Question

SELECT
SUM(IF(fruit='Apel', 1,0)) AS apelfruit,
SUM(IF(fruit='Orange', 1,0)) AS orangfruit,
SUM(IF(fruit='Manggo', 1,0)) AS manggofruit
FROM fruit_table WHERE fruit IN ('Apel', 'Orange', 'Manggo')
;

如何在不使用第三个变量的情况下交换它们？输出应该像

*a=10
*b=20

Answer 1

不确定面试官是否在寻找其他东西的异或，但似乎您可以简单地使用+，-和x。如果a较大或为负也应该起作用。

*a+=*b
*b-=*a
*b=*b x -1
*a-=*b

在您的示例中，这将给我们：

*a+=*b --> *a = 30
*b-=*a --> *b = -10
*b=*b x -1 --> *b = 10
*a-=*b --> *a = 20

Answer 2

这是执行此操作的简单代码：

#include <stdio.h>
#include <stdlib.h>

void usingXOR(int** x, int** y){

unsigned long long a = (unsigned long long)*x;
unsigned long long b = (unsigned long long)*y;

a = a^b;
b = a^b;
a = a^b;

*x = (int*)a;
*y = (int*)b;
}

void main(){

int x=5;
int y=10;

int* a = &x;
int* b = &y;

//If you only want to swap the values the pointers are pointing to
//Here the addresses the pointers are holding dont get swapped
(*a) = (*a)+(*b);
(*b) = (*a)-(*b);
(*a) = (*a)-(*b);

//If you want to swap addresses in the pointers
//printf("Before swap address a: %p\n", a);
//printf("Before swap address b: %p\n", b);

//usingXOR(&a,&b);

printf("a: %d\n", *a);
printf("b: %d\n", *b);

//printf("After swap address a: %p\n", a);
//printf("After swap address b: %p\n", b);
}

如何在不使用第三个变量的情况下交换两个指针值

2 个答案: