Double forced into an int

is this true?

int myInt;
double myDouble;

scanf("%lf", &myDouble);

myInt = myDouble

if(myInt == myDouble)
{
proceed............

would they equal if the input was say 123.00 but not equal if you input 123.01

i am pretty sure that my asumption is right but let me know if i am wrong.

Quote:

---

Originally posted by Buy2easy.com

int myInt;
double myDouble;

scanf("%lf", &myDouble);

---

2 questions about that...

I'm not sure about this but is there such a thing as Only 'Double'? isn't it double int or something?


and why are you putting %lf? it's that used for long float?

-Emo

if you're not sure about something like this, you could always test it by executing it :)

Code:

---

void testIntDouble(){
        int i =  123;
        double f = 123.01;
        cout << (i == f) << endl; //displays 0 (false)
        f = 123.00;
        cout << (i == f) << endl; //displays 1 (true)
}

---

No, there is no double int. double is a type of its own. int means an integer, a whole number, not a real number as float or double.

Equality tests of floating point values (float or double) are always risky. If the double is the result of a calculation then it might be just a little inaccurate - enough that an equality test fails.

Code:

---

double d = sqrt(16);
if(d ==4.0)
  cout << "Exact" << endl;
else
  cout << "Inexact" << endl;

---

This should print Exact, but it might not.

i was just woundering if a double type could be placed into a int type. like:

int = double

and if the double was 120.000 thenthe int would be 120 but 120.000 is equal to 120 but if the input was a 120.25 then the int would just be 120 because you would lose the .25 so then the double would be larger.

this was for a program that i had due for class so there are reasons that it had to be checked aginst each other.

Not directly. Double is floating point, int, long, short are purely binary numbers.

Type conversion is what this is called. In C++ you can sometimes overload operatiors and create implicit data conversions, but in C
you have to explicitly convert datatypes.


C example

Code:

---

int dbl_to_int(double z){
    char tmp[14];
    sprintf(tmp,"%.0f",z);
    return atoi(tmp);
}
double int_to_dbl(int z){
    char tmp[14];
    sprintf(tmp,"%d",z);
    return atof(tmp);
}

---

Huh? I don't know about your compiler, but ANSI C dictates that float and double get implicitly converted to int with rounding method set to floor, optionally issuing a warning about data loss.
Thus:
double d = 31.999;
int i = d;

will compile with i then having the value 31. It will probably issue a warning which can be muted by explicit casting:

int i = (int)d;

In both cases it should generate code that instructs the FPU to generate an integer from the floating point number.

i was about to say. I just did it and it works fine. Just like i had anticipated. The reason i was woundering is because i had to write a program that wouldreject anything that was not a whole number and i thought that might be a good way to check. There is probably onouther way to do it but that one is the best one i could think of.