long-integer

问题 What is the Python equivalent of these C types: float double long I want to represent a float in the above three forms. 回答1: I believe you mean types , not functions. In C, a double is just a double-precision float (meaning smaller numbers are represented more accurately). In Python, it's just a float . Similarly, a long is (was) just an int that stored larger numbers. In Python 2, there is int , long , and float . However, Python automatically promotes int to long when it grows above sys

问题 I have the following JPQL query: SELECT il FROM InsiderList il WHERE ( il.deleteFlag IS NULL OR il.deleteFlag = '0' ) AND il.clientId = :clientId AND ( LOWER( il.name ) LIKE :searchTerm OR il.nbr LIKE :searchTerm OR LOWER( il.type ) LIKE :searchTerm OR LOWER( il.description ) LIKE :searchTerm ) The customer wants us to be able to search be the nbr field, which is a java.lang.Long . Q : How do you perform a LIKE search on a java.lang.Long using JPQL? 回答1: You can use the CAST in HQL: SELECT il

问题 I have the following JPQL query: SELECT il FROM InsiderList il WHERE ( il.deleteFlag IS NULL OR il.deleteFlag = '0' ) AND il.clientId = :clientId AND ( LOWER( il.name ) LIKE :searchTerm OR il.nbr LIKE :searchTerm OR LOWER( il.type ) LIKE :searchTerm OR LOWER( il.description ) LIKE :searchTerm ) The customer wants us to be able to search be the nbr field, which is a java.lang.Long . Q : How do you perform a LIKE search on a java.lang.Long using JPQL? 回答1: You can use the CAST in HQL: SELECT il

问题 This is my code with int j : void solve(){ unsigned long long n; cin>>n; unsigned long long sum = 0; int j = 1; for(int i=3;i<n+1;i+=2){ sum += ((4*i)-4)*(j); j++; } cout<<sum<<"\n"; } Input: 499993 Output: 6229295798864 but it is giving wrong output, and here is my code with long long j which is working fine: void solve(){ int n; cin>>n; unsigned long long sum = 0; long long j = 1; for(int i=3;i<n+1;i+=2){ sum += ((4*i)-4)*(j); j++; } cout<<sum<<"\n"; } Input: 499993 Output:

问题 This is my code with int j : void solve(){ unsigned long long n; cin>>n; unsigned long long sum = 0; int j = 1; for(int i=3;i<n+1;i+=2){ sum += ((4*i)-4)*(j); j++; } cout<<sum<<"\n"; } Input: 499993 Output: 6229295798864 but it is giving wrong output, and here is my code with long long j which is working fine: void solve(){ int n; cin>>n; unsigned long long sum = 0; long long j = 1; for(int i=3;i<n+1;i+=2){ sum += ((4*i)-4)*(j); j++; } cout<<sum<<"\n"; } Input: 499993 Output:

问题 This question already has answers here : C++ convert hex string to signed integer (8 answers) Closed 6 years ago . I need help in programming. The program has to accept a string that will eventually be turned to unsigned long. But here's the catch, there must be an error catcher that when you enter combination of hex and symbols like a!!!!!! will produce an error and the unsigned long variable must be able to accept and store the input greater that 4294967295 which is FFFFFFFF . I've tried

问题 I have a bunch of hex strings, one of them, for example is: d1bc4f7154ac9edb which is the hex value of "-3333702275990511909". This is the same hex you get if you do Long.toHexString("d1bc4f7154ac9edb"); For now, let's just assume I only have access to the hex string values and that is it. Doing this: Long.parseLong(hexstring, 16); Doesn't work because it converts it to a different value that is too large for a Long. Is there away to convert these unsigned hex values into signed longs? Thanks

问题 This question already has answers here : Why does Java implicitly (without cast) convert a `long` to a `float`? (4 answers) Closed 4 years ago . if you call the following method of Java void processIt(long a) { float b = a; /*do I have loss here*/ } do I have information loss when I assign the long variable to the float variable? The Java language Specification says that the float type is a supertype of long . 回答1: Yes, this is possible: if only for the reason that float has too few

问题 I know about weird stuff with precision errors, but I can't fathom, Why is (long)9223372036854665200d giving me 9223372036854665216 ? 回答1: 9223372036854665200d is a constant of type double . However, 9223372036854665200 does not fit in a double without loss of precision. A double only has 52 bits of mantissa, whereas the number in question requires 63 bits to be represented exactly. The nearest double to 9223372036854665200d is the number whose mantissa equals 1

问题 I know about weird stuff with precision errors, but I can't fathom, Why is (long)9223372036854665200d giving me 9223372036854665216 ? 回答1: 9223372036854665200d is a constant of type double . However, 9223372036854665200 does not fit in a double without loss of precision. A double only has 52 bits of mantissa, whereas the number in question requires 63 bits to be represented exactly. The nearest double to 9223372036854665200d is the number whose mantissa equals 1