CODEWITHSUNDEEP
ctype-conversionprintf
idk.mp3
idk.mp3

Reputation: 43

If I have one variable as int, and use it for calculating other of type float, why is result shown as rounded-down float?

I'm doing an equation solver. The code goes like this:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    printf("Equation solver of type ax + by = 0, for y\n");
    int a, b, x;
    float y;
    printf("Enter value of a as integer: ");
    scanf("%d", &a);
    printf("Enter value of b as integer: ");
    scanf("%d", &b);
    printf("Enter value of x as integer: ");
    scanf("%d", &x);
    y = a*x/b; //We have equation ax + by = 0, from that: ax = by, and that is: y = ax/b
    printf("Solution for y of equation: %dx + %dy = 0, for x = %d, is: %.2f", a, b, x, y);
    return 0;
}

When I enter a = 3, b = 5 and x = 3, the output is supposed to be 1.80, but I get 1.00. Why is this? Wouldn't expression a*x/b be converted into type of float? I'm just beginner and we only mentioned type conversion with our professor, maybe I got it wrong?

Upvotes: 3

Views: 82

Answers (2)

Ted Lyngmo
Ted Lyngmo

Reputation: 117298

In a*x/b all variables are ints so it's a pure integer calculation.

If you cast one of the first operands used to float, the other will also be implicitly converted to float

y = (float) a * x / b;

More about what implicit conversions there are can be found here: Implicit conversions

Broken down according to operator precedence:

Precedence Operator Description Associativity
3 * / % Multiplication, division, and remainder Left-to-right
14 = Simple assignment Right-to-left 
y = a*x/b; // with a = 3, b = 5 and x = 3

According to operator precedence, a*x will be performed first and as can be read in the Usual arithmetic conversions chapter on Implicit conversions: "The arguments of the following arithmetic operators [*, /, %, +, -, ...] undergo implicit conversions for the purpose of obtaining the common real type, which is the type in which the calculation is performed"

If "both operands are integers. Both operands undergo integer promotions". Since both a and x are int - no promotion is necessary and the common type is therefore int and the calculation is done using ints:

3 * 3 => 9

Next according to operator precedence comes the result of a*x divided by b. Again, b is an int and 9 is an int so the same procedure as above gives us int as the common type.

9 / 5 => 1

The last step according to operator precedence is the simple assignment y = 1. In the chapter "Conversion as if by assignment" we can read: "In the assignment operator, the value of the right-hand operand is converted to the unqualified type of the left-hand operand."

This means that it's only now that the result of the calculation, 1, becomes 1.f (float)

1 => 1.f

and that's what's assigned to y.

By converting a to float, like in y = (float) a * x / b;, you can follow the rules for finding the common type and see that x will be converted to float and then b, so the calculation becomes 3.f * 3.f / 5.f and the result will be 1.8f.

If you are sure that a * x will not cause integer overflow (~produce a value that is too large or small to store in an int), you can delay the conversion until the division:

y = a * x / (float) b;

It will now do an integer multiplication and only then will the result of that multiplication be converted to float and divided by 5.f, as if done like so:

y = (float) (3 * 3) / 5.f;

Upvotes: 3

dbush
dbush

Reputation: 223937

Wouldn't expression a*x/b be converted into type of float?

The result of that expression gets converted to float before being assigned to y. Since each operand has type int, the math is done with that type.

Specifically, a*x and b both have type int, so integer division is performed which truncates the fractional part.

If you cast one of the variables to type float, then floating point division will be performed.

y = (float)a*x/b;

Upvotes: 3

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