4 ms·

> This means that it cannot be used safely in portable incantations of variadic functions that expect pointer arguments. Why only variadic functions? The answe

• _kst_ 7y ago
  For non-variadic functions, the argument is implicitly converted to the parameter type (if possible). For variadic functions, the compiler doesn't know the parameter type. So to print a null pointer, you need to do this: printf("NULL = %p\n", (void*)NULL); or printf("null pointer = %p\n", (void*)0); The %p format requires an argument of type void*, so you have to convert to that type if necessary.
• colonwqbang 7y ago
  In a fixadic function the integer literal would be promoted to a pointer, because the caller knows that the argument should be a pointer. In a variadic function call, the caller guesses the type of each argument based on what is passed. A variadic function needs to know exactly the size of its arguments and sizeof(int) != sizeof(pointer) on many systems. This works: void f(char *p); f(0); /* integer literal auto promoted to pointer */ This probably doesn't, at least not as the number of arguments to f is increased: void f(...); f(0); /* assume type of function is void f(int) */
• deleted 7y ago
  [deleted]
• loeg 7y ago
  colonwqbang hit the nail on the head. For non-variadic functions, the function declaration provides the correct parameter type, and caller arguments are coerced to the correct type. > In C, isn't the input to anything coerced to the data type it will represent, in actual complete disregard of the input type? In short: no. Implicit casts between incompatible types are warnings/errors.