// How does this program work with ternary operator
#include <stdio.h>
int main()
{
int x = 2, y = 5;
(x & y) ? printf("True ") : printf("False "); // How do we get output
(x && y) ? printf("True ") : printf("False ");
return 0;
有人能解释一下这种行为吗?
function checkSignsWeird(a,b){
var output = "";
if(a^b < 0){
output = "The "+a+" and "+b+" have DIFFERENT signs.";
}else{
output = "The "+a+" and "+b+" have the SAME sign.";
}
console.log(ou
我正在尝试使用布尔掩码从两个不同的数据帧中获取匹配。使用
使用逻辑OR运算符:
x = df[(df['A'].isin(df2['B']))
or df['A'].isin(df2['C'])]
Output:
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
但是,如果使用按位OR运算符,则会成功返回结果。
x = df[(df['A'
我有这样的代码,我使用“按位和”(&)尝试并更改x
#include <stdio.h>
int main (void){
unsigned long x = 0;
printf("x = %lu\n", x);
x &= 1;
printf("x = %lu\n", x);
return 0;
}
我编译并执行代码以获得输出:
x = 0
x = 0
但我料到:
x = 0
x = 1
为什么?