Pointers in C: Allocating memory with value assignment VS allocating memory without value assignment

Question

Char pointer's memory allocation via calloc works as expected only when I do not assign some value to the pointers.

I am trying to get some simple examples working to better my understanding of pointers / double pointers / triple pointers / ... . While writing the code for double pointers, I stumbled upon a really weird behavior.

#define PRINT(X, ...) printf("Address of " X "\n", ##__VA_ARGS__);

...
...

    int i = 0;
    char buf[12];
    char **two_star;

    two_star = calloc(10, sizeof(char*));
    for (i = 0; i < 10 ; ++i){
        two_star[i] = calloc(10, sizeof(char));
        PRINT("two_star[%d]   = %p", i, two_star[i]);
    }

    for (i = 0; i < 10 ; ++i){
        two_star[i] = calloc(10, sizeof(char));
        snprintf(buf, 12, "pre_%d_suff", i);
        two_star[i] = buf;
        PRINT("two_star[%d]   = %p   two_star[%d]   = %s", i, two_star[i], i, two_star[i]);
    }

I get the following results for the above two for loops:

For Loop 1:

Address of two_star[0]   = 0xbcc090
Address of two_star[1]   = 0xbcc0b0
Address of two_star[2]   = 0xbcc0d0
Address of two_star[3]   = 0xbcc0f0
Address of two_star[4]   = 0xbcc110
Address of two_star[5]   = 0xbcc130
Address of two_star[6]   = 0xbcc150
Address of two_star[7]   = 0xbcc170
Address of two_star[8]   = 0xbcc190
Address of two_star[9]   = 0xbcc1b0

For Loop 2:

Address of two_star[0]   = 0x7ffcd2238ab0   two_star[0]   = pre_0_suff
Address of two_star[1]   = 0x7ffcd2238ab0   two_star[1]   = pre_1_suff
Address of two_star[2]   = 0x7ffcd2238ab0   two_star[2]   = pre_2_suff
Address of two_star[3]   = 0x7ffcd2238ab0   two_star[3]   = pre_3_suff
Address of two_star[4]   = 0x7ffcd2238ab0   two_star[4]   = pre_4_suff
Address of two_star[5]   = 0x7ffcd2238ab0   two_star[5]   = pre_5_suff
Address of two_star[6]   = 0x7ffcd2238ab0   two_star[6]   = pre_6_suff
Address of two_star[7]   = 0x7ffcd2238ab0   two_star[7]   = pre_7_suff
Address of two_star[8]   = 0x7ffcd2238ab0   two_star[8]   = pre_8_suff
Address of two_star[9]   = 0x7ffcd2238ab0   two_star[9]   = pre_9_suff

The issue here is evident. The pointers being allocated in the second for loop all have the same value of address. Meaning, that two_star[0] and two_star[9] end up being the same value as the same memory address ends up being updated again and again.

memcpy instead of two_star[i] = buf;

fas
– fas

2019-09-11 21:51:20 +00:00
Commented Sep 11, 2019 at 21:51 — fas
– fas, Commented Sep 11, 2019 at 21:51

fas · Accepted Answer · 2019-09-11 21:56:50Z

2

Use strcpy: strcpy(two_star[i], buf); instead of two_star[i] = buf;, because in your case you don't copy string, just reassign pointer two_star[i] to buf (and here is also memory leak, because you've lost pointer to allocated memory).

edited Sep 11, 2019 at 21:56

answered Sep 11, 2019 at 21:53

fas

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1 Comment

Abhishek Mangal Over a year ago

This works. Will accept it as the correct answer when time permits.

chux · Accepted Answer · 2019-09-12 02:09:09Z

1

Using strcpy(two_star[i], buf); is insufficient.

    char buf[12];
    two_star[i] = calloc(10, sizeof(char));
    // snprintf(buf, 12, "pre_%d_suff", i);
    strcpy(two_star[i], buf);  // bad, trying to put 12 characters into 10
    two_star[i] = buf;

The destination is too small.

Instead, code could right size.

   int size_needed = snprintf(NULL, 0, "pre_%d_suff", i)  + 1;
   two_star[i] = malloc(size_needed);
   snprintf(two_star[i], size_needed, "pre_%d_suff", i);

Note: Error checking omitted for brevity.

answered Sep 12, 2019 at 2:09

chux

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