Thread: char ** and argv memory question

Originally Posted by simo_mon

hi there i am super confused, i am writing a program to launch processess and on my travels i was getting some wierd results from command line arguments that i was attempting to parse ( none of that here ) but the wierdest thing ever happens below....???

i have removed all of the other code, all other functions etc i have a blank utility.c folder ( where i keep all functions that i call in main, i deleted everything except for 1 line comment),i have removed any function calls from main .... what you see below is all that is running ??

note that there is no malloc for the pointer char **progNameAndArgs1;, nor is there a malloc for an element within that ie progNameAndArgs1[0] = malloc ....

Code:

int main(int argc, char *argv[])
{

	int test1 = 0;
	int numberOfProgs;
	int i =0;
	
	char **progNameAndArgs1;
	char **progNames;
	char *progNames2[2];
	
	char **progNameAndArgs2 = malloc(sizeof(char*)*(2));
	
	
	printf(" main progNameAndArgs1 %s \n ",progNameAndArgs1[0]);
	printf(" main progNameAndArgs1 %s \n ",progNameAndArgs1[1]);

so my question is how do i get the output

Code:

 main progNameAndArgs1 ./driver
  main progNameAndArgs1 ./foo

when i pass ./driver ./foo at the command line ????

Answer:
The output is undefined. You may get garbage. You may get a crash. You may even get the same data which you passed as command line.
There's no real explanation as to what happened. Just don't do it.

from my thinking

1 . char **progNameAndArgs1 is pointing to the same memory as argv ??? is this standard??

No. Compiler dependant or chance. Behavior is undefined.

couple of questions

1 when a program completes, and then i run it again am i accessing the same area of memory ??

Maybe. Maybe not. It's up to the OS to handle that.
Sometimes you do end up with the same memory address, sometimes not.

2 if i create a string array and add an entry

Code:

  char **stringArray = malloc(sizeOf(char*)*2)
  stringArray[0] = malloc(sizeof(char)*5)
  strcpy(stringArray[0],"ted");

if i do the right thing then when i finish up i free(stringArray) or the individual elements....

free(stringArray[0]);
free(stringArray);
Every pointer you allocate via malloc must have a free.
You can't just try to free the entire array and expect it to work.

but if i don't free it up is it left somehow in memory ??

Yes, until the app exits (but then again, this is also OS dependant, so don't do it).

so if i ran my program for a 2nd time but changed the code to

Code:

  char **stringArray ;

  printf(" %s \n",stringArray[0] )

and then printed stringArray[0] would i get ted again ?????

Again, undefined. See the first example.

> char **progNameAndArgs2 = malloc(sizeof(char*)*(2));
A rough guess, the run-time code did the same thing to generate an argv, then freed that memory before calling main.

As a result, the memory you allocate looks remarkably like an argv as well.

Try it with another compiler, I'm sure you'll get a different answer.
I just get a core file for my trouble.

Which is pretty much what you would expect, what with all the uninitialised data.

Originally Posted by Salem

> char **progNameAndArgs2 = malloc(sizeof(char*)*(2));
A rough guess, the run-time code did the same thing to generate an argv, then freed that memory before calling main.

As a result, the memory you allocate looks remarkably like an argv as well.

Try it with another compiler, I'm sure you'll get a different answer.
I just get a core file for my trouble.

Which is pretty much what you would expect, what with all the uninitialised data.

Eh, no. The OP prints progNameAndArgs1.Since the pointer is never initialized, it points at some arbitrary "random" location in memory, which in this case happens to be a copy of argv - there are possobly copies of argv on the stack when main is called. But add an "int x" in front of your progNameAndArgs1 variable, and it will [highly likely] contain something else, which may well lead to a crash.

--
Mats