[lldb-dev] Is there a way to know whether a variable have been init?

Enrico Granata egranata at apple.com
Sun Aug 31 11:23:09 PDT 2014


There is no obvious and general way to know what you want to know
Consider the following 

int x;
...
if (foo)
  x = 1;
else
  ;
// is x written to here or not?

If you wanted to pursue this, you could of course track the memory locations of all variables, and step through the code, and figure out when a write to those locations occur. From that point on, your variable would be initialized.
I expect this to be unusably slow for anything but very very small portions of code.

I guess the next question is "what are you really trying to do"?

Sent from my iPhone

> On Aug 30, 2014, at 12:58 AM, Kun LIng <lingcc at lingcc.com> wrote:
> 
> Hi all,
>     I am currently using python interface to run a non-interactively script based on LLDB to trace the variable changing during each step during the program running. 
> 
>     C allows a variable to be declared at first without initialization.  And the value of the variable after declaration could be a random value before initialized according to my test using LLDB with the following code compiled by clang+llvm. 
> 
> ----Example Code ----
> #include <stdio.h>
> #include <stdlib.h>
> int g = 42;
> float u = 5;
> struct node {
>   int field;
>   struct node* next;
> };
> void f() {
>     g = g + 1;
>     printf("Hello world!\n");
> }
> 
> int main(int argc, char *argv[])
> {
>   int x;
>   int y=10;
>   x  = y+1;
>   struct node* root = (struct node*)malloc(sizeof(struct node));
>   root->field = 3;
>   root->next = (struct node*)malloc(sizeof(struct node));
>   root->next->field = 2;
>   free(root->next);
>   free(root);
> 
>   int i = 0;
>   i = i + 1;
>   f();
>   return 0;
> }
> 
> -----End Example Code ----
> 
>    My question is:
>    In LLDB, is there a way to check whether a variable is currently be declared, and have not been initialized? So that my non-interactively script could check this, and ignore to show it before initialized ?
> 
>    Thanks very much.
> 
> Best,
> Kun Ling
>  
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