; row < ((mode == TEST) ? NO_OF_TESTS : ROWS); row++)
//                 ^                                   ^

But I am more interested in the way the faulty implementation behaved.

Here is a complete piece of code (does not make sense in itself, as it is ripped out of context, but it demonstrates the problem).

#include <stdio.h>
#include <stdlib.h>

#define TEST 0
#define INTERACTIVE 1

#define ROWS 2
#define NO_OF_TESTS 3
#define MAX_FRUIT_LEN 50

int main(void)
{
    char test_cases[NO_OF_TESTS][MAX_FRUIT_LEN] =
    {{"Orange"},
     {"Apple"},
     {"Pineapple"}};

    int mode = TEST;
    int row = 0;

    //This fails - but in a strange way
    //Uncomment this `for` loop and comment the other one to see the effects

    //for(int row = 0; row < (mode == TEST) ? NO_OF_TESTS : ROWS; row++)

    //With the parantheses, obviously, it works.
    for(row = 0; row < ((mode == TEST) ? NO_OF_TESTS : ROWS); row++)
    {
        printf("Working:\tIn row %d: Mode: %d condition_eval: %d\n"
        , row , mode, row < ((mode == TEST) ? NO_OF_TESTS : ROWS));

        printf("Not Working:\tIn row %d: Mode: %d condition_eval: %d\n"
        , row, mode, row < (mode == TEST) ? NO_OF_TESTS : ROWS);

        printf("Row: %d \tFruit Name: %s\n",row, test_cases[row]);
    }
    printf("\nTerminating conditional evaluation (at row %d):\n", row);

    printf("Working:\tIn row %d: Mode: %d condition_eval: %d\n"
    , row , mode, row < ((mode == TEST) ? NO_OF_TESTS : ROWS));

    printf("Not Working:\tIn row %d: Mode: %d condition_eval: %d\n"
    , row, mode, row < (mode == TEST) ? NO_OF_TESTS : ROWS);

    return 0;
}

Looking at the output and the (wrong) conditional

row < (mode == TEST) ? NO_OF_TESTS : ROWS

it appears that the compiler interprets this as:

   (row < (mode == TEST)) ? NO_OF_TESTS : ROWS
// ^                    ^

The question is: Why?

This expression:

(mode == TEST)

could be interpreted as either being the right operand to the < operator, or as the left operand to the ? operator. (But not both at the same time, I guess.)

Which rules apply? Is it a matter of operator precedence? Do sequence points play a role? What is the order of evaluation, and why?

I'm quite confused; any help is greatly appreciated.

Answer 1

The ternary conditional operator has a low precedence.

So

row < (mode == TEST) ? NO_OF_TESTS : ROWS

is grouped as

(row < (mode == TEST)) ? NO_OF_TESTS : ROWS

Folk like to think in terms of operator precedence tables , but really the groupings are hardwired into the language grammar.

Answer 2

The question is: Why?

Is it a matter of operator precedence ?

yes !

The precedence of the ternary operator is quite dangerous , you are not the first doing that error, I did myself

The best way to be quiet in expressions is to add the ()

Answer 3

Yes, it is a matter of precedence - relational operators (as well as bitwise, logical, arithmetic, unary, and postfix operators) have higher precedence than the ternary operator, so the expression a < b ? c : d a < b ? c : d is parsed as (a < b) ? c : d (a < b) ? c : d .

Same if the leftmost expression is a && b , a == b , a * b , etc.