[英]Explanation needed for this pre-processor directive C/C++
I tried to solve this problem in some test, but later when i ran it at home, it gave unexpected answer. 我尝试通过一些测试解决此问题,但后来当我在家中运行它时,它给出了意外的答案。 I am not able to understand this code: 我无法理解此代码:
#include <stdio.h>
#include <conio.h>
#define swap(a,b) temp=a; a=b; b=temp;
int main()
{
int i, j, temp;
i=5;
j=10;
temp=0;
if( i > j) //evaluates to false
swap( i, j );
printf( "%d %d %d", i, j, temp); //expected output: 5 10 0
getch();
return 0;
}
Output i am getting is: 10 0 0 我得到的输出是:10 0 0
Please someone explain how is it working. 请有人解释它如何工作。
Code below 下面的代码
if( i > j) //evaluates to false
swap( i, j );
Becomes 成为
if( i > j) //evaluates to false
temp=i; i=j; j=temp;
which is equivalent to 相当于
if( i > j) //evaluates to false
{temp=i;} i=j; j=temp;
If condition is false, there would be unexpected results as below 如果条件为假,则将出现以下意外结果
i=5;
j=10;
temp=0;
i=j; /* i becomes 10 */
j=temp; /* j becomes 0 */
Learnings 学问
{}
尝试在{}
内放置块(如果是,则为做某事,同时做) #define swap(a,b) do { temp=a; a=b; b=temp; } while(0)
Note that there is no terminating semicolon after while(0) 请注意, while(0)之后没有终止分号。
Expanding the macro, you get: 扩展宏,您将得到:
if (i > j)
temp = i;
i = j;
j = temp;
This is why seasoned c-programmers wrap macro bodies in do{...}while(0)
. 这就是为什么经验丰富的c程序员在do{...}while(0)
包装宏体的原因。
#define swap(a, b) do{temp=a; a=b; b=temp;}while(0)
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