how printf(100) works in C

[aswal]

The prototype of printf() is

printf(const char *__format, ...);

and i typed printf(100) and it run successfully printing a string. How it happened why it doesn't show a type mismatched error.

[weaknessforcats]

printf has a char* as the first argument. If your compiler accepts 100 as valid, then I would get another compiler.

printf(100) does not compile without a cast using Visual Studio 2013.

[aswal]

I am using turbo C++ when i run it as a C program then it works but when i run it as a C++ program then it shows the error which should be shown.

[donbock]

Was there a prototype for printf in scope?
That is, did you include <stdio.h> before calling printf(100)?

[aswal]

No because in C we don't have to include header files. On including those it is showing the error. But I want to know why it works when we did not gave it's prototype and the output it gave.

Also, was the approach of function prototype was present in C or it came in C++

[weaknessforcats]

In C you so not need to have a function prototype. You don't need to include stdio.h.

If you do this, then printf will be considered an external function that returns an int. If your makefile includes the printf library, which is probably does, then the 100 is the char* argument and your build will be OK. When you run the program, you crash trying to access memory location 100 which is not in your process.

Stuff like this is why C++ was started in the first place.

[androidapp]

Format is a pointer, pointer accepts integer value.

[donbock]

The original question was why didn't printf(100) provoke a compiler error. The answer is that without a prototype, the compiler had no reason to think there was anything wrong with printf(100).

The C Standard allows you to call a function without using a prototype. You can also hike into the desert without any water. The only thing stopping you from doing either of those things is a common-sense concern for the consequences.