问题描述
int _tmain(int argc, _TCHAR* argv[])
{
int a=5;
printf("%d,%d,%d,%d",a++,++a,a--,--a);//4,5,4,5
printf("\na++-%d",a++);//5
printf("++a-%d",++a);//7
printf("a---%d",a--);//7
printf("--a-%d",--a);//5
return 0;
}
在同一行中单独打印时输出如何变化?输出在注释部分给出
How the output varies when in same line and printed seperately?? Output is given in commented part
Guidance will be helpful
推荐答案
x = 6;
a = x++;最终以包含6的"a"和包含7的"x"结尾.
"++ x"递增"x"的值，然后返回新值.
所以
ends up with "a" containing 6, and "x" containing 7.
"++x" increments the value of "x" and then returns the new value.
So
x = 6;
a = ++x;以包含7的"a"和包含7的"x"结尾.
尽管printf的每个参数都是单独评估的，但不必按照您编写它们的顺序从左到右进行评估，因此，第一行中的奇数值.
其他很明显:a输入为5，在使用该值后将第一个添加为1，将第二输入为6，并在使用该值并添加"7"之前将其添加.然后，它以7的形式输入第三位，并在使用该值后递减，因此以6的形式输入最后的1，在使用该值之前将其递减，以便将其打印为"5".
ends up with "a" containing 7, and "x" containing 7.
Although each parameter for printf is evaluated separately, they do not have to be evaluated in the left-to-right order you wrote them, hence the odd looking values in your first line.
The others are obvious: a enters as 5, is incremented the the first one after the value is used, enters the second as 6 and is incremented before the value is used and "7" printed. It then enters the third as 7 and is decremented after the value is used, so enters the final one as 6, which is decremented before it is used so it is printed as "5".
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