Ehm, no

and once I remove them, it crashes right away

well you need them if you don't want "undefined function" warnings

so what is the difference between the two different delcarations?

"Undefined function" is a compilation error, isn't it?

Return value type

no, it's a warning

why would you remove it ?

note: include the header <stdio.h> or explicitly provide a declaration for 'snprintf' warning: implicitly declaring library function 'vsnprintf' with type 'int (char *, unsigned long, const char *, struct __va_list_tag *)' [-Wimplicit-function-declaration]

C is so ugly

so the pointer lives in the stack but hold the value to the memory address that is stored in the heap in this case ?

In this case yes

what happens to memory on the stack during runtime?

It returns one The function is not a pointer

something i found in a book

Has nothing to do with your question

Has nothing to do with your question

U asked about the difference between int* func() and int func()

They have the same function name

Those are just different return types

Yes that is correct

Yes that is correct

Some people are weird. Just because u can do something, doesn't mean u should

U can do int *****p

U can do int *****p

U can do int *****p

What the hell...

Some people are weird. Just because u can do something, doesn't mean u should

Much love <3

Much love <3

That’s why, in C, function names are also pointer variables. When you create a function called go_to_warp_speed(int speed), you are also creating a pointer variable called go_to_warp_speed that contains the address of the function.

A stack is a special area of computer's memory which stores temporary variables created by a function. In stack, variables are declared, stored and initialized during runtime. It is a temporary storage memory. When the computing task is complete, the memory of the variable will be automatically erased

char ** it says they use to point to an array of strings , technically is an character array of arrays

Do stdlib.h/stdio.h do some kind of initialization of IO? I need to get rid of these includes and define all functions/structs manually, but fseek() crashes once I remove the includes.

note: include the header <stdio.h> or explicitly provide a declaration for 'snprintf' warning: implicitly declaring library function 'vsnprintf' with type 'int (char *, unsigned long, const char *, struct __va_list_tag *)' [-Wimplicit-function-declaration]

what happens to memory on the stack during runtime?

That’s why, in C, function names are also pointer variables. When you create a function called go_to_warp_speed(int speed), you are also creating a pointer variable called go_to_warp_speed that contains the address of the function.

they are saying function names are pointers variables that point to the location in memory where the memory is located right?

like int *somefunction(), or int somefunction(), or int ** somefunction()?

so returns an int, one returns an int pointer, and one returns an pointer, pointing to another pointer ?

one question , why would you point to another pointer , and why would you return that in a function?

U can do int *****p

A pointer points to some memory It's not a "variable"

It is a variable

Don't do it, ever...

No you can't.

why can't you do int ****p; ?

Because you must check and fix your keyboard first, or further will be worse!

Function name literal also called identifier is a constant of type "pointer to function with there corresponding signature".

Google banned you?

where is this defined? / does this come from? I strongly believe this is not true. If this would be the case, how would the compiler be able to detect and warn if a function type is used as operand of sizeof ? and further, why should it be ill-formed. it is true that in many cases a function designator implicitely converts to a function pointer. In the below example, foo is a function type and not a pointer to function void foo() {} src: n1570, 6.5.3.4#1, 6.3.2.1#9

where is this defined? / does this come from? I strongly believe this is not true. If this would be the case, how would the compiler be able to detect and warn if a function type is used as operand of sizeof ? and further, why should it be ill-formed. it is true that in many cases a function designator implicitely converts to a function pointer. In the below example, foo is a function type and not a pointer to function void foo() {} src: n1570, 6.5.3.4#1, 6.3.2.1#9

where is this defined? / does this come from? I strongly believe this is not true. If this would be the case, how would the compiler be able to detect and warn if a function type is used as operand of sizeof ? and further, why should it be ill-formed. it is true that in many cases a function designator implicitely converts to a function pointer. In the below example, foo is a function type and not a pointer to function void foo() {} src: n1570, 6.5.3.4#1, 6.3.2.1#9