error function returns address of local variable

Question

I m beginner with C and I am learning on my own   I am creating the following function   char  foo int x        if x  lt  0           char a 1000           char b    blah           x   x - 1          char  c   foo x           strcpy a  b           strcat a  c           return a              blah         I am basically trying to return an appended string  but I get the following error    error  function returns address of local variable   any suggestions  how to fix this

User · Answer

All the answer explain the problem really good.

However, I would like to add another information.

I faced the same problem at the moment I wanted the output of a function to be a vector.

In this situation, the common solution is to declare the output as an argument of the function itself. This way, the alloc of the variable and the physical space necessary to store the information are managed outside the function. Pseudocode to explain the classical solution is:

void function(int input, int* output){
    //...
    output[0] = something;
    output[1] = somethig_else;
    //...
    return;
}

In this case, the example code within the question should be changed in:

void foo(int x, char* a){
     if(x < 0){
        char b = "blah";
        //...
        strcpy(a, b);
        //..
        return;
      }
    //..
}

User · Answer

This line   char b    blah     Is no good - your lvalue needs to be a pointer   Your code is also in danger of a stack overflow  since your recursion check isn t bounding the decreasing value of x   Anyway  the actual error message you are getting is because char a is an automatic variable  the moment you return it will cease to exist   You need something other than an automatic variable

User · Answer

char b    blah      should be   char  b    blah

User · Answer

The local variables have a lifetime which extends only inside the block in which it is defined  The moment the control goes outside the block in which the local variable is defined  the storage for the variable is no more allocated  not guaranteed   Therefore  using the memory address of the variable outside the lifetime area of the variable will be undefined behaviour   On the other hand you can do the following    char  str to ret   malloc  sizeof  char    required size                return str to ret    And use the str to ret instead  And when returning str to ret  the address allocated by malloc will be returned  The memory allocated by malloc is allocated from the heap  which has a lifetime which spans the entire execution of the program  Therefore  you can access the memory location from any block and any time while the program is running   Also note that it is a good practice that after you have done with the allocated memory block  free it to save from memory leaks  Once you free the memory  you can t access that block again

User · Answer

a is an array local to the function Once the function returns it does not exist anymore and hence you should not return the address of a local variable  In other words the lifetime of a is within the scope      of the function and if you return a pointer to it what you have is a pointer pointing to some memory which is not valid  Such variables are also called automatic variabels because their lifetime is automatically managed you do not need to manage it explicitly         Since you need to extend the variable to persist beyond the scope of the function you You need to allocate a array on heap and return a pointer to it        char  a   malloc 1000      This way the array a resides in memory untill you call a free   on the same address  Do not forget to do so or you end up with a memory leak

User · Answer

I came up with this simple and straight-forward  i hope so  code example which should explain itself    include  lt string h gt   include  lt stdio h gt   include  lt stdlib h gt      function header definitions    char  getString                           lt - with malloc  good practice  char   getStringNoMalloc        lt - without malloc  fails  don t do this   void getStringCallByRef char  reference      lt - callbyref  good practice      the main    int main int argc  char argv                       calling with malloc     char   a   getString        printf  MALLOC     a    s  n   a        free a                    calling without malloc     char   b   getStringNoMalloc        printf  NO MALLOC     b    s  n   b     this doesnt work  question to yourself  WHY        HINT  the warning says that a local reference is returned            NO free here                   call-by-reference     char c 100       getStringCallByRef c       printf  CALLBYREF     c    s  n   c        return 0       WITH malloc char  getString          char   string      string   malloc sizeof char  100        strcat string   bla        strcat string            strcat string   blub         printf  string     s  n   string        return string       WITHOUT malloc  watch how it does not work this time  char  getStringNoMalloc           char string 100              strcat string   bla         strcat string             strcat string   blub           INSIDE this function  string  is OK      printf  string     s  n   string         return string    but after returning   it is NULL             and the call-by-reference way to do it  prefered  void getStringCallByRef char  reference         strcat reference   bla        strcat reference            strcat reference   blub          INSIDE this function  string  is OK     printf  string     s  n   reference         OUTSIDE it is also OK because we hand over a reference defined in MAIN        and not defined in this scope  local   which is destroyed after the function finished     When compiling it  you get the  intended  warning   me box    gcc -o example o example c  example c  In function    getStringNoMalloc     example c 58 16  warning  function returns address of local variable  -Wreturn-local-addr           return string    but after returning   it is NULL                             basically what we are discussing here   running my example yields this output    me box      example o  string    bla blub  MALLOC     a   bla blub  string    bla blub  NO MALLOC     b    null   string    bla blub  CALLBYREF     c   bla blub    Theory   This has been answered very nicely by User  phoxis  Basically think about it this way  Everything inbetween   and   is local scope  thus by the C-Standard is  undefined  outside   By using malloc you take memory from the HEAP  programm scope  and not from the STACK  function scope  - thus its  visible  from outside   The second correct way to do it is call-by-reference  Here you define the var inside the parent-scope  thus it is using the STACK  because the parent scope is the main       Summary   3 Ways to do it  One of them false  C is kind of to clumsy to just have a function return a dynamically sized String  Either you have to malloc and then free it  or you have to call-by-reference  Or use C

User · Answer

Neither malloc or call by reference are needed  You can declare a pointer within the function and set it to the string array you d like to return   Using  Gewure s code as the basis   char  getStringNoMalloc void       char string 100            char  s ptr   string       strcat string   bla        strcat string            strcat string   blub          INSIDE this function  string  is OK     printf  string     s  n   string        return s ptr       works perfectly   With a non-loop version of the code in the original question   char  foo int x           char a 1000       char  a ptr   a      char  b    blah               strcpy a  b        return a ptr

User · Answer

a is defined locally in the function  and can t be used outside the function  If you want to return a char array from the function  you ll need to allocate it dynamically   char  a   malloc 1000     And at some point call free on the returned pointer   You should also see a warning at this line  char b    blah    you re trying to assign a string literal to a char

[c] error: function returns address of local variable

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