I am writing a code in C. It is in an environment with precoded functions, so I can't explain it completely. I wanted to iterate over a variable but this wasn't working. I eventually figured out this was because the variable was not defined globally but in a function, and was being redefined every time the function was called.
Now, at the top of my code, globally, I want to write the following code.
int killing_time = 20000;
int killing_period;
killing_period = killing_time;
The compiler gives me the following errors:
data definition has no type or storage class
Don't I clearly define it to be an integer?
initializer element is not constant
If I define killing_time as const int killing_time = 20000 it still gives the same error:
type defaults to ‘int’ in declaration of ‘killing_period'
I could of course define killing_period to be 20000 and just start iterating over that, but I want to know what is going on.
I hope we can figure this out together.
killing_period = killing_time; is not a valid statement in the global scope.
You can use assignment on declaration, but assigned variable (initializer element) must be constant:
const int killing_time = 20000;
int killing_period = killing_time;
Anyway, you shouldn't do it like that.
There's more than one way to do this, one of them is to pass the address of your iterator to the function where you use it, that way the changes made in the scope of the function are permanent:
void f1(int* i){
(*i)++; //will increment i
}
void f2(int* i){
(*i)++; //will increment i
}
int main ()
{
int i = 0;
f1(&i); //pass the address of i
f2(&i); //pass the address of i
printf("%d", i); // i = 2
return 0;
}
killing_period = killing_time; is executable code and you simply can't have executable code outside functions.
Nor can you do this:
int killing_time = 20000;
int killing_period = killing_time; // wrong
Because these are file scope variables and therefore have static storage duration. And variables with static storage duration can only be initialized by what C calls a constant expression. 20000 above is an integer constant expression, but killing_time isn't.
The simple fix is to do something like this instead:
#define KILLING_TIME 20000
...
int killing_time = KILLING_TIME;
int killing_period = KILLING_TIME;
killing_period = killing_time; is your problem. It is an executable statement and not allowed in global space... it must be within a function.
Still not sure what you are after, but would the following help?
int killing_time = 20000;
int killing_period = -1; // negative if not yet set
int your_function(void)
{
if ( killing_period < 0 ) {
/* one-time initialization on the first call */
killing_period = killing_time;
}
return killing_period;
}
If you need multiple functions setting killing_period it may be be best to burry that in a separate function dedicated to the test and set, or perhaps a macro.
Related
I work with the Aurix MUC, I tried to read the contents of the memory after the execution of a program, to see what he wrote in the memoir
I noticed that when I use a global variable in a function, the new value of this global variable after processing in the function, is not written in memory.
Here is an example:
int a = 100;
void plus (int a)
{
a = a + 17;
}
int main (void)
{
plus(a);
return 0;
}
when I display the contents of the memory I find the value 100 of a
and I do not find the new value of a which is normally 117.
I tried to declare the variable a as volatile, but it does not change anything
on the other hand if I do the calculation directly in the main like this
int a = 100;
int main (void)
{
a = a + 17
return 0;
}
like that I find the value 117 in the memory.
so I need to understand where are there save variables values used in the call functions?
and why the new variable of a is not written in memory,
and why the variables declared in local are not also written in the memory?
In this code:
int a = 100;
void plus(int a)
{
a = a + 17;
}
The int a in void plus(int a) declares a new instance of a that hides the previous a. Then, in a = a + 17;, this new instance of a is used. To have plus change the a declared outside the function, use:
int a = 100;
void plus(void)
{
a = a + 17;
}
Also, in main, change plus(a); to plus();.
In C, each identifier has a scope, where is where in the source code is visible. A declaration outside of any function has file scope, meaning the identifiers declared are visible throughout the file (technically a translation unit). In a function definition, a declaration of a parameter has block scope associated with the function—it is visible only within the function.
Because the int a = 100; has file scope, you do not need to redeclare it in plus in order to use it—it is visible inside the function, so you can just use it without a new declaration.
I have recently run into some trouble while trying to perform the following logic:
static const int size = getSize();
int getSize() {
return 50;
}
The error I have received is initialiser element is not constant
Having read online I understand that this issue is because the compiler evaluates the static const expression at compilation and therefore cannot know what the value is supposed to be.
My question is how do I get around this?
If I have a library that contains many functions but they all require this logic how are they supposed to use it without having to calculate it each time?
And even if they have to, what if the logic itself can change throughout runtime but I only ever want the first value I receive from the function?
Perhaps I should clarify that the logic in getSize is just an example, it could also contain logic that retrieves the file size from a specific file.
Unlike in C++ you cannot initialize global variables with the result of a function in C, but only with real constants known at compile time.
You need to write:
static const int size = 50;
If the constant must be computed by a function you can do this:
Dont declare static const int size = ... anymore, but write this:
int getSize()
{
static int initialized;
static int size;
if (!initialized)
{
size = SomeComplexFunctionOfYours();
initialized = 1;
}
return size;
}
int main(void)
{
...
int somevar = getSize();
...
That way SomeComplexFunctionOfYours() will be called only once upon the first invocation of getSize(). There is a small price to be paid: each time you invoke getSize(), a test needs to be performed.
Or you can initialize it explicitely like this, but then size cannot be const anymore:
static int size;
void InitializeConstants()
{
size = SomeComplexFunctionOfYours();
}
int main(void)
{
InitializeConstants();
...
int somevar = size;
...
The compiler needs to know the value of your constant variable at the compilation time, because its a constant.
Also you can't initialize a variable with a function.
You should do something like this :
#define SIZE 50
static const int size = SIZE;
If I convert const int to int inside a void it works.
But when I create an extern const int it doesn't.
void ReadFromEpprom()
{
int Start = 343;
const int End=Start; //This works
}
Example 2
Header File
extern const int End;
Source file
const int End;
void ReadFromEpprom()
{
int Start = 343;
End=Start; //This doesn't work
}
In second situation I get error:
(364) attempt to modify object qualified const
How can I solve this?
Should I make it with another way?
When you declare a constant variable this means the variable will not change.So it makes sense that constant variables must be initialized immediately.You are doing this in your first example which is correct.In your second example you have a constant variable and then you are trying to modify it's value.This is incorrect since the variable is already constant.
The extern here is a red herring.
If you use const int End then you need to initialise End at the point of this declaration. That's because it's const.
So const int End = Start; works fine, but const int End; is not syntactically viable.
In C you can't arrange things so you have a const at global scope and a function that sets the value at run-time. But what you could do is embed the value in a function (as as static), and call that function to initialise and subsequently retrieve the value.
Initialiazation versus assignment:
A const object can be initialized (given a value at the declaration), but not assigned (given a value later).
This is initialization, and "works". It is initialization the local variable End that exist inside ReadFromEpprom().
void ReadFromEpprom()
{
...
const int End=Start; //This works
End=Start; attempts to assigned End that exist at file scope, outside of ReadFromEpprom(). const objects cannot be assigned.
const int End;
void ReadFromEpprom()
{
...
End=Start; //This doesn't work
}
How can I solve this?
Let external code to read localEnd via a function ReadEnd(), yet allow local code to write localEnd.
static int localEnd;
int ReadEnd() {
return localEnd;
}
void ReadFromEpprom() {
int Start = 343;
localEnd = Start;
}
I agree with all the other answers.
It seems that you want to initialize a const value with a value that will be determined at run time, which is impossible.
What you can do is two workarounds.
Remove const. Add a comment near its definition that the variable is set only once in the very beginning.
Use a function instead of a const int variable; implement the "run code only once" idea inside this function.
Header file:
int CalcEnd();
Source file:
int CalcEnd()
{
static int init_done = 0;
static int result;
if (!init_done)
{
result = 343; // or any complex calculation you need to do
init_done = 1;
}
return result;
}
Note how the function uses static variables and logic to do initialization only once.
If you use the function idea, and don't want to remember typing the parentheses in the function call (CalcEnd()), you can define a macro:
#define End MyCalcEnd()
This will make it appear as if End were const int variable, when actually it's a function call. This usage of macros is controversial, use it only if you are sure it will not lead to confusion later.
I know that If a function has no argument & only return type (say int), then I can change my int variable by assigning the function to my variable as below,
main()
{
int var_name;
var_name = func();
printf("My variable value is updated as : %d", a);
}
func()
{ return 100; }
Also I know that If I have my function's return type as void, with no arguments, then I can only print the value inside the function itself and cannot return anything in turn.
But, my doubt is, is there anything else that I can do to update my var_name by calling a function with no arguments & no return type ?
ie., void func(void); by using something like pointer concepts ??
I could not able to find the exact answer for the same by my searches among so many websites.. I will be very grateful if someone can help me out finding whether I can do it by this way or not,.
Thanks,.
It is possible to modify a local variable in main, from a function with no arguments and no return value, if there's a global pointer to it:
#include <stdio.h>
int *p;
void func() {
*p = 6;
}
int main() {
int a = 5;
p = &a;
func();
printf("a = %d\n", a); // prints: a = 6
return 0;
}
There's no good way to do that. If you want the function to modify a local variable, you should probably change the function so it either returns a value that you can assign to the variable, or takes the variable's address as an argument.
But if you don't mind writing some ugly code, you can define a global (file-scope) pointer variable, assign the local variable's address to the global pointer, and then use that to modify the variable inside the function.
An example:
#include <stdio.h>
int *global_pointer;
void func(void) {
*global_pointer = 42;
}
int main(void) {
int local_variable = 0;
global_pointer = &local_variable;
func();
printf("local_variable = %d\n", local_variable);
}
It's very easy to shoot yourself in the foot his way. For example, if you refer to the pointer after the calling function has terminated (and the local variable no longer exists), you'll have undefined behavior.
This technique can actually be useful if you need to make a quick temporary change in a body of code in which you can't make major interface changes. Just don't do it in code that will be maintained by anyone else -- and wash your hands afterward.
You can have global variable
int var_name;
void func();
int main()
{
func();
printf("%d\n",var_name);
}
void func()
{
var_name = 20;
}
But if your variable is local to main() then this can't be done.
There are two ways to modify the value of var_name.
Make changes in the calling function and return the value.( which you have already shown)
Pass the address of the var_name to the function and have pointer as arguement in the func(int *p) and modify the value inside the func()
Thats it!! No other way this can be done.
I have a variable that I want to initialize at runtime before the rest of the program runs. After initialization, I don't want the value of the variable to change. Is there any C language construct to go about doing this ?
Let my main C program is contained in a file Prog.c
//Contents of Prog.c
//...includes.. and variable initialization
int main(..)
{
//Initialize variables here
//.. Huge program after this where I don't want these constants to change
}
You can do it indirectly through const pointers, at least:
typedef struct {
int answer;
} state;
const state * state_init(void)
{
static state st;
st.answer = 42; /* Pretend this has to be done at run-time. */
return &st;
}
int main(void)
{
const state *st = state_init();
printf("the answer is %d, and it's constant\n"," st->answer);
}
This way, all main() has is a const pointer to some state which it cannot modify.
A constant global should work, yes?
const int val = 3; // Set before main starts
// const, so it will never change.
int main(void)
{
printf("%d\n", val); // using val in code
}
However, if the value isn't known at compile-time, you can set it at run-time this way:
const int const* g_pVal;
int main(void)
{
static const int val = initialize_value();
g_pVal = &val;
printf("%d\n", *g_pVal);
}
The only way I can think of is if you read your value as a non constant and then call a function which take a constant value and pass your variable to it. Inside the function you make all the wished operations.