long-integer

问题 I was working with numbers of 200 digits in python. When finding the square root of a number using math.sqrt(n) I am getting a wrong answer. In[1]: n=9999999999999999999999999999999999999999999999999999999999999999999999 999999999999999999999999998292000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000726067 In[2]: x=int(math.sqrt(n)) In[3]: x Out[1]: 10000000000000000159028911097599180468360808563945281389781327

问题 Okay, I have to do simple task for my c++ class. Two functions, first is Fibonacci sequence, second some random sequence (finding e ). It looks like this: #include <stdio.h> #include <cstdlib> #include <string> #include <math.h> void fib(int number) { int a=0, b=1; printf("%d\n", a); for (; number>0; number--) { printf("%d\n", b); b+=a; a = b-a; } } void e_math(unsigned int number) { for (double n = 1; number>0; number--, n++) { printf("%f\n", pow((1+1/n), n)); } } int main(int argc, char**

问题 Okay, I have to do simple task for my c++ class. Two functions, first is Fibonacci sequence, second some random sequence (finding e ). It looks like this: #include <stdio.h> #include <cstdlib> #include <string> #include <math.h> void fib(int number) { int a=0, b=1; printf("%d\n", a); for (; number>0; number--) { printf("%d\n", b); b+=a; a = b-a; } } void e_math(unsigned int number) { for (double n = 1; number>0; number--, n++) { printf("%f\n", pow((1+1/n), n)); } } int main(int argc, char**

问题 I have a problem with bit shifting and unsigned longs. Here's my test code: char header[4]; header[0] = 0x80; header[1] = 0x00; header[2] = 0x00; header[3] = 0x00; unsigned long l1 = 0x80000000UL; unsigned long l2 = ((unsigned long) header[0] << 24) + ((unsigned long) header[1] << 16) + ((unsigned long) header[2] << 8) + (unsigned long) header[3]; cout << l1 << endl; cout << l2 << endl; I would expect l2 to also have a value of 2147483648 but instead it prints 18446744071562067968. I assume

问题 I have a problem with bit shifting and unsigned longs. Here's my test code: char header[4]; header[0] = 0x80; header[1] = 0x00; header[2] = 0x00; header[3] = 0x00; unsigned long l1 = 0x80000000UL; unsigned long l2 = ((unsigned long) header[0] << 24) + ((unsigned long) header[1] << 16) + ((unsigned long) header[2] << 8) + (unsigned long) header[3]; cout << l1 << endl; cout << l2 << endl; I would expect l2 to also have a value of 2147483648 but instead it prints 18446744071562067968. I assume

问题 On my 64 bit computer the long long type has 64 bits. print(sizeof(long long)) # prints 8 I need to use 128 bit integers and luckily GCC supports these. How can I use these within Cython? The following doesn't work. Compiling foo.pyx containing just cdef __int128_t x = 0 yields $ cython foo.pyx Error compiling Cython file: ------------------------------------------------------------ ... cdef __int128_t x = 0 ^ ------------------------------------------------------------ foo.pyx:2:5: '__int128

问题 In the C++/CLI project I have the method void DoSomething(long x); . If I want to use it in any unit-test written in C# , the method parameter x shows up as type int . Why do I have to change the signature to void DoSomething(long long x); to use it with parameters of type long in my unit-tests (C#)? 回答1: In C# long is a 64 bit data type. In C++ All we know about long is that it has to hold as much or more than an int and it is at least 32 bits. If you use a long long in c++ that is

问题 I have the following data set id<-c(1,1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4) s02<-c(001,002,003,004,001,002,003,004,005,001,002,003,004,005,006,007,001,002,003,004,005,006,007,008,009,010,011,012,013,014,015,016,017,018,019,020,021,022,023,024,025,026,027,028,029) dat1<-data.frame(id,s02) I would wish to create a data set based on this dat1. I would wish to have an R code that creates n s02 automatically as s02__0, s02__1, s02__2, s02__3, s02_

问题 This question already has answers here : Casting variables in Java (5 answers) Closed 4 years ago . What do (long) and (int) in the following statements before the expression mean? long years = (long) (minutes / minutesInYear); int remainingDays = (int) (minutes / 60 / 24) % 365; 回答1: (minutes / minutesInYear) is unclear as to what variable type it will return. If you want it in an int variable, you must write (int) before it. This is called "typecasting", and it basically converts one format

问题 This question already has answers here : Casting variables in Java (5 answers) Closed 4 years ago . What do (long) and (int) in the following statements before the expression mean? long years = (long) (minutes / minutesInYear); int remainingDays = (int) (minutes / 60 / 24) % 365; 回答1: (minutes / minutesInYear) is unclear as to what variable type it will return. If you want it in an int variable, you must write (int) before it. This is called "typecasting", and it basically converts one format