CODEWITHSUNDEEP
cdebuggingscopeundefined-behavior
stackypancakes
stackypancakes

Reputation: 13

Why does my C code print out different values when referring to the same variable?

Here's goes my first attempt at a question on stackoverflow -

I have the following implemented in C:

When running my program, I have realized that the value of hashstring printed out differs in the two following functions:

void foo() {
    ...
    Obj *obj = Obj(args);
    char *hashstring = getHash(obj);
    printf("My value of hashstring: %s\n", hashstring);
    ...
}

char* getHash(Obj *obj) {
    ...
    int hashint = getHashInt(args);
    // I am trying to do a conversion from int to string here
    char hashstring[12]; 
    sprintf(hashstring, "%d", hashint);
    printf("My value of hashstring: %s\n", &hashstring);
    return &hashstring;
}

Here's what gets printed out:
My value of hashstring: 1001 // printing in getHash: this is the expected
My value of hashstring: 0
My value of hashstring: 2002 // printing in getHash: this is the expected
My value of hashstring: 0
My value of hashstring: 3003 // printing in getHash: this is the expected
My value of hashstring: 0

I have tried debugging by printing the pointer of hashstring instead, in foo(). It seems like in every single call, I am getting the same address for hashstring.

Here's what gets printed out:
0x7ffee1992f54
0x7ffee1992f54
0x7ffee1992f54

I guessed that this probably meant that I am somehow referencing the same hashstring variable in getHash(), which is strange since I am creating a new variable each time. I have also tried other variants like snprintf(hashstring, 12, "%d", hashint) in getHash(), or creating a char *pointerToHash and passing in *pointerToHash to sprintf.

Any ideas?

Upvotes: 1

Views: 536

Answers (1)

Vlad from Moscow
Vlad from Moscow

Reputation: 311126

This function

char* getHash(Obj *obj) {
...
int hashint = getHashInt(args);
// I am trying to do a conversion from int to string here
char hashstring[12]; 
sprintf(hashstring, "%d", hashint);
printf("My value of hashstring: %s\n", &hashstring);
return &hashstring;
}

invokes undefined behavior because it returns a pointer to the local array

 char hashstring[12];

that will not be alive after exiting the function.

Moreover the return type of the function char * and the type of the returned expression char ( * )[12] are different and there is no implicit conversion from one type to another.

To avoid the problem you should allocate the array dynamically like for example

char *hashstring = calloc( 12, sizeof( char ) );

and in the return statement write

return hashstring;

Then in the caller you should free the allocated memory when the allocated array will not be needed any more.

Upvotes: 1

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