pass by reference in C

xeon123 · 10-22-2007, 05:30 PM

Why, in this example, I can't print the ptr value?

What am I doing wrong?

Code:

void call1(char* ptr);
void call2(char* ptr);

void teste()
{
  char* ptr;
  call1(ptr);
  printf("%s\n", ptr);

  call2(ptr);
  printf("%s\n", ptr);
}

void call1(char* ptr)
{
  ptr = "OLA";
}

void call2(char* ptr)
{
  ptr = (char*) malloc(strlen("HELLO"));
  strcpy(ptr,"HELLO");
}

int main()
{
  teste();
}

Thanks,
Pedro

xeon123 · 10-22-2007, 05:39 PM

For me, the call2 method sould work, because I'm allocating memory, and put a value inside the memory. Even, if I do a malloc inside a function.

2 - Second questio. If I do a malloc inside a functin, when I return it, the value is deallocated?

Thanks,

xeon123 · 10-22-2007, 05:46 PM

I think that the answer for my question 2 is yes. Here is the code:

Code:

  ptr = (char*) malloc(strlen("HELLO"));
  call3(ptr);
  printf("3-%s\n", ptr);


void call3(char* ptr)
{
  strcpy(ptr,"HELLO");
}

So, I do have a third question:

If when I call a function that would put a value in the parameter char*, and I don't know what value it will be. How should I do it?

For example:

Code:

void attrib(char* ptr)
{
// put a random value in ptr that I don't know what size will have
}


main()
{
char* ptr;
attrib(ptr);//How should I call this method? Like this?
return;
}

xeon123 · 10-22-2007, 05:49 PM

Fourth question:

I the function accepts a char* var[] and a char** var2 parameters, how should I do it?

dmail · 10-22-2007, 07:32 PM

Quote:

Why, in this example, I can't print the ptr value?

Ok well call1 is a really bad function and should cause a segmentation fault.
call2: you create a pointer in teste and pass to call2, the value of the pointer could be anything. Inside the function you then allocate memory and copy a string into it, yet when the function returns it will be just junk in the pointer. What is needed is to pass a pointer to pointer like so:

Code:

void teste()
{
  char* ptr;
  call2(&ptr);
  printf("%s\n", ptr);
}


void call2(char** ptr)
{
  *ptr =  malloc(strlen("HELLO"));
  strcpy(ptr,"HELLO");
}

Quote:

2 - Second questio. If I do a malloc inside a functin, when I return it, the value is deallocated?

No see above, but you do have a memory leak.

Quote:

If when I call a function that would put a value in the parameter char*, and I don't know what value it will be. How should I do it?

I don't think I fully understand this question, you have already shown you know how to get a string length.

Quote:

I the function accepts a char* var[] and a char** var2 parameters, how should I do it?

You will have seen this before many times in a main function.
int main ( int argc, char* argv[] )
or
int main ( int argc, char** argv )