char ** and argv memory question

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:

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main progNameAndArgs1 ./driver
  main progNameAndArgs1 ./foo

---

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

from my thinking

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


couple of questions

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

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....

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

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 ?????

Quote:

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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.

Quote:

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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.

Quote:

---

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.

Quote:

---

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.

Quote:

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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).

Quote:

---

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.

Quote:

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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.

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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.

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Mats