argument data type confusion

Hi,

suppose if i write a test program like

Code:

---


void function1(unsigned int var1);
int main(void)
{
function1(-3);
}

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,
Satya

%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.

i didn't read the page fully , but it offers an explanation to your question here
.....
in particular,the conversion part of the article