I have two int values that I want to combine into a decimal number. So for example, I have A = 1234 and B = 323444. Both are int and I do not want to change it if possible.
I want to combine them to get 1234234233.323444.
My initial method was to divide b by 1e6 and add it to A to get my value.
I assigned
int A = 1234234233;
int B = 323444;
double C;
A = 1234;
B = 323444;
C = A + (B/ 1000000);
printf("%.6f\n", C);
I get 1234234233.000000 as a result. It rounds my C and I do not want that as I want 1234234233.323444
how can I solve this?
Try like this:
C = A + (B/ 1000000.0);
ie, make the denominator as double so that when integer by integer division is made it does not return weird results like you are getting.
NOTE:-
Integer/Integer = Integer
Integer/Double = Double
Double/Integer = Double
Double/Double = Double
B is an integer and dividing an integer by another integer (10000 here) will always give an integer and that's why you are getting unexpected result. Changing 10000, which is of type int, to 10000.0 (double type) will solve this problem. It seem that 10000 and 10000.0 are integer by mathematical definition but both are of different type in programming languages, former is of type int while latter is of type double.
C = A + (B/ 1000000.0);
or
C = A + ((double)B/ 1000000);
to get the expected result.
Related
I have an expression which does the same calculation. When I try to do the whole calculation in a single expression and store it in variable "a", the expression calculates the answer as 0. When I divide the equations in two different parts and then calculate it, it gives the answer -0.332087. Obviously, -0.332087 is the correct answer. Can anybody explain why is this program misbehaving like this?
#include<stdio.h>
void main(){
double a, b, c;
int n=0, sumY=0, sumX=0, sumX2=0, sumY2=0, sumXY=0;
n = 85;
sumX = 4276;
sumY = 15907;
sumX2 = 288130;
sumY2 = 3379721;
sumXY = 775966;
a = ((n*sumXY) - (sumX*sumY)) / ((n*sumX2)-(sumX*sumX));
b = ((n*sumXY) - (sumX*sumY));
c = ((n*sumX2) - (sumX*sumX));
printf("%lf\n", a);
printf("%lf\n", b/c);
}
Output:
0.000000
-0.332097
In your program
a = ((n*sumXY) - (sumX*sumY)) / ((n*sumX2)-(sumX*sumX));
all the variables in the right hand side are of type int, so it will produce a result of type int. The true answer -0.332097 is not a int value, so it will be converted to a valid int value, namely 0. And this 0 is assigned to variable a.
But when you do
b = ((n*sumXY) - (sumX*sumY));
c = ((n*sumX2) - (sumX*sumX));
printf("%lf\n", b/c);
The variable b and c are of type double, so the expression b/c produce a double typed value and the true answer -0.332097 is a valid double value. Thus this part of your code give a right result.
In first equation i.e. a = ((n*sumXY) - (sumX*sumY)) / ((n*sumX2)-(sumX*sumX)) both numerator and denominator will give integer results and the value stored in a will also be integer as integer/integer is integer. In second and third expression as you are solving them individually both b and c will be stored in double and double/double will result in a double i.e. a decimal value.
This problem can be solved by using type casting - or better still using float for the variables.
Add double before your calculation, so after you do your integer calculation in "a", it will convert it to double.
a = (double)((n*sumXY) - (sumX*sumY)) / ((n*sumX2)-(sumX*sumX));
First of all (INTEGER)/(INTEGER) is always an INTEGER. So, you can typecast it like a = (double)(((n*sumXY) - (sumX*sumY)) / ((n*sumX2)-(sumX*sumX)));
OR
We know that any number (n ∈ ℂ), multiplied by 1.0 always gives the same number (n). So, your code shall be like:
a = ((n*sumXY*1.0L) - (sumX*sumY)) / ((n*sumX2)-(sumX*sumX));
Multiplying by 1.0L or 1.0f converts the whole arithmetic operation to long double data type.
Now you can print the number (-0.332097) to stdout.
Non-standard Code
Your code is:
void main()
{
//// YOUR CODE
}
Which is non-standard C. Instead your code should be like:
int main(int argc, char **argv)
{
//// YOUR CODE
return 0;
}
Change all your INTEGERS to DOUBLES. That should solve the problem
The type of 1st expression is int whereas in 2nd expression it is double.
How can I get a float or real value from integer division? For example:
double result = 30/233;
yields zero. I'd like the value with decimal places.
How can I then format so only two decimal places display when used with a string?
You could just add a decimal to either the numerator or the denominator:
double result = 30.0 / 233;
double result = 30 / 233.0;
Typecasting either of the two numbers also works.
As for the second part of the question, if you use printf-style format strings, you can do something like this:
sprintf(str, "result = %.2f", result);
Bascially, the ".2" represents how many digits to output after the decimal point.
If you have an integer (not integer constant):
int i = 20;
int j = 220;
double d = i/(double)j;
This is the simplest way to do what you are trying to achieve, I think..
double result = 30/233.0f;
for iOS development (iPhone/iPad/etc) better to use float type.
float result = 30/233.0f;
I am trying to take the result of div function in c, cast it to an int and then add that int to a greater int value. I get the error of the title all the time, and i can not understand why
out = div(n, 10);
r = (int) out;
a = a + r;
Compiler shows me as an error the second line and out specifically.
Thank you in advance!
A div_t, as returned by div(), is a structure containing two numbers, the quotient and the remainder.
typedef struct {
int quot;
int rem;
} div_t;
If you've used the div() function then you want either r = out.rem or r = out.quot, not clear which from your example.
If all you want is the quotient, though, r = n / 10 is simpler. And if all you want is the remainder, r = n % 10 (for non-negative n). div() is useful in the case where you need both values - the actual divide instruction on many machines can deliver both results from one instruction.
div(x, y) function does both x / y and x % y in one operation. It returns a structure with rem member having the result of x % y and quot having the result of x / y. In your case you would access these values as out.quot and out.rem and both members are already values of type int. Casting a structure containing two integers into an integer does not make any sense.
On many processors there is a single division opcode that always calculates both, so if you need both, then div(x, y) is giving the other one for free. One common instance is converting a number into a decimal string which requires repeatedly taking quotient and remainder with 10; here you can use div efficiently for positive numbers:
res = div(n, 10);
next_digit = res.rem;
// place next_digit into the string
n = res.quot;
I am trying to get a floating variable accurate to just 3 decimal points for a comparison calculation. I am trying the method below, but it doesn't work. I can't see why not, so please can someone tell me where I am going wrong.
Output from this code is b = 10000.050617, bb = 10000050 and fbb = 10000.000. I want fbb to be 10000.050.
int bb; double m,n,p,q,b,t,u,fbb;
m=24.161, n=57.695, p=67.092, q=148.011;
t=p-m; u=q-n;
b=t*t+u*u; bb=b*1000; fbb=bb/1000;
printf("b=%.6lf,bb=%i,fbb=%.3lf\n",b,bb,fbb);
return 0;
When you perform
fbb = bb/1000;
It treats operation as int/int and returns an int
Try
fbb = ((double)bb)/1000.000;
It will be treated as (double)/(double) and return a double.
bb is an int, so bb / 1000 will follow the integer division. Change either or both operand to a double. The simplest way is:
fbb = bb / 1000.0; //type of 1000.0 is double
or
fbb = (double)bb / 1000
When you perform
fbb = bb/1000;
It treats operation as int/int and returns an int. its demotion of value.
Also take long bb; instead of int as int has value 32767 as its high value.
Try
fbb = bb/1000.000;
or
fbb = (double)bb/1000;
bb is int. So bb/1000 is doing a int division, which results again in an int = 1000 (no decimals). That int value is cast to a double.
Use
fbb=(double)bb/1000;
bb is integer and result is integer, and then converted to double
I'm trying to do some arithmetic on integers. The problem is when I'm trying to do division to get a double as a result, the result is always 0.00000000000000000000, even though this is obviously not true for something like ((7 * 207) / 6790). I have tried type-casting the formulas, but I still get the same result.
What am I doing wrong and how can I fix it?
int o12 = 7, o21 = 207, numTokens = 6790;
double e11 = ((o12 * o21) / numTokens);
printf(".%20lf", e11); // prints 0.00000000000000000000
Regardless of the actual values, the following holds:
int / int = int
The output will not be cast to a non-int type automatically.
So the output will be floored to an int when doing division.
What you want to do is force any of these to happen:
double / int = double
float / int = float
int / double = double
int / float = float
The above involves an automatic widening conversion - note that only one needs to be a floating point value.
You can do this by either:
Putting a (double) or (float) before one of your values to cast it to the corresponding type or
Changing one or more of the variables to double or float
Note that a cast like (double)(int / int) will not work, as this first does the integer division (which returns an int, and thus floors the value) and only then casts the result to double (this will be the same as simply trying to assign it to a double without any casting, as you've done).
It is certainly true for an expression such as ((7 * 207) / 6790) that the result is 0, or 0.0 if you think in double.
The expression only has integers, so it will be computed as an integer multiplication followed by an integer division.
You need to cast to a floating-point type to change that, e.g. ((7 * 207) / 6790.0).
Many poeple seem to expect the right-hand side of an assignment to be automatically "adjusted" by the type of the target variable: this is not how it works. The result is converted, but that doesn't affect any "inner" operations in the right-hand expression. In your code:
e11 = ((o12 * o21) / numTokens);
All of o12, o21 and numTokens are integer, so that expression is evaluated as integer, then converted to floating-point since e11 is double.
This like doing
const double a_quarter = 1 / 4;
this is just a simpler case of the same problem: the expression is evaluated first, then the result (the integer 0) is converted to double and stored. That's how the language works.
The fix is to cast:
e11 = ((o12 * o21) / (double) numTokens);
You must cast these numbers to double before division. When you perform division on int the result is also an integer rounded towards zero, e.g. 1 / 2 == 0, but 1.0 / 2.0 == 0.5.
If the operands are integer, C will perform integer arithmetic. That is, 1/4 == 0. However, if you force an operand to be double, then the arithmetic will take fractional parts into account. So:
int a = 1;
int b = 4;
double c = 1.0
double d = a/b; // d == 0.0
double e = c/b; // e == 0.25
double f = (double)a/b; // f == 0.25