CODEWITHSUNDEEP
coperator-precedence
Ghasan غسان
Ghasan غسان

Reputation: 5857

Precedence of assignment within a conditional operator

I've created this simple program to auto-generate sequence of frames to be used in Avisynth scipt:

#include <stdio.h>

int main(void) {

    const int step = 3;
    const int arr[] = {31997, 31998, 32001};
    int i, ii = 0;

    for(i = 32002; i <= 32121; i += step, (sizeof(arr)/sizeof(int) - 1 ) != ii ? ++ii : ii = 0) {

        printf("freezeframe(%d,%d,%d)\n", i, i + step, arr[ii]);

    }

    return 0;
}

Using MinGW with GCC 4.6.2, I get this error: lvalue required as left operand of assignment.

The issue is simply solved by using parenthesis around ii=0. However, I don't get why it is an error. Shouldn't the assignment operator be evaluated first?

Upvotes: 0

Views: 180

Answers (3)

Anoop Vaidya
Anoop Vaidya

Reputation: 46533

It is always advised to use parenthesis, if you write your code in this fashion

(sizeof(arr)/sizeof(int) - 1 ) != ii ? ++ii : ii = 0

like

(((sizeof(arr)/sizeof(int) - 1 ) != ii ? ++ii : ii) = 0)

9 out of 10 times you will stuck.

So make a habit of putting parenthesis and this doesn't make any overhead on the compilers!!!

Upvotes: 0

Pubby
Pubby

Reputation: 53017

Wikepedia has a short section which explains this: http://en.wikipedia.org/wiki/Operators_in_C_and_C++#Notes

The grammar for conditional operator in C is

logical-OR-expression ? expression : conditional-expression

Note that an assignment-expression is not considered a conditional-expression, and a conditional expression cannot be the left of an assignment expression and so it's technically a syntax error. See the grammar: http://www.lysator.liu.se/c/ANSI-C-grammar-y.html

However, GCC is (incorrectly) parsing it as:

((sizeof(arr)/sizeof(int) - 1 ) != ii ? ++ii : ii) = 0

Which is a semantic error since ++i is not an lvalue expression.

Upvotes: 3

ouah
ouah

Reputation: 145829

The conditional operator has higher precedence than the assignment operator in C.

(sizeof(arr)/sizeof(int) - 1 ) != ii ? ++ii : ii = 0

is evaluated as

((sizeof(arr)/sizeof(int) - 1 ) != ii ? ++ii : ii) = 0

For a quick reminder of the operators precedence in C, you can see this:

http://www.kernel.org/doc/man-pages/online/pages/man7/operator.7.html

Upvotes: 4

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