Why the following produce between 0 - 9 and not 10?
为什么是0 - 9而不是10?
My understanding is Math.random() create number between 0 to under 1.0.
我的理解是Math.random()在0到1.0之间创建数字。
So it can produce 0.99987 which becomes 10 by *10, isn't it?
它可以产生0。99987，变成10×10，对吧?
int targetNumber = (int) (Math.random()* 10);
8 个解决方案
#1
17
Casting a double to an int in Java does integer truncation. This means that if your random number is 0.99987, then multiplying by 10 gives 9.9987, and integer truncation gives 9.
在Java中，将double转换为整数，可以进行整数截断。这意味着如果你的随机数是0。99987，那么乘以10得到9。9987，整数截断得到9。
#2
14
From the Math javadoc :
来自数学javadoc:
"a pseudorandom double greater than or equal to 0.0 and less than 1.0"
“伪随机数倍大于或等于0.0，小于1.0”
1.0 is not a posible value with Math.random. So you can't obtain 10. And (int) 9.999 gives 9
1。0和Math.random是不一致的。所以不能得到10。9。999等于9。
#3
2
Cause (Math.random()* 10 gets rounded down using int(), so int(9.9999999) yields 9.
原因(Math.random()* 10使用int()得到整数，所以int(9.9999999)得到9。
#4
1
Because (int) rounds down.(int)9.999 results in 9. //integer truncation
因为(int)轮。(int)9.999 9。/ /整数截断
#5
0
0.99987 which becomes 10 by *10
0。99987变成10 *10。
Not when I went to school. It becomes 9.9987.
当我去学校的时候。它变成了9.9987。
#6
0
Math.floor(Math.random() * 10) + 1
Math.floor(Math.random() * 10) + 1。
Now you get a integer number between 1 and 10, including the number 10.
现在你得到一个整数在1到10之间，包括数字10。
#7
0
You could always tack on a +1 to the end of the command, allowing it to add 1 to the generated number. So, you would have something like this:
您可以在命令的末尾添加一个+1，允许它将1添加到生成的数字中。所以，你会得到这样的东西:
int randomnumber = ( (int)(Math.random( )*10) +1);
This would generate any integer between 1 and 10.
这将生成1到10之间的任何整数。
If you wanted any integer between 0 and 10, you could do this:
如果你想要0到10之间的任何整数，你可以这样做:
int randomnumber = ( (int)(Math.random( )*11) -1);
Hope this helps!
希望这可以帮助!
#8
0
you can add 1 to the equation so the output will be from 1 to 10 instead of 0 to 9 like this :
你可以把1加到等式中，结果是1到10，而不是0到9
int targetNumber = (int) (Math.random()* 10+1);
#1
17
Casting a double to an int in Java does integer truncation. This means that if your random number is 0.99987, then multiplying by 10 gives 9.9987, and integer truncation gives 9.
在Java中，将double转换为整数，可以进行整数截断。这意味着如果你的随机数是0。99987，那么乘以10得到9。9987，整数截断得到9。
#2
14
From the Math javadoc :
来自数学javadoc:
"a pseudorandom double greater than or equal to 0.0 and less than 1.0"
“伪随机数倍大于或等于0.0，小于1.0”
1.0 is not a posible value with Math.random. So you can't obtain 10. And (int) 9.999 gives 9
1。0和Math.random是不一致的。所以不能得到10。9。999等于9。
#3
2
Cause (Math.random()* 10 gets rounded down using int(), so int(9.9999999) yields 9.
原因(Math.random()* 10使用int()得到整数，所以int(9.9999999)得到9。
#4
1
Because (int) rounds down.(int)9.999 results in 9. //integer truncation
因为(int)轮。(int)9.999 9。/ /整数截断
#5
0
0.99987 which becomes 10 by *10
0。99987变成10 *10。
Not when I went to school. It becomes 9.9987.
当我去学校的时候。它变成了9.9987。
#6
0
Math.floor(Math.random() * 10) + 1
Math.floor(Math.random() * 10) + 1。
Now you get a integer number between 1 and 10, including the number 10.
现在你得到一个整数在1到10之间，包括数字10。
#7
0
You could always tack on a +1 to the end of the command, allowing it to add 1 to the generated number. So, you would have something like this:
您可以在命令的末尾添加一个+1，允许它将1添加到生成的数字中。所以，你会得到这样的东西:
int randomnumber = ( (int)(Math.random( )*10) +1);
This would generate any integer between 1 and 10.
这将生成1到10之间的任何整数。
If you wanted any integer between 0 and 10, you could do this:
如果你想要0到10之间的任何整数，你可以这样做:
int randomnumber = ( (int)(Math.random( )*11) -1);
Hope this helps!
希望这可以帮助!
#8
0
you can add 1 to the equation so the output will be from 1 to 10 instead of 0 to 9 like this :
你可以把1加到等式中，结果是1到10，而不是0到9
int targetNumber = (int) (Math.random()* 10+1);