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    <p>Your problems start with your specification. If I understand you correctly, you want to have a function where you pass an array of characters and a length. The size of your array of input characters will be between 1 and <code>len</code>? However, your function has no way of knowing what the size of your array <code>num</code> is. If you wanted this to work, you would need to define your function as </p> <pre><code>const char * extendBinary(char *num, size_t num_len, int len); </code></pre> <p>so that your function doesn't overrun your buffer pointed to by <code>num</code>. Note that I replaced <code>char num[]</code> with <code>char *num</code> as this is the common mechanism for passing a pointer. You cant pass pointers to arrays and then dereference that pointer and get back the original type (that includes its size) -- that's just one thing that C doesn't let you do, so just use a normal pointer and a separate size variable.</p> <p>Finally, you'll have to deal with memory allocation unless you want a memory leak. Thus, you could simply say that whom ever calls extendBinary should free it's return value when done with it.</p> <pre><code>const char * extendBinary(char *num, size_t num_len, int len) { char *ret = malloc(len + 1); int i; memset(ret, '0', len); ret[len] = 0; strncpy(&amp;ret[len - num_len], num, num_len); return ret; } int main(void) { char arr[] = {'1', '0', '1'}; const char *formatted = extendBinary(arr, sizeof(arr), 8); printf("%s\n", formatted); free(formatted); return 0; } </code></pre>
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    1. POReturning arrays and pointers in C?
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    1. USDaniel Santos
    1. USDaniel Santos
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    1. COYour explanation is perfect, and is helping me write the rest of my code. Thank you so much!
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      1. USLaurence
    2. COSo glad to help! However, I omitted some bounds checking in the above code. Specifically, you should make sure that `num_len` is actually smaller than `len` and maybe just allocate a larger buffer if it is (plus one for the null-terminator). Sorry about that, I was trying to keep my answer concise.
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