error in the simplest C code

matiasmorant

the following code

#include<stdio.h>
#include<stdlib.h>

int
main (int argc, char *argv[])
{
int* a;
*a=1;
return 0;
}

compiles ok, but throws "Segmentation fault (core dumped)" at runtime
why is that????

i'm working with gcc in ubuntu. I have been able to compile and run many c programs without any difficulties

chiro

Hey matiasmorant and welcome to the forums.

The simple reason is that a is a pointer that has not been initialized and you are trying to set the contents of the memory of that location to something.

Because of this, you are trying to access memory in the wrong way and it crashes.

Before you work with pointers, you must realize that what you reference must be declared either on the stack or on the heap. Heap elements are created with malloc and other similar statements and are freed using free in C (in C++ it's new and delete).

Things declared on the stack are things like say 'int x = 0;' or 'char r[128];'

If you want to set say x in the above example you can create a pointer that points to this by saying int* xpointer = &x; *xpointer = 3; which will set the value of x to 3. If you want to use the heap then do int* xpointer = malloc(4); (Check to make sure xpointer is not 0 before doing the next part!). *xpointer = 3. This will set a heap variable. (Make sure you call free(xpointer); when you are finished).

Just out of curiosity, how long have you been learning? Is this your first time?

matiasmorant

thanks a lot! it's great to receive such a quick response

I started learning programming about 10 months ago, I already finished the programming course, and I did plenty of programs with pointers, pointers to pointers, pointers to functions, etc. Then I started working with java and forgot some of the c language. Now, I was trying to learn openGL with java, but couldn't make jogl work, so i came back to c, where it works perfect up to now. That's my programming biography ;)

now, what about the following code? (i hope this will be the last question)

#include<stdio.h>
#include<stdlib.h>

void SWAP(double *f1, double *f2){
double *temp;
temp=f1;
f1=f2;
f2=temp;
printf("inside %f %f\n",*f1,*f2);
}

int
main (int argc, char *argv[])
{
double *f1, *f2;

f1=(double*)malloc(sizeof(double));
f2=(double*)malloc(sizeof(double));
*f1=1;
*f2=2;
printf("before %f %f\n",*f1,*f2);
SWAP(f1,f2);
printf("after %f %f\n",*f1,*f2);

return 0;
}

i expected the answer to be

before 1.000000 2.000000
inside 2.000000 1.000000
after 2.000000 1.000000

but it was

before 1.000000 2.000000
inside 2.000000 1.000000
after 1.000000 2.000000

why doesn't the swap function..... swap?

(i need to swap the references and not the contents, since i will be working with matrices)

thanks in advance!

chiro

The reason for this is because you are passing the pointers by value and not by reference.

Passing by value means that you give the values to the function more or less a kind of copy, whereas passing the values by reference passes the actual pointer to the object and allows you to modify either the actual corresponding pointer or object.

What you are actually doing is passing a copy of the pointer and just setting stuff inside the routine that has no effect on anything outside the scope of the function. This is why when you print the two variables, you get the swap inside the function but not outside.

What you need to do is pass a pointer to the pointer of the function and modify the pointer itself to change by reference *pointer where you pass double** pointertopointer. This will modify the actual pointer itself even outside the function and everything will act as expected.

rcgldr

You would need the code to look like this:

matiasmorant

thanks friend! that was the last question