[ACCEPTED]-Default argument promotions in C function calls-promotions

Accepted answer

Score: 41

Upvoted AProgrammer's answer—those 17 are the real goods.

For those of you who 16 are wondering why things are this way: in the 15 dark ages before 1988, there was no such 14 thing as a function prototype in classic 13 "K&R" C, and the default argument promotions 12 were instituted because (a) there were 11 essentially "free", as it costs no more 10 to put a byte in a register than to put 9 a word in a register, and (b) to cut 8 down on potential errors in parameter passing. That 7 second reason never quite cut it, which 6 was why the introduction of function prototypes 5 in ANSI C was the single most important 4 change ever in the C language.

As to when 3 default promotions kick in: default argument promotions are used exactly when the expected type of the argument is unknown, which is to 2 say when there's no prototype or when the 1 argument is variadic.

Score: 37

(Non variadic) parameters to functions with 25 a prototype are converted to the corresponding 24 type, which can be char, short, float.
Parameters 23 to functions without prototype and variadic 22 parameters are subject to default argument 21 promotions.

If you define a function with 20 a prototype and use it without the prototype 19 or vise versa and it has parameters of type 18 char, short or float, you'll probably have 17 a problem at run-time. You'll have the 16 same kind of problems with variadic functions 15 if the promoted type doesn't match what 14 is used when reading the argument list.

Example 13 1: problem when defining a function with 12 a prototype and using it without.

definition.c

void f(char c)
{
   printf("%c", c);
}

use.c

void f();

int main()
{
   f('x');
}

can 11 fail because an int will be passed and the 10 function expect a char.

Example 2: problem 9 when defining a function without a prototype 8 and using it with one.

definition.c

void f(c)
   char c;
{
   printf("%c", c);
}

(This 7 is kind of definition is very old fashioned)

use.c

void f(char c);

int main()
{
   f('x');
}

can 6 fail because an int is expected but a char 5 will be passed.

Note: you'll remark that 4 all functions from the standard library 3 have types which result from default promotions. So 2 they didn't cause problem during the transition 1 when prototypes were added.

Score: 17

Your confusion stems from a very slight 11 misunderstanding of the terminology - both 10 declarations and definitions can include 9 prototypes (or not):

void func(int a, char b, float c);

That is a function declaration that 8 includes a prototype.

void func(int a, char b, float c) { /* ... */ }

That is a function 7 definition that includes a prototype.

"Prototyped" and 6 "non-prototyped" are just attributes of 5 a function type, and both declarations and definitions 4 introduce the type of the function.

So you 3 can have a declaration without a prototype:

void func();

or 2 you can have a definition without a prototype 1 (K&R C style):

void func(a, b, c)
    int a;
    char b;
    float c;
{ /* ... */ }

Source: stackoverflow.com

[ACCEPTED]-Default argument promotions in C function calls-promotions

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