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expr_precedence.c revision 1.10
      1  1.10  rillig /*	$NetBSD: expr_precedence.c,v 1.10 2022/08/25 19:03:48 rillig Exp $	*/
      2   1.1  rillig # 3 "expr_precedence.c"
      3   1.1  rillig 
      4   1.1  rillig /*
      5   1.1  rillig  * Tests for the precedence among operators.
      6   1.1  rillig  */
      7   1.1  rillig 
      8   1.1  rillig int var;
      9   1.1  rillig 
     10   1.1  rillig /*
     11   1.1  rillig  * An initializer needs an assignment-expression; the comma must be
     12   1.1  rillig  * interpreted as a separator, not an operator.
     13   1.1  rillig  */
     14   1.1  rillig /* expect+1: error: syntax error '4' [249] */
     15   1.1  rillig int init_error = 3, 4;
     16   1.1  rillig 
     17   1.1  rillig /* expect+1: error: non-constant initializer [177] */
     18   1.1  rillig int init_syntactically_ok = var = 1 ? 2 : 3;
     19   1.1  rillig 
     20   1.1  rillig /*
     21   1.1  rillig  * The arguments of __attribute__ must be constant-expression, as assignments
     22   1.1  rillig  * don't make sense at that point.
     23   1.1  rillig  */
     24   1.1  rillig void __attribute__((format(printf,
     25   1.3  rillig     /*
     26   1.3  rillig      * Inside of __attribute__((...)), symbol lookup works differently.  For
     27   1.3  rillig      * example, 'printf' is a keyword, and since all arguments to
     28   1.3  rillig      * __attribute__ are constant expressions, looking up global variables
     29   1.3  rillig      * would not make sense.  Therefore, 'var' is undefined.
     30   1.3  rillig      *
     31   1.8  rillig      * See lex.c, function 'search', keyword 'in_gcc_attribute'.
     32   1.3  rillig      */
     33   1.9  rillig     /* expect+1: error: syntax error '=' [249] */
     34   1.1  rillig     var = 1,
     35   1.1  rillig     /* Syntactically ok, must be a constant expression though. */
     36   1.1  rillig     var > 0 ? 2 : 1)))
     37   1.1  rillig my_printf(const char *, ...);
     38   1.4  rillig 
     39   1.4  rillig void
     40   1.4  rillig assignment_associativity(int arg)
     41   1.4  rillig {
     42   1.4  rillig 	int left, right;
     43   1.4  rillig 
     44   1.5  rillig 	/*
     45   1.5  rillig 	 * Assignments are right-associative.  If they were left-associative,
     46   1.5  rillig 	 * the result of (left = right) would be an rvalue, resulting in this
     47   1.5  rillig 	 * error message: 'left operand of '=' must be lvalue [114]'.
     48   1.5  rillig 	 */
     49   1.4  rillig 	left = right = arg;
     50   1.4  rillig 
     51   1.4  rillig 	left = arg;
     52   1.4  rillig }
     53   1.6  rillig 
     54   1.6  rillig void
     55   1.6  rillig conditional_associativity(_Bool cond1, _Bool cond2, int a, int b, int c)
     56   1.6  rillig {
     57   1.6  rillig 	/* The then-expression can be an arbitrary expression. */
     58   1.6  rillig 	var = cond1 ? cond2 ? a : b : c;
     59   1.6  rillig 	var = cond1 ? (cond2 ? a : b) : c;
     60   1.6  rillig 
     61   1.6  rillig 	/* The then-expression can even be a comma-expression. */
     62   1.6  rillig 	var = cond1 ? cond2 ? a, b : (b, a) : c;
     63   1.6  rillig 
     64   1.6  rillig 	var = cond1 ? a : cond2 ? b : c;
     65   1.6  rillig 	/*
     66   1.6  rillig 	 * In almost all programming languages, '?:' is right-associative,
     67   1.6  rillig 	 * which allows for easy chaining.
     68   1.6  rillig 	 */
     69   1.6  rillig 	var = cond1 ? a : (cond2 ? b : c);
     70   1.6  rillig 	/*
     71   1.6  rillig 	 * In PHP, '?:' is left-associative, which is rather surprising and
     72   1.6  rillig 	 * requires more parentheses to get the desired effect.
     73   1.6  rillig 	 */
     74   1.6  rillig 	var = (cond1 ? a : cond2) ? b : c;
     75   1.6  rillig }
     76