Your code isn’t C, but this is: #include <stdio.h> int main () { float add = 0; int count[] = { 3, 2, 1, 3, 1, 2, 3, 3, 1, 2, 1 }; for (int i = 0; i < 11; i++) { add += 1 / ((float) count[i] + 1); } printf(“%f\n”, add); return … Read more
Floating point calculation is reasonably complex subject. This has to do with the binary representation of a floating point number, that doesn’t guarantee every possible number (obviously), to have an exact representation, which can lead to errors in operations, and yes, these errors can propagate. Here is a link on the subject, although it isn’t … Read more
The code for negatives is incorrect. You cannot just blindly negate the result when the base m is negative. , but . Also, you’re not printing anything if m is zero! And ints are signed by default so signed int is noise. floats are signed too; but here you could as well use a double … Read more
Q: How do I fix problem 1 … ? 10 ^ i is 10 exclusive-or’ed with i. It is not 10i. To find the integer portion of log10(num) int i_portion = 0; if (num <= 0) Handle_elsewhere(); else { int i_portion = 0; int n = num; while (n >= 10) { n /= 10; … Read more
The results of pow are often irrational, fractional or too large to store as a long, If you want to use powers of integers you can use BigInteger bi = BigInteger.valueOf(100); BigInteger bi2 = bi.pow(20); // 10^40 Another reason maybe that Math has many function which are taken from C, and also from common CPU … Read more
Suppose x is the number you wish to reduce the precision of and bits is the number of significant bits you wish to retain. When bits is sufficiently large and the order of magnitude of x is sufficiently close to 0, then x * (1L << (bits – Math.getExponent(x))) will scale x so that the … Read more
Operations on doubles have their own bytecode instructions: dadd, ddiv, dmul, etc. So you can search for those operations in the class files for your project. I’m sure there are tools to do that, so search for those. But if you can’t find any suitable, it’s not too hard to build your own. On a … Read more
This should work: for (int i=1; i<=4; i++) { for (float j=i; j<=std::max(1.6, i); j+=0.5) { cout << j << endl; } } The above code will print the desired sequence. 6 solved int and float increment condition in for loop [closed]
Something like as follows. Hope that its understandable. Output:https://www.ideone.com/EnP40j #include <iostream> #include <iomanip> #include <cmath> int main() { float num1 = 95.345f; float num2 = 95.344f; num1 = roundf(num1 * 100) / 100; //rounding two decimals num2 = roundf(num2 * 100) / 100; //rounding two decimals std::cout << std::setprecision(2) << std::fixed << num1 << ” … Read more
What is precision? It refers to how closely a binary floating point representation can represent a real value. Real values have infinite precision and infinite range. Digital values have finite range and precision. In practice a single-precision IEEE-754 can represent real values of a precision of 6 significant figures (decimal), while double-precision is good for … Read more
Got it, friend of mine has had the same issue. _setmode(_fileno(stdout), _O_TEXT); And cout go back to normal behavior. He’s telling me something about a sh***y VC++ compiler and openGL old version. (Init openGl world screwed stdout in some sort of way, if i understood) 0 solved Cout + Float, no debug error, crash on … Read more
So from the comments: if col is float col[H][W];, your trying to index vx/vy via a float. You would have to cast to int again: int vx2 = vx[static_cast<int>(col[iposy][iposx])]; int vy2 = vy[static_cast<int>(col[iposy][iposx])]; Be careful: There is no implicit index checking, so if your floats are out of range (negative or > WIDTH/HEIGHT), you most … Read more
As far as I know flaoting numbers(for single precision) are stored in memory as follows: sign s (denoting whether it’s positive or negative) – 1 bit mantissa m (essentially the digits of your number – 24 bits exponent e – 8 bits For example: 3.14159 would be represented like this: 0 10000100 11001001000011111100111 ^ ^ … Read more
The function call func(1.14, 3.33) is ambiguous because 1.14 and 3.33 are doubles and both can be converted to either int or float. Therefore the compiler does not know which function to call. You can fix this by explicitly specifying the type of the constants func(float(1.14), float(3.33)) or by changing the overload from func(float, float) … Read more
80-bit long double (not sure about its size in MSVS) can store around 18 significant decimal digits without loss of precision. 1300010000000000000144.5700788999 has 32 significant decimal digits and cannot be stored exactly as long double. Read Number of Digits Required For Round-Trip Conversions for more details. 8 solved Blatant floating point error in C++ program