?：将一个表达式留空时的三元条件运算符行为
我正在编写一个控制台应用程序，它会尝试通过反复试验来“猜测”一个数字，它工作得很好，但它让我对我心不在焉地写的某个部分感到疑惑，
代码是：
#include <stdio.h>
#include <stdlib.h>
int main()
{
int x,i,a,cc;
for(;;){
scanf("%d",&x);
a=50;
i=100/a;
for(cc=0;;cc++)
{
if(x<a)
{
printf("%d was too big\n",a);
a=a-((100/(i<<=1))?:1);
}
else if (x>a)
{
printf("%d was too small\n",a);
a=a+((100/(i<<=1))?:1);
}
else
{
printf("%d was the right number\n-----------------%d---------------------\n",a,cc);
break;
}
}
}
return 0;
}
更具体地说，是让我困惑的部分 我是否
a=a+((100/(i<<=1))?:1);
//Code, code
a=a-((100/(i<<=1))?:1);
使用 ((100/(i<<=1))?:1) 来确保 100/(i<<=1) 返回0（或 false）整个表达式将计算为 1 ((100/(i<<=1))?:***1***)，并且我留下了如果它是 true 空 ((100/(i<<=1))? _this space_ :1) ，它似乎可以正常工作，但是离开该部分是否有任何风险条件为空？
I was writing a console application that would try to "guess" a number by trial and error, it worked fine and all but it left me wondering about a certain part that I wrote absentmindedly,
The code is:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int x,i,a,cc;
for(;;){
scanf("%d",&x);
a=50;
i=100/a;
for(cc=0;;cc++)
{
if(x<a)
{
printf("%d was too big\n",a);
a=a-((100/(i<<=1))?:1);
}
else if (x>a)
{
printf("%d was too small\n",a);
a=a+((100/(i<<=1))?:1);
}
else
{
printf("%d was the right number\n-----------------%d---------------------\n",a,cc);
break;
}
}
}
return 0;
}
More specifically the part that confused me is
a=a+((100/(i<<=1))?:1);
//Code, code
a=a-((100/(i<<=1))?:1);
I used ((100/(i<<=1))?:1) to make sure that if 100/(i<<=1) returned 0 (or false) the whole expression would evaluate to 1 ((100/(i<<=1))?:***1***), and I left the part of the conditional that would work if it was true empty ((100/(i<<=1))? _this space_ :1), it seems to work correctly but is there any risk in leaving that part of the conditional empty?
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这是一个 GNU C 扩展（参见 ?: wikipedia 条目），因此为了可移植性您应该明确声明第二个操作数。
在“true”的情况下，它返回条件的结果。
以下语句几乎是等效的：
唯一的区别是在第一个语句中，
x始终计算一次，而在第二个语句中，如果为 true，x将计算两次。所以唯一的区别是评估 x 时会产生副作用。不管怎样，我认为这是语法的微妙使用......如果您对维护代码的人有任何同情心，您应该明确说明操作数。 :)
另一方面，对于常见用例来说，这是一个很好的小技巧。
This is a GNU C extension (see ?: wikipedia entry), so for portability you should explicitly state the second operand.
In the 'true' case, it is returning the result of the conditional.
The following statements are almost equivalent:
The only difference is in the first statement,
xis always evaluated once, whereas in the second,xwill be evaluated twice if it is true. So the only difference is when evaluatingxhas side effects.Either way, I'd consider this a subtle use of the syntax... and if you have any empathy for those maintaining your code, you should explicitly state the operand. :)
On the other hand, it's a nice little trick for a common use case.
这是 C 语言的 GCC 扩展 。当
?:之间没有出现任何内容时，则在真实情况下使用比较的值。This is a GCC extension to the C language. When nothing appears between
?:, then the value of the comparison is used in the true case.