关于 C 中运算符优先级的问题

Question

I'm starting to learn to program in c, and I thought I was already pretty confident with the precedence of operators, until I did this:

a > b ? c = a : (c = b);

Of course at the first time I didn't use parenthesis on the last sentence, but since that ended up causing compiling issues I searched how to solve that problem on this forum and I read that adding parenthesis could do the job. However, I thought that the expressions inside parenthesis get executed before anything else written in the same line, which would mean that the c = b sentence was executed first and then the ternary operator. I did something similar but easier to read in order to get a better idea of what was happening with this operator precedence thing and tried executing this line:

printf("Number is %d", i) + (i = 5);

I know this expression returns returns a value, but since I don't need it and this isn't a line that I will keep for more than 5 seconds, I won't store it in any variable. What gets my attention in this case is that, when I execute the code, I doesn't show up on the screen with the value 5, but instead it uses the previous value, which means that the computer is just reading it from left to right. When I do:

(i = 5) + printf(Numer is %d, i);

it first does the assignment of i and only after that the printf function is executed. My question is: how does the computer execute an expression that uses operators of different orders of precedence? It clearly doesn't run first the operator with the highest precedence, because in the first printf the value stored wasn't the one assigned on the parenthesis, but it also doesn't just read from left to right because in that case there would be no operator precedence. How does it work?

Answer 1

Parenthesis and operator precedence only dictate how operands are grouped. It does not dictate the order of evaluation .

In this expression:

a > b ? c = a : (c = b);

The three parts of the ternary operator are a > b , c = a , and c = b respectively. This operator also has the property that only one of the second and third clause are evaluated, based on the result of the first. Formally speaking, there is a sequence point between the evaluation of the first clause and of either the second or third. So a > b is first evaluated. If it is nonzero, c = a is evaluated, otherwise c = b is evaluated.

In this expression:

printf("Number is %d", i) + (i = 5); 

There is nothing that dictates whether printf("Number is %d", i) or i = 5 is evaluated first. Unlike the ternary operator, there is no sequence point between the evaluation of the operands of the + operator. This expression also has a problem: i is both read and written in the same expression without a sequence point. Doing so triggers undefined behavior . This is also true for:

(i = 5) + printf(Numer is %d, i);

On a side note, this:

a > b ? c = a : (c = b);

Can be more clearly written as:

c = a > b ? a : b;