wanna divide 2 numbers and get the result like this:
5 / 2 = 2.50
But it only outputs 2.
I don\'t now what i\'m doing wrong.
Here my code:
You have to use float
or double
variables, not int
(integer) ones. Also note that a division between two integers will lead to an integer result, meanwhile a division between a float
/double
and an integer will lead to a float result. That's because C implicitly promote this integer to float
.
For example:
5/2 = 2
5/2.0f = 2.5f
Note the .0f
, this actually means that we are dividing with a float.
To avoid the typecast in float you can directly use scanf with %f flag.
float a;
float b;
float c;
printf("First number\n");
scanf("%f", &a);
printf("Second number\n");
scanf("%f", &b);
c = a / b;
printf("%f", c);
The '/' - sign is for division. Whenever in C language, you divide an integer with an integer and store the data in an integer, the answer as output is an integer. For example
int a = 3, b = 2, c = 0;
c = a/b; // That is c = 3/2;
printf("%d", c);
The output received is: 1
The reason is the type of variable you have used, i.e. integer (int
)
Whenever an integer is used for storing the output, the result will be stored as integer and not a decimal value.
For storing the decimal results, C language provide float
, double
, long float
and long double
.
Whenever you perform an operation and desires an output in decimal, then you can use the above mentioned datatypes for your resultant storage variable. For example
int a = 3, b = 2;
float c = 0.0;
c = (float)a/b; // That is c = 3/2;
printf("%.1f", c);
The output received: 1.5
So, I think this will help you to understand the concept.
Remember: When you are using float
then the access specifier is %f
. You need to convert your answer into float
, just as I did, and then the answer will be reflected.
In C, only an int type number is displayed. 5/2 gives a floating point type number. So, the compiler compiles it only with the integer value.
You need a double
variable to store the result. int
stores only integers. Additionally, you have to typecast the other variables also before performing the division.
Do something like this
double c;
.
.
.
c = (double)a / (double)b;
printf("%f", c);
NOTE:
You do not need the &
in printf()
statements.