Why do I have to specify data type each time in C?

Question

As you can see from the code snippet below, I have declared one char variable and one int variable. When the code gets compiled, it must identify t

Answer 1

Because there's no portable way for a variable argument functions like scanf and printf to know the types of the variable arguments, not even how many arguments are passed.

See C FAQ: How can I discover how many arguments a function was actually called with?

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This is the reason there must be at least one fixed argument to determine the number, and maybe the types, of the variable arguments. And this argument (the standard calls it parmN, see C11(ISO/IEC 9899:201x) §7.16 Variable arguments ) plays this special role, and will be passed to the macro va_start. In another word, you can't have a function with a prototype like this in standard C:

void foo(...);

Answer 2

The reason why the compiler can not provide the necessary information is simply, because the compiler is not involved here. The prototype of the functions doesn't specify the types, because these functions have variable types. So the actual data types are not determined at compile time, but at runtime. The function then takes one argument from the stack, after the other. These values don't have any type information associated with it, so the only way, the function knows how to interpret the data is, by using the caller provided information, which is the format string.

The functions themselves don't know which data types are passed in, nor do they know the number of arguments passed, so there is no way that printf can decide this on it's own.

In C++ you can use operator overloading, but this is an entire different mechanism. Because here the compiler chooses the appropriate function based on the datatypes and available overloaded function.

To illustrate this, printf, when compiled looks like this:

 push value1
 ...
 push valueN
 push format_string
 call _printf

And the prototype of printf is this:

int printf ( const char * format, ... );

So there is no type information carried over, except what is provided in the format string.

Answer 3

Compiler may be smart, but functions printf or scanf are stupid - they do not know what is the type of the parameter do you pass for every call. This is why you need to pass %s or %d every time.

Answer 4

Because in the printf you're not specifying data type, you're specifying data format. This is an important distinction in any language, and it's doubly important in C.

When you scan in a string with with %s, you're not saying "parse a string input for my string variable." You can't say that in C because C doesn't have a string type. The closest thing C has to a string variable is a fixed-size character array that happens to contain a characters representing a string, with the end of string indicated by a null character. So what you're really saying is "here's an array to hold the string, I promise it's big enough for the string input I want you to parse."

Primitive? Of course. C was invented over 40 years ago, when a typical machine had at most 64K of RAM. In such an environment, conserving RAM had a higher priority than sophisticated string manipulation.

Still, the %s scanner persists in more advanced programming environments, where there are string data types. Because it's about scanning, not typing.

Answer 5

printf and scanf are I/O functions that are designed and defined in a way to receive a control string and a list of arguments.

The functions does not know the type of parameter passed to it , and Compiler also cant pass this information to it.