Global variables not working in my function - c

My Main file
The file where I store global variables
The 2 files compile well but nothing is printed as I want.

The problem is your function sum(int a, int b) has argument names that shadow the global variables you are trying to access. Also you need to call the sum() function, not just declare it.
Try this instead:
#include <stdio.h>
extern int a;
extern int b;
void sum() {
printf("Sum is : %d", a + b);
}
int main() {
sum();
}

Related

Cannot print called float variables

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);
}

How to share global array between c files

I am trying to share a global array between two files as below:
main.c:
#include <stdio.h>
#include "another.h"
int n=10;
int a[n];
void main(){
printf("Enter value:\n");
scanf("%d",&a[0]);
display();
}
another.h:
#include <stdio.h>
extern int n;
extern int a[n];
void display(){
printf("%d",a[0]);
return;
}
However, I am getting the following error:
variable modified 'a' at file scope
I'm not able to understand why I'm getting this error. Does anyone have an explanation?
The following appears to be an attempt at creating a variable length array. But regardless what it is actually, the 2nd of the following two lines is not legal:
extern int n;
extern int a[n];
n is not yet defined at the time it is used to create a.
In case you were thinking to create a VLA...
by definition VLAs are only created with automatic storage duration on the stack. This makes them unusable for use as extern or for any type of globally scoped variable.
However you can declare the size variable as an extern in a header file so that it is project global for any source file that includes that file, and once it is defined it can be used to size any array within project scope:
<some.h>
extern size_t array_size;
extern int gArray[100];
<some.c>
#include "some.h"
...
size_t array_size = 20;
int gArray[100] = {0};//Not a VLA
...
int main(void)
{
int array[array_size] = {0};
for(int i=0;i<sizeof(gArray);i++
{
//populate each element of gArray here
}
...
<someother.c>
#include "some.h"
int otherFunc(void)
{
//gArray is visible here just as it is in main(), with exactly the same values.
int local array[array_size] = {gArray[0],gArray[1],gArray[2],...,gArray[19]};
...

Is variable declaration necessary in C?

See the comments to see what is being referred as declaration.
If the whole variable declaration part was missing, what would be the problem?
Appears that variable definition and initialization either simultaneously or separately like in the example would suffice.
#include <stdio.h>
// Variable declaration:
extern int a, b;
extern int c;
extern float f;
int main () {
/* variable definition: */
int a, b;
int c;
float f;
/* actual initialization */
a = 10;
b = 20;
c = a + b;
printf("value of c : %d \n", c);
f = 70.0/3.0;
printf("value of f : %f \n", f);
return 0;
}
If the declaration was missing then it would create no problem in main function since the locally defined variables i.e. a,b,c,f will be used in the functionality of main till its scope ends.
The declaration merely tells that the definition lies elsewhere (in some other .c file) or the definition lies after the function main in the same .c file.
There will be no problem here if the mentioned declaration is missing.
// Variable declaration:
extern int a, b;
extern int c;
extern float f;
This tells the compiler that these variables are defined somewhere else(in another file).
/* variable definition: */
int a, b;
int c;
float f;
This is where you define variables but they are not the same as the external variables you declared since they are in the inner scope of the main function.
The scope is the place where variables live. extern keyword notes that the scope is global.
You can define variables with the same name in an inner scope and access only them as you did in the main function but it's not a good practice.
void foo()
{
int a = 5;
printf("%d\n", a); // 5
// Creating an inner scope
{
int a = 20;
printf("%d\n", a); // 20
}
printf("%d\n", a); // 5
}
The correct way to use the extern keyword with variables is like this.
//h1.h
extern int global_var; // Declaration of the variable
//c1.c
#include h1.h
int global_var = 0; // Definition of the global var. Memory is allocated here.
//main.c
#include h1.h
int main()
{
printf("global var value is %d\n", global_var); // use of the var defined and
// initialized in c1.c
return 0;
}
This program will print 0 since the variable is defined and initialized in c1.c.
Extern extends the visibility of the C variables and C functions. so that lets the compiler know that there is another place that those vars are declared and memory was allocated for them elsewhere.
for example in another c file.
if you compile a c file containing a global var for example:
int c = 5;
and you create a function on you c file that uses this c var, for example:
int someFunc(void){
return c;}
if you run someFunc in your main and print its return value, you will get 5. but you must compile both c files together.
in your program, you only use the locally allocated var declared in your main function.
When it comes to simple variables, there is really no difference between the declaration and definition. There is a difference when it comes to structs and functions. Here is an example:
// Declarations
struct myStruct;
int foo();
int main()
{
...
}
// Definitions
struct myStruct {
int a, b;
};
int foo() {
return 42;
}
In your case, you are hiding the previous declarations so that they are not accessible before the end of the scope. This is commonly called shadowing. It's basically the same thing as this:
int main()
{
int i=0;
printf("i: %d\n", i);
{
int i=42; // Now the previous i is inaccessible within this scope
printf("i: %d\n", i);
}
// And now it is accessible again
printf("i: %d\n", i);
}

what is the difference between the two codes?

this is the first code
#include <stdio.h>
void a();
int i;
int main()
{
printf("%d\n",i);
a();
return 0;
}
int i=5;
void a()
{
printf("%d\n",i);
}
output
5
5
the second code is
#include <stdio.h>
void a();
extern int i;
int main()
{
printf("%d\n",i);
a();
return 0;
}
int i=5;
void a()
{
printf("%d\n",i);
}
ouput
5
5
what is the difference between the codes if their outputs are same...
and if they are same then what is the use of extern
extern is something like saying to the Linker that you have defined the variable in some other file( even global variables can be made extern inside other functions in same file) don't throw error now you will find the variable later then link it .
using extern your only declaring the variable not defining it(no memory allocated).
In Your second program for first declartion
extern int i;
No memory is allocated but later when you declare
int i = 5 ;
the memory is allocated for i. while linker searches for i in printf if i is not declared extern it throws a error since its extern linking takes place when it finds the defintion of i.
extern means that you are declaring the variable but not defining it (allocating memory for it). Have a look at this link for a good explanation of extern: http://www.geeksforgeeks.org/understanding-extern-keyword-in-c/

C Undefined reference

I got some problem with the following codes particularly in header.c where i can't access the extern int x variable in header.h... Why? Does extern variable in .h not global? How can i use this on other files?
===header.h===
#ifndef HDR_H
#define HDR_H
extern int x;
void function();
#endif
===header.c===
#include <stdio.h>
#include "header.h"
void function()
{
printf("%d", x); //****undefined reference to x, why?****
}
===sample.c===
int main()
{
int x = 1;
function();
printf("\n%d", x);
return 0;
}
The declaration
extern int x;
tells the compiler that in some source file there will be a global variable named x. However, in the main function you declare a local variable x. Move that declaration outside of main to make it global.
The extern keyword say the variable exists but does not create it. The compiler expects that another module will have a global variable with that name, and the linker will do the right thing to join them up.
You need to change sample.c like this:
/* x is a global exported from sample.c */
int x = 1;
int main()
{
function();
printf("\n%d", x);
return 0;
}
extern declares a variable, but does not define it. It basically tells the compiler there is a definition for x somewhere else. To fix add the following to header.c (or some other .c file but only one .c file):
int x;
Note that in main() the local variable x will hide the global variable x.
Indeed extern int x; means x will be defined in another place/translation unit.
The compiler expects to find a definition of x in the global scope somewherelse.
I would reorganise/modify your code like this and get rid of header.c
===sample.h===
#ifndef SAMPLE_H
#define SAMPLE_H
extern int x;
void function();
#endif
===sample.c===
#include <stdio.h>
#include "sample.h"
int x;
void function()
{
printf("%d", x);
}
int main()
{
x = 1;
function();
printf("\n%d", x);
return 0;
}

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