i tried to find the area of the circle and found a garbage value. I m using a gcc compiler
area of a circle
#include<stdio.h>
void main()
{
int a=1;
float c;
c = 22/7 * a * a;
printf("%f is the area" , c);
}
3.142857 is the area should be the output. but it is showing 3.000000 is the area
22/7 is integer division so will truncate to 3. You need something like:
float c = 22.0 / 7 * a * a;
And, as an aside, 22/7 isn't that good a representaion of PI. 355/133 would be better but much better than that would be (on systems where it is available):
#include <math.h>
// Now use M_PI
or (on systems where it is not):
#define M_PI 3.14159265358979323846264338327950288
There's also very little reason to use float types when double variables have more range and precision. Single precision float variables can be useful if you have a lot of them and want to reduce storage space but, in most other cases, it's worth just using double precision.
It's not a garbage value, the operation you are performing on all the integers, take a look
22/7 * a * a
here variable a is integer, as well as 22 and 7, so the complete operation's end result is int and it get converted into the float c variable which results 3.0000.
to solve this, write:
22/7.0 * a * a
3.142857 is the area should be the output. but it is showing 3.000000 is the area
You might be thinking because you have a float data type on the left-hand side of your assignment operator, the computation on the right-side of your assignment operator should be a float computation.
However, C programming language won't promote the variables to float implicitly. It will do integer division and multiplication and after computing the result, the result (which will be an integer) will be implicitly casted to the data type on the LHS of your assignment operator (i.e., float).
To learn more about integer division, check this link. It says:
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded.88) If the quotient a/b is re-presentable, the expression (a/b)*b + a%b shall equal a.
So, you were getting 3.0000 because according to integer division, 22/7 is 3 (fractional part discarded) and after multiplying with a twice, you will get an integer of value 3. To assign the computed value of the expression to your variable c, language does the implicit conversion to float having a value of 3.000.
To solve your problem, you need to explicitly use variables or constants of type float on RHS of your assignment operator or you can use explicit casting operator.
float c = 22.0f / 7 * a * a;
Because of higher associativity of division operator, (22.0 / 7) will be calculated first and their result will be a float as one of operand (22.0 and 7) is a float constant. Then, the result of the expression will be calculated and integer operand will be promoted to float data type. At the last, your result (which is float) will be assigned to your variable c.
See this link to learn about the difference between 22.0 and 22.0f.
