how C calculate int and float data type mixed-CodePudding

[SOLVED]

I'm learning C and I encountered some problems during my learning process.

When I'm trying to calculate an int datatype and float datatype in one equation, I ended up getting 2 different outputs. and I'm confused because logically the result should be the same.

here is my code version 1 and version 2. The difference lies in the variable z.

the first case: the value of z is 1
the second case: the value of z is 0

All the input is 11

UPDATE 1

thank you all for your comments and answers. but I'm still confused about what role the variable pplays in case 1 and how could it led to the variable z to be 1 rather than 0?

Case 1

int main()
{
    float a, fee, p;
    int x, y,z;
    scanf("%f", &a);
    fee = 0.6 * a;
    x = (int)(fee);
    y = (int)(2 * fee - 2 * x);
    p = fee * 10.0;
    z = (int)(p - 10 * x - 5 * y);
    printf("%f %d %d %d", fee, x, y, z);
    return 0;
}

Case 2

int main()
{
    float a, fee, p;
    int x, y,z;
    scanf("%f", &a);
    fee = 0.6 * a;
    x = (int)(fee);
    y = (int)(2 * fee - 2 * x);
    z = (int)(fee * 10.0 - 10 * x - 5 * y);
    printf("%f %d %d %d", fee, x, y, z);
    return 0;
}

CodePudding user response：

1) z = (int)(p - 10 * x - 5 * y);
             F    I   I   I   I            // Float or Int

2) z = (int)(fee * 10.0 - 10 * x - 5 * y); // semicolon removed
              F     D      I   I   I   I   // Float, Double or Int

All calculations in 1) are done in float precision, in 2) in double precision.

Suggestion: always prefer double when using floating-point values.

CodePudding user response：

if you cast a floating point number less then 1 like 0.9f to INT, it will become zero.