为什么可以将另一种类型的参数传递给函数？

Question

Consider the following piece of code:

void function (char arg)
{
    // function's content
}

int main(void)
{
    long double x = 1.23456;

    function(x);
}

I'm giving the function an argument it is not supposed to get. Why does it not cause an error?

Answer 1

It's converted implicitly.

In the context of an assignment, argument passing, a return statement, and several others, an expression of any arithmetic type is implicitly converted to the type of the target if the target type is also arithmetic. In this case, the double argument is implicitly converted to char . (That particular conversion rarely makes sense, but it's valid as far as the language is concerned.)

Note that this implicit conversion is not done for variadic arguments (for example arguments to print after the format string), because the compiler doesn't know what the target type is. printf("%d\\n", 1.5) doesn't convert 1.5 from double to int ; rather it has undefined behavior.

There are also some rules for evaluating expressions with operators of different types. I won't go into all the details here, but for example given:

int n = 42;
double x = 123.4;

if you write n + x the value of n is promoted (implicitly converted) from int to double before the addition is performed.

Answer 2

在您的示例中， double类型被隐式转换为char 。