Arithmetic conversions

This is a discussion on Arithmetic conversions within the C Programming forums, part of the General Programming Boards category; Hi all, Im a beginner with no previous programming experience. Anyway, Im taking an online course in ANSI C programming, ...

  1. #1
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    Arithmetic conversions

    Hi all,

    Im a beginner with no previous programming experience. Anyway, Im taking an online course in ANSI C programming, and need some help.

    This is a test question, and Im not asking for the answer. Just some help in a way for me to find/understand the answer.

    the question is:

    Given the following declarations and initialization:


    int i = 1;
    int x = 2.0;

    What is the value and type of the expression i/x ?



    My book explains usual arithmatic conversions, and it says:

    "if either operand is of type double, the other operand is converted to double."

    I wrote a sample program to see how the computer evaluates this.


    Code:
    #include <stdio.h>
    
    int i = 1;
    int x = 2.0;
    int y;
    
    int main(void)
    {
    	y=i/x;
    	printf("y = %d as decimal", y);
    
    return 0;
    }
    When I run this I get 0 as result, however when I run:

    Code:
    #include <stdio.h>
    
    int i = 1;
    int x = 2.0;
    int y;
    
    int main(void)
    {
    	y=i/x;
    	printf("y = %f as float", y);
    
    return 0;
    }
    I get a runtime error.

    Any suggestions that I could try to help me understand this better?


    Thanks,

    Darren

  2. #2
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    you need to use floats for your types if they aren't integers. Change all your "int"s to "float"

  3. #3
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    Two pointers;

    1) With the initialisation "int x = 2.0;", what value do you expect x to wind up with. And what type is it?

    2) The reason for a runtime error in your second code example is that you're passing an int (y) to printf() which has been told to expect a double. Using the %f specifier doesn't magically turn y into a double.

  4. #4
    cwr
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    With:
    Code:
    int i = 1;
    int x = 2.0;
    Both variables are of type int. Just because you used 2.0 doesn't make it a floating point type. The 2.0 will be converted to an int because x is an int. Therefore, when you do a division, you are just doing an integer divide, since neither type is a float or double.

    With your printf, you are trying to print an integer variable (y) using the %f double format specifier. This doesn't make any sense, so you must use %d, or cast the y to a double.

    Of course, you will still get an output of 0, because y is an int type, but at least it will work. Even if you change y to be a double type, you will still get 0, because both the other variables are ints.

  5. #5
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    Thankyou for your responses,

    I think I know the answer.


    Darren

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