Why passing string to scanf() compiles fine in C?

Question

I recently wrote a simple program where I by mistake use scanf() instead of printf() for displaying a message on console. I was expecting to get compile time error, but it compiles fine without warnings & crashes at runtime. I know that scanf() is used for taking input from keyboard. Shouldn't I get an error in following program?

#include <stdio.h>
int main()
{
    scanf("Hello world");   // oops, It had to be printf()
    return 0;
}

Is it invokes undefined behavior(UB)? Is there any mention about this in C standard? Why it isn't checked at compile time whether proper & valid arguments are passed to scanf() function or not?

Answer 1

The code is behaving correctly. Indeed, scanf is declared

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

Luckily, your format does not contain any % for which there would be no correspondence in ..., which would invoke UB.
Further, format is a string literal which allows the compiler to go through it ensuring you passed the right type of parameters in regards to the format specifiers, as part of sanity checks enabled with higher warning levels. (-Wformat family)

Answer 2

scanf does take a string paramater. That's why it compiles fine, but it appears that your parameter does not match what it expects, so it crashes (or just exits) at runtime.

Answer 3

The compiler cannot read your mind.

Both printf("Hello world") and scanf("Hello world") are well-formed.

Answer 4

Passing a string to scanf compiles fine, because scanf expects a string as it first parameter. See the documentation or programming manual for detailed scanf description.