Thread: argument data type confusion

  1. #1
    Registered User
    Join Date
    Feb 2012

    argument data type confusion


    suppose if i write a test program like
    void function1(unsigned int var1);
    int main(void)
    void function1(unsigned int var1)
    printf("%d", var1);
    The output is -3. how it happens the argument is unsigned but iam passing signed but still prints the signed value. My bigger question is how the arguments are handled if the passing parameters are different types compared to declaration. Can someone please explain me in detail.

    Thanks and regards,

  2. #2
    Registered User
    Join Date
    Jun 2005
    %d tells printf() to treat the corresponding argument as if it is a signed int. Providing an argument that is not signed int results in undefined behaviour.

    It is the programmer's responsibility to ensure that the types specified in the format string match the supplied arguments. You have not fulfilled your responsibility.
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy or unhelpful in reply to you, or tell you you need to demonstrate more effort before you can expect help, it is likely you deserve it. Suck it up, Buttercup, and read this, this, and this before posting again.

  3. #3
    Registered User
    Join Date
    Nov 2012
    i didn't read the page fully , but it offers an explanation to your question here
    in particular,the conversion part of the article
    Last edited by africanwizz; 06-12-2014 at 03:58 AM. Reason: meh..

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