Embarassing, but necessary question

[cnewbie1]

Sorry, this is basic, so I refuse to not understand it:

main() {
	int number = 5;
	printf("%i devided by two is %d\n", tall, devider(tall));	

}

double devider(int x) {
	return x/2;
}

This gives the warning:

test.c:7: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘double’

I don't understand this. I'm sending an in value to a function which returns a double. And the printf has %d for the returned double value.
Still the compiler insists that %d expects an int variable?

.

[Fordy]

%d isn't for double - printf - C++ Reference

[cnewbie1]

Even more embarassing. Thanks :-)

[Bayint Naung]

You have a good compiler. What compiler is it?
It's good habit to pay attention to compiler warnings.

[KBriggs]

where do you declare tall?


His compiler is actually quite poor if that is the only warning he gets when he tried to compile that. He should get:

silly.c: In function ‘main’:
silly.c:3: warning: incompatible implicit declaration of built-in function ‘printf’
silly.c:3: error: ‘tall’ undeclared (first use in this function)
silly.c:3: error: (Each undeclared identifier is reported only once
silly.c:3: error: for each function it appears in.)
silly.c: At top level:
silly.c:7: error: conflicting types for ‘devider’
silly.c:3: note: previous implicit declaration of ‘devider’ was here

Even assuming he includes the necessary headers, it certainly should not compile without tall being declared.

[Elysia]

Where is main's return type?

[Imanuel]

Indeed, %d is same as %i for integer integral values. Check your textbook to see what symbol you need to print a float or double value.