#include <stdio.h>
int main()
{
void foo(), f();
f();
}
void foo()
{
printf("2 ");
}
void f()
{
printf("1 ");
foo();
}
Output:
1 2
How the declaration is working here?
And if I define F() before foo() I am getting error.
Wrong!
#include <stdio.h>
int main()
{
void foo(), f();
f();
}
void f()
{
printf("1 ");
foo();
}
void foo()
{
printf("2 ");
}
ERROR
> main.c: In function 'f': main.c:21:13: error: incompatible implicit
> declaration of function 'foo'
> foo();
> ^ main.c:7:18: note: previous implicit declaration of 'foo' was here
> void foo(), f();
^
Why is this happening?
This is a matter of scope. In the first example, foo and f are known to main because you declared them. f() knows foo because it's declared before it.
In your second example, the declaration of f and foo being local to main, f() doesn't know foo because it wasn't declared before it.
The error message my compiler gives is "implicit declaration".
You declared the functions f() and foo() as void, in the scope of main. You then use them outside of that scope (namely, you call foo() from inside f() - and f is declared outside of main).
The compiler treats this encouter with foo(); in the second line of f() as "the first time I heard of this function" - since it is no longer in the scope of main, it has forgotten everything it was told while it was in the main scope (including "the return type of foo() and f() will be void). Absent any information, it will assume that foo() returns an int. When it finally comes across the definition of foo, lower down in the code, it realizes that it was wrong about its assumption. But rather than quietly fixing it, it complains.
That's C for you.
If you put the declaration before main(), your problem goes away:
#include <stdio.h>
void foo(), f();
int main(void)
{
f();
}
void f()
{
printf("1 ");
foo();
}
void foo()
{
printf("2 ");
}
Thus - the problem is not the "multiple declaration" of your function (as you implied in the title of your question) - it is the scope of the declaration that is causing trouble.
It's called "forward declaration". Check google for the details, but in your case:
void f();
void foo();
void f() { foo(); }
void foo() {}
int main { f(); return 0; }
You can redeclare something as many times as you want as long as they have the exact same declaration. For example the following is valid:
void foo();
void foo();
void foo();
However, void foo(); and foo(); aren't the same. The latter implicitly defaults to int foo();. This is why you need to define foo(); before you call it or it will get treated as a redeclaration/new declaration.
Related
I cannot print float variables when calling my functions. int variables print but floats won't print their proper value when passed to my function.
I tried to change the float to int and that worked
int main() {
int foo = 6;
call(foo);
}
void call(int bar) {
printf("%d", bar);
}
This worked and it does indeed print 6.
But, doing the same but with floats only prints out 0.00000:
int main() {
float foo = 6;
call(foo);
}
void call(float bar) {
printf("%f", bar);
}
How do I correctly call and then print float variables?
you need a forward declaration of call
void call(float integerrrr);
int main(){
float integerrrr=6;
call(integerrrr);
}
void call(float integerrrr){
printf("%f", integerrrr);
}
your compiler probably warned you about this
You could simply define call above main instead of below it. The compiler must have seen the declaration of functions when they are used, so a forward declaration like pm100 suggests is one way. Moving the whole definition above main is the another (that does not require a forward declaration).
#include <stdio.h>
void call(float integerrrr){
printf("%f", integerrrr);
}
int main(void){
float integerrrr = 6;
call(integerrrr); // now the compiler knows about this function
}
INT type variables i can print but float just does not
If your program actually compiles as-is it will use an old (obsolete) rule that makes an implicit declaration of undeclared functions when they are used. The implicit declaration would be int call(); - which does not match your actual function. The program (even the one that seems to be working) therefore had undefined behavior.
the compiler of c work from top to bottom so line by line,
so u will have to call the first function void call() then your main function:
void call(float integerrrr){
printf("%f", integerrrr);
}
int main(){
float integerrrr=6;
call(integerrrr);
}
This question already has answers here:
Error "initializer element is not constant" when trying to initialize variable with const
(8 answers)
Closed 2 years ago.
Hi i have a test code for calling malloc as below:
#include <stdio.h>
#include <stdlib.h>
int *p;// = (int*)malloc(sizeof(int));
int main() {
//...
}
Of course this code will be fail when compile with the error: initializer element is not constant and i have referenced this question: Malloc function (dynamic memory allocation) resulting in an error when it is used globally. They said that we have to use malloc() in side a function. But if i change my code to:
#include <stdio.h>
#include <stdlib.h>
int *p;
static int inline test_inline(int *x) {
printf("in inline function \n");
x = (int*)malloc(sizeof(int));
return x;
}
test_inline(p);
int main(){
//...
}
As the definition of inline function: "Inline Function are those function whose definitions are small and be substituted at the place where its function call is happened. Function substitution is totally compiler choice." So this mean we can substitute the inline function test_inline in above example with the code inside it and it means we have call malloc() in global ? Question 1: is this wrong about inline or malloc() ?
Question 2: In the link i give about malloc function dynamic there is an answer said that "Not only malloc, u can't call any function as you have called here. you can only declare function as global or local there" but i see that we still can call function in global and in global we can initialization not only declaration as below:
#include <stdio.h>
#include <stdlib.h>
int b;
b = 1;
int test() {
printf("hello");
}
test();
int main() {
//...
}
So this mean in the global we still can declaration and initialization and call function. But when we compile the above code it has a warning that warning: data definition has no type or storage class So why we have this warning with variable b ? I do not see any thing which inconsequential here. And with the line test(); i have call a function outside main(), i know this make no sense because we never run test() but i have no problem, stil build success. So back to question 1 about the malloc(), i think with the answer that "we can not call a function in global or can not initialize", i think it is not true. Is there any explain more reasonable?
Please refer to the comments.
#include <stdio.h>
#include <stdlib.h>
int b;
b = 1; //this is only allowed, because the previous line is a tentative definition. [1]
int test() {
printf("hello");
}
test(); // this is taken as a function declaration, not a function call [2]
int main() {
//...
}
Case [1]:
Change you code to
int b = 5; // not a tentative defintion.
b = 1; // this assignment is not valid in file scope.
you'll see an error.
Case [2]:
If the signature of the function differs, you'll again see an error. Example: try the below:
float test( int x ) {
printf("hello");
return 0.5;
} //return changed to float, accepts an int as paramater.
test(); //defaults to int and no parameter - conflict!!
this will produce the error for conflicting types.
So, bottom line, no assignment, function call - all in all, no code that needs to execute at runtime, can be put into file scope. The reason behind that being, unless it's contained in a function that's called from main(), there's no way to know when / how to execute it.
You're not calling functions "globally".
Taking your example:
#include <stdio.h>
#include <stdlib.h>
int b;
b = 1;
int test() {
printf("hello");
}
test();
int main() {
//...
}
In C types default to int.
So the lines
int b;
b = 1;
are basically
int b;
int b = 1;
and the lines
int test() {
printf("hello");
}
test();
are just
int test() {
printf("hello");
}
int test(); // -> this is just a matching declaration
Have a look at:
https://godbolt.org/z/3UMQAr
(try changing int test() { ... to char test() { ... and you get a compiler error telling you that those types don't match)
That said, you can't call functions there. Functions are called at runtime by your program (especially malloc, which is asking your OS to allocate memory for you). I'm not a C expert here but as far as I know C doesn't have constexpr functions, which would be the only "exception".
See: Compile-Time Function Execution
Question 1: is this wrong about inline or malloc()
kind of: malloc does have to be called in a function, but the variable it works on can be declared global. i.e. int *pointer = NULL;//global scope
then pointer = malloc(someByteCount);//called within function. Now, pointer is still global, but also has a memory address pointing to someByteCount bytes of memory.
Question 2: In C, all functions are defined on the same level of a .c file, just like main(void){...return 0}, but all functions (except main(void)) must be called within the {...} of other functions, so in short, functions cannot be called from global space.
Illustration for Q2:
//prototypes
void func1(void);
void func2(void);
void func3(void);
int main(){
int val = test_inline(p);//...
}
int main(void)
{
//legal
func1();
func2();
func3();
return 0;
}
//not legal
func1();
func2();
func3();
//definitions
void func1(void)
{
return 0;
}
void func2(void)
{
return 0;
}
void func3(void)
{
return 0;
}
Errors in syntax of your example (see comments):
int *p = NULL;//initialize before use
static int inline test_inline(int *x) {
printf("in inline function \n");
x = (int*)malloc(sizeof(int));
printf("%p\n", x);
return 0;
//return x;//function returns int, not int *
}
//... test_inline(p);//must be called in a function
int main(void){
int val = test_inline(p);//function declaration returns int, not pointer
return 0;
}
This code compiles, and runs, but as noted in comments, usefulness may be lacking.
Question 1: is this wrong about inline or malloc() ?
Neither. Your understanding of inline is incorrect. The function call may be replaced with an inline expansion of the function definition. First, let's fix the function definition because the return type int doesn't match the type of what you're actually returning:
static inline int *test_inline( int *x )
{
printf( "in inline function\n" );
x = malloc( sizeof *x );
return x; // x has type int *, so the return type of the function needs to be int *
}
If you call this function like so:
int main( void )
{
int *foo = test_inline( foo );
...
}
what the compiler may do is substitute the function call with the assembly language equivalent of the following:
int main( void )
{
int *foo;
do
{
printf( "in inline function\n" );
int *x = malloc( sizeof *x );
foo = x;
} while( 0 );
...
}
Nothing's happening "globally" here. The substitution is at the point of execution (within the body of the main function), not at the point of definition.
Question 2: In the link i give about malloc function dynamic there is an answer said that "Not only malloc, u can't call any function as you have called here. you can only declare function as global or local there" but i see that we still can call function in global and in global we can initialization not only declaration as below:
In the code
int test() {
printf("hello");
}
test();
the line test(); is not a function call - it's a (redundant and unnecessary) declaration. It does not execute the function.
Here are some excerpts from the language definition to clarify some of this:
6.2.4 Storage durations of objects
...
3 An object whose identifier is declared without the storage-class specifier
_Thread_local, and either with external or internal linkage or with the storage-class
specifier static, has static storage duration. Its lifetime is the entire execution of the
program and its stored value is initialized only once, prior to program startup.
Bold added. Any variable declared outside the body of a function (such as p in your first code snippet) has static storage duration. Since such objects are initialized before runtime, they cannot be initialized with a runtime value (such as the result of a function call).
6.7.4 Function specifiers
...
6 A function declared with an inline function specifier is an inline function. Making a
function an inline function suggests that calls to the function be as fast as possible.138)
The extent to which such suggestions are effective is implementation-defined.139)
138) By using, for example, an alternative to the usual function call mechanism, such as ‘‘inline
substitution’’. Inline substitution is not textual substitution, nor does it create a new function.
Therefore, for example, the expansion of a macro used within the body of the function uses the
definition it had at the point the function body appears, and not where the function is called; and
identifiers refer to the declarations in scope where the body occurs. Likewise, the function has a
single address, regardless of the number of inline definitions that occur in addition to the external
definition.
139) For example, an implementation might never perform inline substitution, or might only perform inline
substitutions to calls in the scope of an inline declaration
All this means is that the inlined code behaves like it was still a single function definition, even if it's expanded in multiple places throughout the program.
I am calling one function in which values are printed, but when void is added before the call, the function is not giving correct (or any) output.
I tried various methods
#include <stdio.h>
void func1();
void func2();
void func1()
{
printf("Inside func1()\n");
}
void func2()
{
printf("Inside func2()\n");
}
int main()
{
void func1();
void func2();
printf("Inside main()\n");
return 0;
}
Output is :-
Inside main
When void is removed before calling func1 and func2, the output is changed.
#include <stdio.h>
void func1();
void func2();
void func1()
{
printf("Inside func1()\n");
}
void func2()
{
printf("Inside func2()\n");
}
int main()
{
func1();
func2();
printf("Inside main()\n");
return 0;
}
Output is :-
Inside func1
Inside func2
Inside main
Can anyone explain how void is affecting desired output?
The statement void func1(); within main is a function declaration. This is a no-op at runtime.
func1(); actually calls the function.
It's how the language grammar works, that's all. It's rather clever if you think about it.
In this case, void func1(); is not a call to the function, it just means: somewhere there may be a function with this signature, I want to use it even if it is not declared forward. Usually we call this a function declaration and we don't use it inside main but at the beginning of the file.
A function declaration is the action to tells the compiler about a function's name, return type, and parameters.
eg:-void func1();
A function definition provides the actual body of the function.
eg:-
void func1()
{
printf("Inside func1()\n");
}
To call a function, you simply need to pass the required parameters along with the function name, and if the function returns a value, then you can store the returned value.
eg:-func1();
A declaration always begins with a type name. When you write
int func();
you are declaring to the compiler that you have a function that takes an unspecified number of parameters and returns int (so the compiler knows about it).
When you write
func();
you are telling the compiler to execute function func(); and discard its result, in a normal execution statement.
Always you have a C statement beginning with a type identifier
, that's not to be executed, but to inform the compiler of the existence of such thing (to declare it)
In your example, you changed two declarations at the beginning of main() into to execution statements. As such, the compiler generated code to execute those, resulting in the output you got on execution.
#include <stdio.h>
void m();
void n() {
m();
}
void main() {
void m() {
printf("hi");
}
}
On compiling, an error
"undefined reference to m"
is shown. Which m is being referred to?
First, let me declare clearly,
Nested functions are not standard C. They are supported as GCC extension.
OK, now, in your code, m() is a nested function inside main(). It is having block scope for main() only. Outside main() other functions cannot see the existence of m() ,neither can call m() directly. m() can be called only inside main().
In your case, the call to m() inside n() is causing the issue. Even if you provided the forward declaration as void m();, linker won't be able to find the definition of m() and throw error.
Solution: Move the definition of m() outside main(), then you can use it from any other function.
Also note, the recommended signature of main() is int main(void).
As has been explained elsewhere, C doesn't support nested functions as a rule (gcc does as an extension, but almost no other compiler that I know of does).
You need to move the definition of m outside of main. Preferably you should define m before it is used by n:
#include <stdio.h>
void m()
{
printf("hi\n");
}
void n()
{
m();
}
int main( void ) // void main() is not a valid signature for main
{
n(); // call n, which calls m, which prints "hi"
return 0;
}
In script languages, such as Perl and Python, I can change function in run-time. I can do something in C by changing the pointer to a function?
Something like:
void fun1() {
printf("fun1\n");
}
void fun2() {
printf("fun2\n");
}
int main() {
fun1 = &fun2;
fun1(); // print "fun2"
return 0;
}
No. You can't do that.
You can regard fun1 as a placeholder for the fixed entry point of that function.
The semantic you are looking for is that from fun1=&fun2; point on every call to fun1 causes fun2 to be called.
fun1 is a value not a variable. In the same way in the statement int x=1; x is a variable and 1 is a value.
Your code makes no more sense than thinking 1=2; will compile and from that point on x=x+1; will result in x being incremented by 2.
Just because fun1 is an identifier doesn't mean it's a variable let alone assignable.
Yes; You can and this is the simple program which will help you in understanding this
#include <stdio.h>
void fun1() {
printf("fun1\n");
}
void fun2() {
printf("fun2\n");
}
int main() {
void (*fun)() = &fun1;
fun(); // print "fun1"
fun = &fun2;
fun(); // print "fun2"
return 0;
}
Output
➤ ./a.exe
fun1
fun2
You can't change fun1, but you can declare a function pointer (not a function, only a pointer to a function).
void (*fun)();
fun = fun1;
fun(); /* calls fun1 */
fun = fun2;
fun(); /* calls fun2 */
As you might have noticed it is not necessary to take the address of fun1/fun2 explicitely, you can omit the address-of operator '&'.