my question deals with creating variables that are visible throughout the program file. In other words, a file-local variable.
Consider this example
#include <stdio.h>
struct foo
{
char s[] = "HELLO";
int n = 5;
};
struct foo *a;
int main()
{
puts("Dummy outputs!\n");
printf("%s\n",a->s);
printf("%d\n",a->n);
return 0;
}
Now, this code snippet won't run.
Why?
Because the structure pointed to be pointer variable a will not get allocated as the statement never executed.
Now, how do you get it allocated without changing the scope of this variable a?
#include <stdio.h>
struct foo {
char const *s;
int n;
};
/* static for file-local */
static struct foo a = { "HELLO" , 5 };
int main(void) {
printf("%s\n", a.s);
printf("%d\n", a.n);
return 0;
}
Now, how do you get it allocated without changing the scope of this variable a?
I am sure there a lot of ways to solve your problem. Here's my suggestion.
Change the definition of struct foo to contain a fixed number of characters in s.
Create a as an object instead of a pointer. Initialize it with the necessary values.
Make a a static variable so its use is limited to the file only.
Use the object a instead of the pointer a in rest of the file.
#include <stdio.h>
struct foo
{
char s[20];
int n;
};
static struct foo a = {"HELLO", 20};
int main()
{
puts("Dummy outputs!\n");
printf("%s\n",a.s);
printf("%d\n",a.n);
return 0;
}
This:
struct foo
{
char s[] = "HELLO";
int n = 5;
};
Is not valid C code. You first declare the type:
struct foo
{
char s[10];
int n;
};
Then define a variable of that type:
static struct foo a = { "HELLO", 5 };
The static keyword allows this variable to have file local scope.
You can now use it like this:
static struct foo a = { "HELLO", 5 };
void other()
{
puts("Dummy outputs!\n");
printf("%s\n",a.s);
printf("%d\n",a.n);
}
int main()
{
puts("Dummy outputs!\n");
printf("%s\n",a.s);
printf("%d\n",a.n);
other();
return 0;
}
Note that a is accessible from both functions. It will not however be viewable from functions defined in other files because it is declared as static.
As for using a pointer vs the struct directly, you can take the address of this structure at any time you need to use it in that way:
some_function(&a);
well, i need to use a pointer instead of a structure directly
Try this:
#include <stdio.h>
#include<string.h>
#include<stdlib.h>
struct foo{
char s[20];
int n;
};
int main(void){
struct foo *a;
a = malloc(sizeof(struct foo));
puts("Dummy outputs!\n");
strcpy(a->s, "HELLO");
a->n = 5;
printf("%s\n",a->s);
printf("%d\n",a->n);
free(a);
return 0;
}
Output:
Dummy outputs!
HELLO
5
Related
So I am working on a project in C that requires that I pass pointers to a struct into functions. The project is structured as follows:
struct structName {
unsigned short thing2;
char thing1[];
};
void function_1(struct structName *s) {
strcpy(s->thing1, "Hello");
printf("Function 1\n%s\n\n", s->thing1); // prints correctly
}
void function_2(struct structName *s) {
// can read thing2's value correctly
// thing1 comes out as a series of arbitrary characters
// I'm guessing it's an address being cast to a string or something?
printf("Function 2\n%s\n\n", s->thing1); // prints arbitrary characters ('É·/¨')
}
int main() {
struct structName s;
function_1(&s);
printf("Main\n%s\n\n", s.thing1);
function_2(&s);
printf("Main 2\n%s\n\n", s.thing1);
}
This code outputs the following:
Function 1
Hello
Main
Hello
Function 2
É·/¨
Main 2
É·/¨
Obviously, the program has more than just what I've written here; this is just a simplified version; so if there's anything I should check that might be causing this let me know. In all honesty I reckon it's probably just a stupid rookie error I'm making somewhere.
[EDIT: Seems like s.thing1 is being mutated in some way in the call to function_2(), since the odd value is replicated in main() - I should point out that in my program the printf()s are located right before the function call and in the first line of the function, so there's no chance that it's being written to by anything I'm doing. I've updated the example code above to show this.]
Thanks in advance!
The structure contains a flexible member at its end, if you declare a static object with this type, the length of this member will be zero, so strcpy(s->thing1, "Hello"); will have undefined behavior.
You are supposed to allocate instances of this type of structure with enough extra space to handle whatever data you wish to store into the flexible array.
Here is an example:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct pstring {
size_t length;
char data[];
} pstring;
pstring *allocate_pstring(const char *s) {
size_t length = strlen(s);
pstring *p = malloc(sizeof(*p) + length + 1);
if (p != NULL) {
p->length = length;
strcpy(p->data, s);
}
return p;
}
void free_pstring(pstring *p) {
free(p);
}
int main() {
pstring *p = allocate_pstring("Hello");
printf("Main\n%.*s\n\n", (int)p->length, p->data);
free_pstring(p);
return 0;
}
I have learned how to use functions and structs and pointers. I want to combined them all into one. But the code that I write doesn't seem to work. The compiler tells me the test is an undeclared identifier. Here is the code:
#include <stdio.h>
#include <stdlib.h>
struct character
{
int *power;
};
void test (use_power)
int main ()
{
test (use_power)
printf("%d\n",*power);
return 0;
}
void test ()
{
int use_power = 25;
struct character a;
a.power = &use_power;
}
Your code has many mistakes it can't even compile
Multiple missing semicolons.
Implicit declaration of test() here
test (use_power)
with a missing semicolon too.
power is not declared in main().
This line
void test use_power()
does not make sense and is invalid, and also has no semicolon.
The a instance in test() defined at the end is local to test() and as such will be deallocated when test() returns. The use_power int, has exactly the same problem and trying to extract it's address from the function is useless because you can't access it after the function has returned.
I have no idea what you were trying to do, but this might be?
#include <stdio.h>
#include <stdlib.h>
struct character {
int *power;
};
/* Decalre the function here, before calling it
* or perhaps move the definition here
*/
void test(struct character *pointer);
/* ^ please */
int
main(void) /* int main() is not really a valid signature */
{
struct character instance;
test(&instance);
if (instance.power == NULL)
return -1;
printf("%d\n", *instance.power);
free(instance.power);
return 0;
}
void
test(struct character *pointer)
{
pointer->power = malloc(sizeof(*pointer->power));
if (pointer->power != NULL)
*pointer->power = 25;
}
Your code seems to be wrong. Your definition for test contains no arguments as
void test ()
{
int use_power = 25;
struct character a;
a.power = &use_power;
}
but your prototype contains one argument
void test (use_power)
which is wrongly put. First there are no semicolons; at the end of your prototype declaration, secondly by looking at your code, use_power is a variable and not a datatype so it cannot be present solely in a function declaration.
You will get an argument mismatch error.
You have used the line in main()
printf("%d\n",*power);
which is absolutely wrong. you cannot access any member of a structure without a structure variable.
And again, you have not mentioned the; after your call to the incorrect test()before this line
As you have not put your question so properly, I must figure out what you wish to achieve. I bet you want to hold the address of a integer in the pointer member of a structure and then print its value.
Below is a code snippet which will work as you desire.
#include <stdio.h>
#include <stdlib.h>
struct character
{
int *power;
};
struct character a; //define a structure variable
void test ();
int main ()
{
test ();
printf("%d\n",*(a.power)); // print the member of structure variable a
return 0;
}
void test ()
{
int use_power = 25;
a.power = &use_power;
}
example
#include <stdio.h>
struct character {
int *power;
};
void test(struct character *var);
int main (void){
struct character use_power;
int power = 5;
use_power.power = &power;
test(&use_power);
printf("%d\n", power);
return 0;
}
void test(struct character *var){
int use_power = *var->power;
*var->power = use_power * use_power;
}
I've got a struct called members which contains a bunch of char arrays and integers. The struct has been declared in Header.h and defined it by "struct members pt" in source.c, inside main. From here a for-loop is being runned 5 times and adding variables to the character arrays and ints in pt[x].
Now I need to be able to access this from a function called void search(int a); (Should probably not be a void since I want it to return a value. But I'll fix that later)
What void search is supposed to do is basicly
int willReturn[10];
int b = 0;
for(int x = 0; x<a; x++)
{
if(pt[x].hasPayed == 0)
{
willReturn[b] = x;
b++;
}
}
There might be something wrong about that code, but the thing that I need to know is how I can access pt[x].hasPayed.
Any ideas?
I do not want to use any global variables.
Thank you in advance.
Below sample code might help you.
header.h
struct members {
int hasPayed;
};
main.c
#include <stdio.h>
#include <string.h>
#include "main.h"
typedef struct members MEMBERS;
void print_member(MEMBERS *pt) {
int i;
for(i =0 ; i< 10; i++)
{
printf(" %d\n",pt[i].hasPayed);
}
}
void main () {
MEMBERS pt[10];
int i;
for(i =0 ; i< 10; i++)
{
pt[i].hasPayed= i;
}
print_member(pt);
}
Instead of print_member code your search logic.
Pass a pointer to pt as a paraamter to search. e.g void search(int a, struct members *pt).
Another way if you don't want to pass pointers. Place pt in a function as a static variable.
struct members** ____get_pt(){
static struct members pt[ /* size */ ];
/*
or for dynamical size,
static struct members* pt;
*/
return &pt;
}
// Define a macro for convenience, above where you want to use 'pt'.
#define PT (*____get_pt())
Then you can use PT[x].hasPayed everywhere, without global variables.
However, this could do not improve your code ...
I am trying to change my C functions of nested structures to operate on pointers instead of passing and making copies of the structures which are quite large in reality.
here is a simplified version of what I want to do passing the structures around....
struct InnerStruct
{
int int1;
int int2;
};
struct OuterStruct
{
struct innerStruct inner1;
int outerResult;
};
void main (void)
{
struct OuterStruct outer1;
outer1 = get_outer ();
}
struct OuterStruct get_outer (void)
{
struct OuterStruct thisOuter;
thisOuter.inner1 = get_inner (void);
thisOuter.outerResult = get_result (thisOuter.inner1);
return thisOuter;
}
struct InnerStruct get_inner (void)
{
struct InnerStruct thisInnner;
thisInner.int1 = 1;
thisInner.int2 = 2;
return thisInner;
}
int get_result (struct InnerStruct thisInner)
{
int thisResult;
thisResult = thisInner.int1 + thisInner.int2;
return thisResult;
}
but the structure is quite large in reality and this is a frequent operation, so I'd rather pass around the pointers. Just not sure how the syntax works for nested structures like this. Here is my attempt....
struct InnerStruct
{
int int1;
int int2;
};
struct OuterStruct
{
struct innerStruct inner1;
int outerResult;
};
void main (void)
{
struct OuterStruct outer1;
get_outer (&outer1);
}
void get_outer (struct OuterStruct *thisOuter)
{
get_inner (&(thisOuter->inner1));
thisOuter->outerResult = get_result (&(thisOuter->inner1));
}
void get_inner (struct InnerStruct *thisInner)
{
thisInner->int1 = 1;
thisInner->int2 = 2;
}
int get_result (struct OuterStruct *thisInner)
{
int thisResult;
thisResult = thisInner->int1 + thisInner->int2;
return thisResult;
}
You should really look up more about how pointers work. But here is some sample C++ code. Notice the "&" tells your compiler to "not send the struct itself" to the function but a pointer to it. Just a warning never return a reference to a variable (unless you know what you are doing).
#include <iostream>
struct MyStruct
{
int a;
int b;
};
using namespace std;
void printStruct(MyStruct * mypointer) {
cout << "MyStruct.a=" << mypointer->a << endl;
cout << "MyStruct.b=" << mypointer->b << endl;
}
int main()
{
MyStruct s;
s.a = 2;
s.b = 1;
printStruct(&s);
return 0;
}
This will illustrate an easy way to pass pointers to structs. It is a much more efficient way to pass data around, especially when, as you say, the data can get very large. This illustration uses a compound struct, (struct within struct) with an array and pointer declared to pass around. Comments in code explain things.
This will all build and run so you can experiment with it. i.e., follow the data along with execution.
Here is an easy way: (using my own structs)
typedef struct {
int alfha;
int beta;
} FIRST;
typedef struct {
char str1[10];
char str2[10];
FIRST first;
}SECOND; //creates a compound struct (struct within a struct, similar to your example)
SECOND second[5], *pSecond;//create an array of SECOND, and a SECOND *
SECOND * func(SECOND *a); //simple func() defined to illustrate struct pointer arguments and returns
int main(void)
{
pSecond = &second[0]; //initialize pSecond to point to first position of second[] (having fun now)
SECOND s[10], *pS; //local copy of SECOND to receive results from func
pS = &s[0];//just like above;
//At this point, you can pass pSecond as a pointer to struct (SECOND *)
strcpy(pSecond[0].str2, "hello");
pS = func(pSecond);
// printf("...", pS[0]...);//pseudo code - print contents of pS, which now contains any work done in func
return 0;
}
SECOND * func(SECOND *a) //inputs and outputs SECOND * (for illustration, i.e., the argument contains all
{ //information itself, not really necessary to return it also)
strcpy(a[0].str1, "a string");
return a;
}
Although there is not much going on in func(), when the pointer returns to main(), it contains both the value copied in main, and the value copied in fucn(), as shown here:
Results: (in code)
Contents in pSecond:
If I have a struct in C that has an integer and an array, how do I initialize the integer to 0 and the first element of the array to 0, if the struct is a member another struct so that for every instance of the other struct the integer and the array has those initialized values?
Initialisers can be nested for nested structs, e.g.
typedef struct {
int j;
} Foo;
typedef struct {
int i;
Foo f;
} Bar;
Bar b = { 0, { 0 } };
I hope this sample program helps....
#include <stdio.h>
typedef struct
{
int a;
int b[10];
}xx;
typedef struct
{
xx x1;
char b;
}yy;
int main()
{
yy zz = {{0, {1,2,3}}, 'A'};
printf("\n %d %d %d %c\n", zz.x1.a, zz.x1.b[0], zz.x1.b[1], zz.b);
return 0;
}
yy zz = {{0, {0}}, 'A'}; will initialize all the elements of array b[10] will be set to 0.
Like #unwind suggestion, In C all instances created should initialized manually. No constructor kind of mechanism here.
You can 0-initialize the whole struct with {0}.
For example:
typedef struct {
char myStr[5];
} Foo;
typedef struct {
Foo f;
} Bar;
Bar b = {0}; // this line initializes all members of b to 0, including all characters in myStr.
C doesn't have constructors, so unless you are using an initializer expression in every case, i.e. write something like
my_big_struct = { { 0, 0 } };
to initialize the inner structure, you're going to have to add a function and make sure it's called in all cases where the structure is "instantiated":
my_big_struct a;
init_inner_struct(&a.inner_struct);
Here is an alternative example how you would do things like this with object-oriented design. Please note that this example uses runtime initialization.
mystruct.h
#ifndef MYSTRUCT_H
#define MYSTRUCT_H
typedef struct mystruct_t mystruct_t; // "opaque" type
const mystruct_t* mystruct_construct (void);
void mystruct_print (const mystruct_t* my);
void mystruct_destruct (const mystruct_t* my);
#endif
mystruct.c
#include "mystruct.h"
#include <stdlib.h>
#include <stdio.h>
struct mystruct_t // implementation of opaque type
{
int x; // private variable
int y; // private variable
};
const mystruct_t* mystruct_construct (void)
{
mystruct_t* my = malloc(sizeof(mystruct_t));
if(my == NULL)
{
; // error handling needs to be implemented
}
my->x = 1;
my->y = 2;
return my;
}
void mystruct_print (const mystruct_t* my)
{
printf("%d %d\n", my->x, my->y);
}
void mystruct_destruct (const mystruct_t* my)
{
free( (void*)my );
}
main.c
int main (void)
{
const mystruct_t* x = mystruct_construct();
mystruct_print(x);
mystruct_destruct(x);
return 0;
}
You don't necessarily need to use malloc, you can use a private, statically allocated memory pool as well.