floating-point-conversion

According to http://en.cppreference.com/w/cpp/numeric/math/pow , when std::pow is used with integer parameters, the result is promoted to a double . My question is then the following: How safe is to compare an integer type with the result of a std::pow(int1, int2) ? For example, can the if below evaluate to true? std::size_t n = 1024; if(n != std::pow(2, 10)) cout << "Roundoff issues..." << endl; That is, is it possible that the result on the rhs can be something like 1023.99...9 so when converted to size_t becomes 1023? My guess is that the response in a big NO, but would like to know for

XCode 6.3.1 Swift 1.2 let value: Int = 220904525 let intmax = Int.max let float = Float(value) // Here is an error probably let intFromFloat = Int(float) let double = Double(value) println("intmax=\(intmax) value=\(value) float=\(float) intFromFloat=\(intFromFloat) double=\(double)") // intmax=9223372036854775807 value=220904525 float=2.20905e+08 intFromFloat=220904528 double=220904525.0 The initial value is 220904525. But when I convert it to float it becomes 220904528. Why? This is due to the way the floating-point format works. A Float is a 32-bit floating-point number, stored in the IEEE

Is there a way to round floating points to 2 points? E.g.: 3576.7675745342556 becomes 3576.76 . round(x * 100) / 100.0 If you must keep things floats: roundf(x * 100) / 100.0 Flexible version using standard library functions: double GetFloatPrecision(double value, double precision) { return (floor((value * pow(10, precision) + 0.5)) / pow(10, precision)); } If you are printing it out, instead use whatever print formatting function available to you. In c++ cout << setprecision(2) << f; For rounding to render to GUI, use std::ostringstream Multiply by 100, round to integer (anyway you want),

Possible Duplicate: Floating point comparison I have a problem about the accuracy of float in C/C++. When I execute the program below: #include <stdio.h> int main (void) { float a = 101.1; double b = 101.1; printf ("a: %f\n", a); printf ("b: %lf\n", b); return 0; } Result: a: 101.099998 b: 101.100000 I believe float should have 32-bit so should be enough to store 101.1 Why? paxdiablo You can only represent numbers exactly in IEEE754 (at least for the single and double precision binary formats) if they can be constructed from adding together inverted powers of two (i.e., 2 -n like 1 , 1/2 , 1/4

#include <stdio.h> int main(void){ float a = 1.1; double b = 1.1; if(a == b){ printf("if block"); } else{ printf("else block"); } return 0; } Prints: else block #include <stdio.h> int main(void){ float a = 1.5; double b = 1.5; if(a == b){ printf("if block"); } else{ printf("else block"); } return 0; } Prints: if block What is the logic behind this? Compiler used: gcc-4.3.4 Mysticial This is because 1.1 is not exactly representable in binary floating-point. But 1.5 is. As a result, the float and double representations will hold slightly different values of 1.1 . Here is exactly the difference