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copied!<p>As you are just learning C, I recommend you to really try to understand the <em>differences</em> between arrays and pointers first instead of the <em>common</em> things. </p> <p>In the area of parameters and arrays, there are a few confusing rules that should be clear before going on. First, <strong>what you declare in a parameter list is treated special.</strong> There are such situations where things don't make sense as a function parameter in C. These are</p> <ul> <li>Functions as parameters</li> <li>Arrays as parameters</li> </ul> <h2>Arrays as parameters </h2> <p>The second maybe is not immediately clear. But it becomes clear when you consider that the size of an array dimension is part of the type in C (and an array whose dimension size isn't given has an incomplete type). So, if you would create a function that takes an array by-value (receives a copy), then it could do so only for one size! In addition, arrays can become large, and C tries to be as fast as possible. </p> <p>In C, for these reasons, <strong>array-values</strong> are not existent. If you want to get the value of an array, what you get instead is a pointer to the first element of that array. And herein actually already lies the solution. Instead of drawing an array parameter invalid up-front, a C compiler will <strong>transform</strong> the type of the respective parameter to be a pointer. Remember this, it's very important. The parameter won't be an array, but instead it will be a pointer to the respective element type. </p> <p>Now, if you try to pass an array, what is passed instead is a pointer to the arrays' first element. </p> <h3>Excursion: Functions as parameters</h3> <p>For completion, and because I think this will help you better understand the matter, let's look what the state of affairs is when you try to have a function as a parameter. Indeed, first it won't make any sense. How can a parameter be a function? Huh, we want a variable at that place, of course! So what the compiler does when that happens is, again, to <strong>transform</strong> the function into a <strong>function pointer</strong>. Trying to pass a function will pass a pointer to that respective function instead. So, the following are the same (analogous to the array example):</p> <pre><code>void f(void g(void)); void f(void (*g)(void)); </code></pre> <p>Note that parentheses around <code>*g</code> is needed. Otherwise, it would specify a function returning <code>void*</code>, instead of a pointer to a function returning <code>void</code>.</p> <h2>Back to arrays</h2> <p>Now, I said at the beginning that arrays can have an incomplete type - which happens if you don't give a size yet. Since we already figured that an array parameter is not existent but instead any array parameter is a pointer, the array's size doesn't matter. That means, the compiler will translate all of the following, and all are the same thing:</p> <pre><code>int main(int c, char **argv); int main(int c, char *argv[]); int main(int c, char *argv[1]); int main(int c, char *argv[42]); </code></pre> <p>Of course, it doesn't make much sense to be able to put any size in it, and it's just thrown away. For that reason, C99 came up with a new meaning for those numbers, and allows other things to appear between the brackets:</p> <pre><code>// says: argv is a non-null pointer pointing to at least 5 char*'s // allows CPU to pre-load some memory. int main(int c, char *argv[static 5]); // says: argv is a constant pointer pointing to a char* int main(int c, char *argv[const]); // says the same as the previous one int main(int c, char ** const argv); </code></pre> <p>The last two lines say that you won't be able to change "argv" within the function - it has become a const pointer. Only few C compilers support those C99 features though. But these features make it clear that the "array" isn't actually one. It's a pointer. </p> <h2>A word of warning</h2> <p>Note that all i said above is true only when you have got an array as a <strong>parameter</strong> of a function. If you work with local arrays, an array won't be a pointer. It will <em>behave</em> as a pointer, because as explained earlier an array will be converted to a pointer when its value is read. But it should not be confused with pointers. </p> <p>One classic example is the following:</p> <pre><code>char c[10]; char **c = &c; // does not work. typedef char array[10]; array *pc = &c; // *does* work. // same without typedef. Parens needed, because [...] has // higher precedence than '*'. Analogous to the function example above. char (*array)[10] = &c; </code></pre>

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