integer

问题 Well I can find many big integer libraries but they are for all programming languages except VB.NET. Does anyone know a class/library for Visual Basic .NET which can handle very big numbers? (I'd like to handle 1024 or even more bits). The only calculation support I need it addition, substitution, multiplication, division and rounding. It's ment to be used with this code: Dim Letterz() As Integer = {33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56

问题 Well I can find many big integer libraries but they are for all programming languages except VB.NET. Does anyone know a class/library for Visual Basic .NET which can handle very big numbers? (I'd like to handle 1024 or even more bits). The only calculation support I need it addition, substitution, multiplication, division and rounding. It's ment to be used with this code: Dim Letterz() As Integer = {33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56

问题 I am working with PowerShell to create a renaming script for a number of files in a directory. Two questions here: I have a string variable $strPrefix = "ACV-100-" and an integer counter $intInc = 000001 and I wish to increment the counter $intInc 1 -> 2 and then concatenate the two and store it in a variable $strCPrefix in the following format: ACV-100-000002 . I believe the $intInc will need to be cast in order to convert it once incrementing is complete but I am unsure how to do this.

问题 I am working with PowerShell to create a renaming script for a number of files in a directory. Two questions here: I have a string variable $strPrefix = "ACV-100-" and an integer counter $intInc = 000001 and I wish to increment the counter $intInc 1 -> 2 and then concatenate the two and store it in a variable $strCPrefix in the following format: ACV-100-000002 . I believe the $intInc will need to be cast in order to convert it once incrementing is complete but I am unsure how to do this.

问题 I am writing a program in C. I have two variables one of which is integer and the other one is float. I want to divide the float by the integer and want to get result in float. I am doing the following: int a; float b=0,c=0; scanf("%d",&a); Here I am doing some computations on b so its value is increased randomly i.e 33 etc. c = b/(float)a; printf("c = %2.f\n", c); The problem is I am getting ' c ' as a rounded number (integer) rather than a float value. How can I get ' c ' as a float value.

问题 I am writing a program in C. I have two variables one of which is integer and the other one is float. I want to divide the float by the integer and want to get result in float. I am doing the following: int a; float b=0,c=0; scanf("%d",&a); Here I am doing some computations on b so its value is increased randomly i.e 33 etc. c = b/(float)a; printf("c = %2.f\n", c); The problem is I am getting ' c ' as a rounded number (integer) rather than a float value. How can I get ' c ' as a float value.

问题 What is the equivalent expression in SSRS of the following conversion of a date (@Date) in T-SQL? CONVERT(INT,CONVERT(CHAR,@Date,112)) I need the date parameter value to be converted to an integer in YYYYMMDD format. 回答1: Assuming you have a date parameter called YourDate . You could use the following expression: =Cint(Format(Parameters!YourDate.Value, "yyyyMMdd")) Explanation: Step 1: Format the date to the yyyyMMdd format: Format(Parameters!YourDate.Value, "yyyyMMdd") Step 2: Cast the

问题 What is the equivalent expression in SSRS of the following conversion of a date (@Date) in T-SQL? CONVERT(INT,CONVERT(CHAR,@Date,112)) I need the date parameter value to be converted to an integer in YYYYMMDD format. 回答1: Assuming you have a date parameter called YourDate . You could use the following expression: =Cint(Format(Parameters!YourDate.Value, "yyyyMMdd")) Explanation: Step 1: Format the date to the yyyyMMdd format: Format(Parameters!YourDate.Value, "yyyyMMdd") Step 2: Cast the

问题 I am unable to understand the following: In java, long l = 130L; byte b = (byte)l; If I print the value of b, why do I get -126? What is the bit representation of long l? 回答1: Bytes are signed in Java - so the range of values is -128 to 127 inclusive. The bit pattern for 130 as a long, when simply truncated to 8 bits, is the bit pattern for -126 as a byte. As another example: int x = 255; byte b = (byte) x; // b is now -1 回答2: A byte is a sequence of 8 bits, which makes 2^8 cases = 256. Half

问题 I am unable to understand the following: In java, long l = 130L; byte b = (byte)l; If I print the value of b, why do I get -126? What is the bit representation of long l? 回答1: Bytes are signed in Java - so the range of values is -128 to 127 inclusive. The bit pattern for 130 as a long, when simply truncated to 8 bits, is the bit pattern for -126 as a byte. As another example: int x = 255; byte b = (byte) x; // b is now -1 回答2: A byte is a sequence of 8 bits, which makes 2^8 cases = 256. Half