pow | 易学教程

return value of pow() gets rounded down if assigned to an integer

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问题 I am using the pow function in C and storing the return value in an integer type. see the code snippet below: for (i = 0; i < 5; i++){ val = (int)pow(5, i); printf("%d, ", val); } here i , and val are integers and the output is 1, 5, 24, 124, 624 . I believe this is because a float 25 is treated as 24.99999... which gets rounded down to 24 on assignment to an integer. How can I by pass this if I still need to store the return value in an int ? 回答1: Add 0.5 before casting to int . If your

PHP Table Loop?

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问题 <?php $i=0; while($i < 101){ if($i%2==0){ echo "<tr>".PHP_EOL; } echo "<td>".$i."</td>".PHP_EOL; $i++; if($i%2==0){ echo "</tr>".PHP_EOL; } } ?> This code generates a table with 100 rows and 2 columns. But what I want to do is that show ordered numbers (upp to 100) in the left side of the rowcells and show something else (ex. pow(rownumber) ) in the right side of the rowcells. How can I do that? 回答1: Try this, Will output 100 rows with the number and its power in two columns <table> <?php for

cmath std::pow function giving wrong value when assigned to a variable?

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问题 The method below is keeping track of how many times specific numbers come up from groupings of various sets of numbers void build_prob_distro(const std::vector<Foo>& num_sets, std::map<int, int>& prob_distro){ int key; Foo cur_foo; for(unsigned int foo_num = 0; foo_num<num_sets.size(); foo_num++){ cur_foo = num_sets.at(foo_num); key = 0; int val; for(int cur_foo_num=0; cur_foo_num<cur_foo.get_foo_length(); cur_foo_num++){ std::cout << cur_foo.get_num_at(cur_foo_num)*std::pow(10, cur_foo.get

C# Decimal to double?

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问题 Why can't I use math.pow with numericUpDown value? When I use this code: a = (numericUpDown1.Value); b = (numericUpDown2.Value); m = Math.Pow(a, b); I receive the following error: Severity Code Description Project File Line Suppression State Error CS0266 Cannot implicitly convert type 'decimal' to 'double'. An explicit conversion exists (are you missing a cast?) 回答1: This is because Math.Pow has only one overload - the one accepting two double s. Since you want to work with decimal s, you

Java Math.Pow method

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问题 monPay = (amtFin * amtI)/((1-(1+amtI)*Math.pow(-n,-n))); I am using this code but it is not calculating Math.pow(-n,-n) correctly. It seems like it is Math.pow(-n,-n) is calculating to 0.0. All variables have type double . 回答1: It looks like you're trying to compute monthly loan repayments based on the principal amount etc. However, your formula is wrong: the (1+amtI)*Math.pow(-n,-n) should be Math.pow(1+amtI,-n) . 回答2: What happens is that the value of pow(-n,-n) is too low to be represented

Am I going crazy or is Math.Pow broken?

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问题 I used the base converter from here and changed it to work with ulong values, but when converting large numbers, specifically numbers higher than 16677181699666568 it was returning incorrect values. I started looking into this and discovered that Math.Pow(3, 34) returns the value 16677181699666568, when actually 3^34 is 16677181699666569. This therefore throws a spanner in the works for me. I assume this is just an issue with double precision within the Pow method? Is my easiest fix just to

Why does pow(5,2) become 24? [closed]

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问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . I'm making a simple calculator that you can choose a function, then 2 inputs to get your answer. It's a neat little program and everything is running smoothly except for the powers. Every number works correctly. But according to this: 5^2=24, 5^3=624 . I am using pow(number1,number2) . #include <iostream>

Math.pow yields different results upon repeated calls

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问题 After upgrading to Java 1.8.0_20 our test system reported errors, but the code was not changed. I found out, that Math.pow() called with exactly the same input parameters yields different results upon repreated calls. In Java 1.8.0_11 it behaves as expected and returns always the same value, but with Java 1.8.0_20 and above it sometimes returns slightly different values. This is similar to the question Math.pow yields different result depending on java version, but different because the

Swift's pow() function won't accept Doubles as arguments

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问题 I created this infix operator ^^ as a substitute to using the pow function: infix operator ^^ { associativity left precedence 155 } func ^^ <T: IntegerLiteralConvertible>(left: T, right: T) -> T { return pow(left as Double, right as Double) } I used the IntegerLiteralConvertible protocol as a type constraint for the generics left and right , because from my understanding this diagramm shows, that it basically includes all number types. In order to use the pow function I have to downcast left

Why does this code using Math.pow print “HELLO WORLD”?

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I discovered the following code. I know, it looks less weird/exciting than this one using seemingly random numbers, but it seems to be more complex than this one using bit shifts on a large number: long[] c = {130636800L, -5080148640L, 13802573088L, -14974335980L, 8683908340L, -3006955245L, 651448014L, -89047770L, 7457160L, -349165L, 6998L}; for (int x = 0; x < 11; x++) { long s = 0; for (int i = 0; i < 11; i++) s += c[i] * Math.pow(x, i); System.out.print((char)(s / 1814400)); } Code on Ideone Output: HELLO WORLD How does it work? Is it some form of encryption or did anyone get mad