和＆或运算符在后增量运算符上的工作差异

Question

This post-increment operator usage is confusing.

{int a=0,b=1,c=2,d;
d=a++||b++||c++
printf("%d %d %d %d",a,b,c,d);}

output is

1,2,2,1

value of c did not increase but if I replace it with && operator it increases. Why?

Answer 1

Quoting C11 , chapter §6.5.14, ( emphasis mine )

Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; if the second operand is evaluated, there is a sequence point between the evaluations of the first and second operands. If the first operand compares unequal to 0 , the second operand is not evaluated.

So, in your case,

 d=a++||b++||c++

is the same as

 d= ( (a++ || b++) || c++)      

Then, the statement inside the first parenthesis is evaluated, first a++ (post-increment) evaluates to 0 (side-effect pending), so the RHS of the first || is evaluated, b++ , producing 1 and the result of the || operation is TRUE, yields 1.

That result, 1 , is the LHS of the second || . Hence, the RHS of the second || ( c++ ) is not evaluated anymore and the final result becomes TRUE, again yielding 1 , which gets stored in d .

So, finally,

• a++ is evaluated, becomes 1
• b++ is evaluated, becomes 2
• c++ is not evaluated, remains 2
• the result of || is stored in d , that is TRUE, so stores 1.

On the other hand, for logical AND && operator,

[...] If the first operand compares equal to 0, the second operand is not evaluated.

So, if you replace the last || with && , then for the outer statement, the LHS becomes 1 and the RHS evaluates, making c++ to be evaluated and incremented, as a side effect.