C++ scanf/printf of array

Question

I've written following code:

int main() {
    double A[2];
    printf("Enter coordinates of the point (x,y):\n");
    scanf("%d,%d", &A[0], &A[1]);

    printf("Coordinates of the point: %d, %d", A[0], A[1]);
    return 0;
}

It's acting like this:

Enter coordinates of the point (x,y):

3,5

Coordinates of the point: 3, 2673912

How is it possible, that 5 converts into 2673912??

Solution

You are reading double values using the decimal integer format (%d). Try using the double format (%lf) instead...

scanf("%lf,%lf", &A[0], &A[1])

Comments

[dgnorton]

Don't forget printf("Coordinates of the point: %lf, %lf", A[0], A[1]);

[Radek Simko]

Thank you, that's it! But it's possible to somehow change the type back to double to be able to operate (+,*,-,/,...) with these numbers? What's the type %lf in fact?

[Andrew Stein]

@Radek A[0] and A[1] are doubles. There is no need to "change the type back to double". +, *, ... will would just fine. %lf, tells scanf that the argument is a "long float", that is, a double. Also note dgnorton's comment on using %lf in the printf

[user1579701]

%lf just tells scanf that the target variable is a double. So you can do normal operations on those variables after they are read in. Look at the documentation for printf format specifiers for the full range. Are you having another problem?

[Clifford]

@Radek: %lf and %d are format specifiers not types. Nothing is changing type. The specifiers tell the formatted I/O functions what type the arguments are to be interpreted as (because they can otherwise be any type and the function does not know a priori.