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6

This is a very common problem and the whole point of greedy.c, so let me try and give the definitive answer. Float values with a decimal part (to the right of the decimal, in case you weren't certain) rarely store with exact precision. They have to be converted from base 10 to base 2. Since there is a limit on the number of digits that can be used, it is ...


3

Your problem isn't that round(7.535600) rounds to 8 (It should, and does), but that 7.535600 isn't the correct index value. For that text, I'd expect these values: 96 letter(s) 23 word(s) 1 sentence(s) L: 417.391296 S: 4.347826 Score: 7.455652 Grade 7 Perhaps you are running into problems in your intermediate calculations. Have you made sure that you ...


3

Perhaps you don't understand float imprecision fully enough. If you were to print the value of ch with more digits of precision, you would see what's happening. Try using the following: printf("%0.32f\n", ch); This will show a more accurate value of what is in ch, rather than a rounded off value generated by the default precision resulting from the ...


2

As pointed out by other answers, your example returns true and you should deal with floats carefully because they are NOT precise. The precision problem is caused by the fact that there are some floating-point values that cannot be exactly represented in binary. The decimal numbering system has this problem too with some values. For example, the value 1/3 ...


2

[I posted this on reddit last year but rather than just linking it, I thought I'd repost. In that case, they had printed a few more decimals, but the principle is the same.] A float value has 32 bits to work with: 1 bit to hold the sign 8 bits to hold the offset exponent ( which is the exponent + 127) and 23 bits to hold the significant digits So, in the ...


2

While it is easy to represent integers in a computer, it is different for floating point numbers. Most FP numbers cannot be precisely represented inside a computer. The reason is because numbers are represented in base 2 in a computer, but most base 10 fractions can't be represented by base 2 systems. For a detailed explanation, you might try googling for ...


2

Most of the problem is that no matter what you do, you keep putting the result back into change, a float. Floats have this nasty little problem that they don't store fractional parts of a number, the decimal part, correctly. In fact, your question demonstrates it well. If you store 0.01 in a float, it actually gets stored as 0....


2

The code seems far more complicated than necessary, but this seems to be because you're trying to handle the imprecise float storage problem unsuccessfully. This can be resolved by simply converting dollars and cents in a float to cents only and storing it in an int. It's not that hard. First, remember that round functions round the input to the nearest ...


1

The reason is that if the code is doing division with all integers, it will do integer division, not normal division. Integer division will truncate the fractional part of the division. So, if the result were something like 3.8, the result would be truncated to 3. This would lead to a lot of errors in an entire image. Now, you can reduce the extent of ...


1

I know of two ways, but neither can be done through how printf works. The way I would do it would be to multiply the value by 100 and cast it to an int, which would cut off any remaining decimal values. Then, I would return it to a float and divide by 100. So, printf(%.2f\n", (float((int)(cash * 100))) / 100.0. You might be able to do this without ...


1

The value being coerced has mixed character data (the $ in the case of $208.43). Consider either sanitising the return value or revising the finance.db data type.


1

So you decided to make some functions, why not. The function definition would be like return_type function_name(param1_type param1_name, param2_type param2_name, ...) { // body of the function } for example // function apply_discount takes two parameters of type double // and returns a double itself double apply_discount(double price, double ...


1

I declared n as an int, not a float. That caused the problem.


1

double is one of the types in C, like int and char. It's a double-precision floating point number. So #include <math.h> double round(double x); means that the function takes a double and returns a double. You can pass it a float and it will handle that. So in your usage: int cents = round(change * 100); You are rounding the value of change * ...


1

It's pretty straightforward. In the second one, the code uses a single variable called c. No problems. In the first one, by including float in the while statement, the code is redeclaring the variable c. The first c gets replaced by the second c. At that point, the while statement tries to use the new, second var c. Unfortunately, the value previously ...


1

This is because when a/b is performed the result is o form int type which is 0 since the .625 part of the result is gone due to the expression a/b being an int. When the value a/b (i.e 0) is assigned to float x it stores a value 0 as floating point number 0.000 and displays that. You can try the following line ( float x = (float) a / b ;)


1

answered in my edit - confirmed by Chris G.


1

Line 1: OK Line 2: If you have declared total_float as a float before this code snippet it's OK. Line 3: That is not the correct use of roundf(). I would suggest you use man roundf on a terminal to find the correct use. I could give it to you but that way you are not gonna learn how to solve your problems yourself. I'll give you just the signature of the ...


1

Did you try using round command after multiplying 4.2 by 100? uma1966


1

First things first, your example is returning true. It is difficult to deal with floating point numbers in any programming language unless you know how they are stored/operated on your machine. Although example you gave, returned true on ideone, but it is for sure that there could be any another example for such weird behaviour. float a = 0.1; printf("%....


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