int

问题 In Dart site they said that the integer size is 64 bit , is this the max size of an integer, or this is the size of any integer even if the integer is a small number such as 12? If every integer size is 64 bit , does this affect the performance of the application? 回答1: The answer is: "It depends". First of all, if you compile Dart to JavaScript, all numbers are JavaScript numbers, which means Dart double s. Even the integers, they just happen to be non-fractional doubles. That means that you

问题 While reading the answer to the following question Getting a buffer into a stringstream in hex representation: I did not understand why there is a necessity to cast uint8_t to unsigned (or, as written in comments, even to unsigned char before that), while casting just to int is incorrect. As I understand, no conversions at all would lead to interpreting uint8_t as an underlying type which can (must?) be some of 3 char variations and thus printing it as a character. But what's wrong with

问题 I am trying to compute time lapsed in java using nanoTime. But everytime it gives me different results. Why it is not consistent always ? Sample code : long startTime=System.nanoTime(); String.valueOf(number).length(); long endTime = System.nanoTime(); System.out.println(endTime-startTime); 回答1: nanoTime() and its sister currentTimeMillis() are not exact and depending on the architecture you run your code on they suffer from rounding (see the javadoc for details): This method provides

问题 This question already has answers here : Java String Concatenation with + operator (5 answers) Closed 1 year ago . Consider the below code snippet - public class Student { public static void main(String args[]) { int a = 3; int b = 4; System.out.println(a + b +" "); System.out.println(a + b +" "+ a+b); System.out.println(""+ a+b); } } The output for the above snippet is coming as - 7 7 34 34 It is clear from the output that if we use String at first in the print statement then the integers

问题 This question already has answers here : Division result is always zero [duplicate] (4 answers) Closed 5 years ago . I know you're usually not meant to put out all of your code but this is short and would help with the problem. Can anyone please explain why the output is 0 and how I can change the code to output what should be the volume of a cone. #include <stdio.h> float ConeVolume(int height, int radius); float ConeVolume(int height, int radius) { float pi; pi = 3.14159; float third; third

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 made a statement to a colleague of mine, which was: "chars are automatically promoted to integers in C expressions, and that's fine for performance since CPUs work fastest with their natural word size. I believe char promotion behavior is stated somewhere in the standard due to a char's rank. This is the response I got back: "Characters are not default promoted to an integer. The register size is 32 bit, but multiple byte values in a row can be packed into a single register as a compiler

问题 I made a statement to a colleague of mine, which was: "chars are automatically promoted to integers in C expressions, and that's fine for performance since CPUs work fastest with their natural word size. I believe char promotion behavior is stated somewhere in the standard due to a char's rank. This is the response I got back: "Characters are not default promoted to an integer. The register size is 32 bit, but multiple byte values in a row can be packed into a single register as a compiler

问题 Does calloc() of a double field always evaluate to 0.0 ? Furthermore : Does calloc() of a float field always evaluate to 0.0f ? Does calloc() of an int or unsigned int field always evaluate to 0 ? That is , will the assert() below always succeed on all platforms? double* d = calloc(1, sizeof(double)); assert(*d == 0.0); free(d); 回答1: The calloc sets all bytes of the allocated memory to zero. As it happens, that's also the valid IEEE754 (which is the most common format for floating point