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    <pre><code>CurrentData *AllCurrentData = '\0'; </code></pre> <p>This declares a pointer. This pointer is a variables that holds a number that is interpreted as an address. You initialize the pointer to '\0' (null).</p> <pre><code>getCurrentData(current, AllCurrentData); </code></pre> <p>Here you pass this pointer as parameter to function <code>getCurrentData</code>. As a pointer is a variable, this variable is passed by value, meaning that the function receives a copy of that value (the number that represents an address).</p> <p>When inside the function if you write</p> <pre><code>AllCurrentData = malloc... </code></pre> <p>you modify that copy of the pointer, so outside the function AllCurrentData will still be '\0'. You need to pass a pointer to that pointer (Inception, I know).</p> <pre><code>getCurrentData(current, &amp;AllCurrentData); getCurrentData(.., CurrentData **p_AllCurrentData) { *p_AllCurrentData = malloc(...); } </code></pre> <hr> <p><strong>Let me explain this concept on a simpler example:</strong></p> <pre><code>int *v = NULL; // here v is a pointer. This pointer has value 0 (invalid address) v = malloc(10 * sizeof(int)); // here the pointer is assigned a valid address, lets say 0x41A0 f(v); void f(int *x) { // here the pointer x receives a copy of the value of the pointer v, so x will be, like v, 0x41A0 x[0] = 4; /*this is ok as you modify the memory at the address 0x41A0, so this modification is seen from outside the function, as v points to the same address.*/ x = malloc(...); /* this is NOT ok, as x receives a new address, lets say 0xCC12. But because x has a copy of the value from v, v will still be unmodified.*/ } </code></pre> <p>so if you want to allocate a pointer inside a function this is not ok:</p> <pre><code>int *v = NULL; // here v is a pointer. This pointer has value 0 (invalid address) f(v); void f(int *x) { // here the pointer x receives a copy of the value of the pointer v, so x will be, like v, NULL x = malloc(...); /*this is NOT ok, as x receives a new address, lets say 0xCC12. But because x has a copy of the value from v, v will still be unmodified, it will still have NULL.*/ } </code></pre> <p>The right way to do is:</p> <pre><code>int *v = NULL; // here v is a pointer. This pointer has value 0 (invalid address) // as v is a variable (of type pointer to int), v has an address, lets say 0xAAAA. f(&amp;v); void f(int **p_x) { /* here the p_x is a pointer to (a pointer of int), so this pointer receives a copy of the value the address of p, so p_x is 0xAAAA.*/ *p_x = malloc(...); /* lets say malloc returns the address 0xBBBB. This will be the address of our vector. *p_x= says we modify the value at the address p_x. So the value found at the adress 0xAAAA will be 0XBBBB. But because 0xAAAA is the address of v, we effectively modified the value of v to 0xBBBB. So now v has the address of our starting vector.*/ // if you want to modify values inside this vector you need: (*p_x)[0] = ... } </code></pre>
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    1. POc - Access violation when reading array
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    1. USbolov
    1. USbolov
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    1. POc - Access violation when reading array
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      1. VTAcceptedByOriginator
    1. COLet me know how confusing this is :))
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      1. USbolov
    2. COOn a scale from 0 to 9? :)
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      1. USMartin James
    3. COOn a scale from 0 to 9, i vote 8. Pointers give me headaches for days. I threw some *s in and BAMM, now i love you.
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      1. USTinMan
 

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