![\"Polynomial](\"https:\/\/www.modernescpp.com\/wp-content\/uploads\/2018\/01\/Polynomial_expansion.png\")
<\/p>\nToday’s post is about expressions. You should avoid complicated expressions, know the precedence rules for arithmetic or logical expressions, and know the order of evaluation of expressions. The main reasons for undefined behavior are having the wrong precedence rules for expressions in mind or assuming an evaluation order for expressions that is just wrong or not guaranteed. I know that’s a lot to digest. Let’s start.<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
Here are the four rules for today.<\/p>\n
The rules for the precedence and the evaluation are not as easy as it sounds. They even change with C++17; therefore, we should start simple.<\/p>\n
What does complicated mean? Here is the original example of the guidelines:<\/p>\n
<\/p>\n
\/\/ bad: assignment hidden in subexpression (1)<\/span>\r\nwhile<\/span> ((c =<\/span> getc()) !=<\/span> -<\/span>1<\/span>)\r\n\r\n\/\/ bad: two non-local variables assigned in a sub-expressions (1)<\/span>\r\nwhile<\/span> ((cin >><\/span> c1, cin >><\/span> c2), c1 ==<\/span> c2)\r\n\r\n\/\/ better, but possibly still too complicated (1)<\/span>\r\nfor<\/span> (char<\/span> c1, c2; cin >><\/span> c1 >><\/span> c2 &&<\/span> c1 ==<\/span> c2;)\r\n\r\n\/\/ OK: if i and j are not aliased (2)<\/span>\r\nint<\/span> x =<\/span> ++<\/span>i +<\/span> ++<\/span>j; \r\n\r\n\/\/ OK: if i != j and i != k (2)<\/span>\r\nv[i] =<\/span> v[j] +<\/span> v[k];\r\n\r\n\/\/ bad: multiple assignments \"hidden\" in subexpressions (1)<\/span>\r\nx =<\/span> a +<\/span> (b =<\/span> f()) +<\/span> (c =<\/span> g()) *<\/span> 7<\/span>;\r\n\r\n\/\/ bad: relies on commonly misunderstood precedence rules (1)<\/span>\r\nx =<\/span> a &<\/span> b +<\/span> c *<\/span> d &&<\/span> e ^<\/span> f ==<\/span> 7<\/span>;\r\n\r\n\/\/ bad: undefined behavior (3)<\/span>\r\nx =<\/span> x++<\/span> +<\/span> x++<\/span> +<\/span> ++<\/span>x;\r\n<\/pre>\n<\/div>\n <\/p>\n
I added a few (numbers) to it. First, all expressions having a (1) are bad style and should not pass a code review. For example, do you know what is happening here: x = a & b + c * d && e ^ f == 7;. <\/span>Of course, you have to look up the precedences of the operators. I will come to it in the following rule. The expression (2) may be acceptable if the conditions hold. i and j must be disjunct, and the indices i,j, and i,k must be pairwise disjunct.<\/p>\n(3) is undefined behavior, because it is not defined which x will be evaluated first. Why? The argument is that the final semicolon “;” is a sequence point, and now we have the guarantee that all side effects from the previous evaluations in the sequence are complete.<\/p>\n
With C++17, the rules for operator precedence changed: left-to-right for expressions except for right-to-left in assignments. I will write about it in ES.43.<\/p>\n<\/p>\n
ES.41: If in doubt about operator precedence, parenthesize<\/a><\/h3>\nOn the one hand, the guidelines say: If you doubt operator precedence, use parenthesis (1). On the other hand, they state: You should know enough not to need parentheses here (2):<\/p>\n
<\/p>\n
\nconst<\/span> unsigned<\/span> int<\/span> flag =<\/span> 2<\/span>;\r\nunsigned<\/span> int<\/span> a =<\/span> flag;\r\n\r\nif<\/span> (a &<\/span> flag !=<\/span> 0<\/span>) \/\/ bad: means a&(flag != 0) (1)<\/span>\r\n\r\nif<\/span> (a <<\/span> 0<\/span> ||<\/span> a <=<\/span> max) { \/\/ good: quite obvious (2)<\/span>\r\n \/\/ ...<\/span>\r\n}\r\n<\/pre>\n<\/div>\n <\/p>\n
Okay. For an expert, expression (1) may be obvious, but for a beginner, expression (2) may be a challenge. <\/p>\n
I have only two tips in mind according to the guidelines:<\/p>\n
\n- If in doubt about precedence, use parentheses. Do not forget the beginners!<\/li>\n
- Keep this precedence table from cppreference.com<\/a> under your pillow.<\/li>\n<\/ol>\n
![\"OperatorPrecedence\"](\"https:\/\/www.modernescpp.com\/wp-content\/uploads\/2018\/01\/OperatorPrecedence.png\")
<\/p>\n
I will jump right to the rules ES.43 and ES.44 and write about the rule ES.42 in my next post. With C++17, the order of evaluation of expressions changed.<\/p>\n
ES.43: Avoid expressions with undefined order of evaluation<\/a><\/h3>\nIn C++14, the following expression has undefined behavior. <\/p>\n
\nv[i] =<\/span> ++<\/span>i; \/\/ the result is undefined<\/span>\r\n<\/pre>\n<\/div>\n <\/p>\n
This will not hold for C++17. With C++17, the order of evaluation of the last code snippet is right to left; therefore, the expression has well-defined behavior.<\/p>\n
Here are the additional guarantees we have with C++17:<\/p>\n
\n- Postfix expressions are evaluated from left to right. This includes function calls and member selection expressions.<\/li>\n
- Assignment expressions are evaluated from right to left. This includes compound assignments.<\/li>\n
- Operands to shift operators are evaluated from left to right. <\/li>\n<\/ol>\n
This was the wording of the original proposal<\/a>. They also provided a few examples. Here are they:<\/p>\n\na.b\r\na-><\/span>b\r\na->*<\/span>b\r\na(b1, b2, b3) \/\/ (1)<\/span>\r\nb @<\/span>=<\/span> a\r\na[b]\r\na <<<\/span> b\r\na >><\/span> b\r\n<\/pre>\n<\/div>\n <\/p>\n
How should you read these examples? Quite simple. First, a<\/span> will be evaluated, then b<\/span>, c, and d<\/span>.<\/p>\nExpression (1) is a bit tricky. With C++17, we can only guarantee that the function is evaluated before its arguments, but the order of the evaluation of the arguments is still unspecified.<\/p>\n
I know the last sentence was not easy. Let’s elaborate a little bit more.<\/p>\n
ES.44: Don\u2019t depend on order of evaluation of function arguments<\/a><\/h3>\nIn the last years, I saw many errors because developers assumed that the order of the evaluation of function arguments is left to right. Wrong! You have no guarantees!<\/p>\n
\n#include <iostream><\/span>\r\n\r\nvoid<\/span> func<\/span>(int<\/span> fir, int<\/span> sec){\r\n std::<\/span>cout <<<\/span> \"(\"<\/span> <<<\/span> fir <<<\/span> \",\"<\/span> <<<\/span> sec <<<\/span> \")\"<\/span> <<<\/span> std::<\/span>endl;\r\n}\r\n\r\nint<\/span> main<\/span>(){\r\n int<\/span> i =<\/span> 0<\/span>;\r\n func(i++<\/span>, i++<\/span>);\r\n}\r\n<\/pre>\n<\/div>\n <\/p>\n
Here is my proof. The output from Gcc and clang differs:<\/p>\n
![\"gcc\"](\"https:\/\/www.modernescpp.com\/wp-content\/uploads\/2016\/08\/gcc.png\")
<\/p>\n![\"clang\"](\"https:\/\/www.modernescpp.com\/wp-content\/uploads\/2018\/01\/clang.png\")
<\/p>\n