条件运算,表达式最终的值的类型是?

小水平

表达式1?表达式2:表达式3
这是条件表达式的形式,其中1,2,3式的类型可以不相同,那么最后整个表达式的值的类型是三个表达式中最高型呢?还是不考虑最高型问题,直接取2式或3式的值.
请给个有依据的准确说法.能举出例子更好.

zalem

试了一下，按默认规则cast为最高型别。

          char b=1;
          int a=1;
          double c=1.0;
      
          printf("%d\n",sizeof(b?a:c));

其中的值换来换去看结果就有答案了


p.s.:为什么不自己实验呢？我也没研究过，写几行不就知道答案了？

雨丝风片

原帖由 "小水平" 发表：

请给个有依据的准确说法.

“有依据的准确说法”如下 ：
c99 “6.5.15 Conditional operator”

The first operand is evaluated; there is a sequence point after its evaluation. The second
operand is evaluated only if the first compares unequal to 0; the third operand is evaluated
only if the first compares equal to 0; the result is the value of the second or third operand
(whichever is evaluated), converted to the type described below.92) If an attempt is made
to modify the result of a conditional operator or to access it after the next sequence point,
the behavior is undefined.

If both the second and third operands have arithmetic type, the result type that would be
determined by the usual arithmetic conversions, were they applied to those two operands,
is the type of the result. If both the operands have structure or union type, the result has
that type. If both operands have void type, the result has void type.

If both the second and third operands are pointers or one is a null pointer constant and the
other is a pointer, the result type is a pointer to a type qualified with all the type qualifiers
of the types pointed-to by both operands. Furthermore, if both operands are pointers to
compatible types or to differently qualified versions of compatible types, the result type is
a pointer to an appropriately qualified version of the composite type; if one operand is a
null pointer constant, the result has the type of the other operand; otherwise, one operand
is a pointer to void or a qualified version of void, in which case the result type is a pointer to
an appropriately qualified version of void.

小水平

原帖由 "zalem" 发表：
试了一下，按默认规则cast为最高型别。

          char b=1;
          int a=1;
          double c=1.0;
      
          printf("%d\n",sizeof(b?a:c));

其中的值换来换去看结果就有答案了


p.s.:..........

真是太棒了,聪明,我不是懒,而是没想到这么好的用法.简单,明了,还说明问题.太感谢了,佩服.