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-rw-r--r--c.html.markdown34
1 files changed, 34 insertions, 0 deletions
diff --git a/c.html.markdown b/c.html.markdown
index a8f71057..3d632eab 100644
--- a/c.html.markdown
+++ b/c.html.markdown
@@ -445,6 +445,17 @@ int main (int argc, char** argv)
for (xx = 0; xx < 20; xx++) {
*(my_ptr + xx) = 20 - xx; // my_ptr[xx] = 20-xx
} // Initialize memory to 20, 19, 18, 17... 2, 1 (as ints)
+
+ // Note that there is no standard way to get the length of a
+ // dynamically allocated array in C. Because of this, if your arrays are
+ // going to be passed around your program a lot, you need another variable
+ // to keep track of the number of elements (size) of an array. See the
+ // functions section for more info.
+ int size = 10;
+ int *my_arr = malloc(sizeof(int) * size);
+ // Add an element to the array
+ my_arr = realloc(my_arr, ++size);
+ my_arr[10] = 5;
// Dereferencing memory that you haven't allocated gives
// "unpredictable results" - the program is said to invoke "undefined behavior"
@@ -530,6 +541,29 @@ swapTwoNumbers(&first, &second);
printf("first: %d\nsecond: %d\n", first, second);
// values will be swapped
*/
+
+/*
+With regards to arrays, they will always be passed to functions
+as pointers. Even if you statically allocate an array like `arr[10]`,
+it still gets passed as a pointer to the first element in any function calls.
+Again, there is no standard way to get the size of a dynamically allocated
+array in C.
+*/
+// Size must be passed!
+// Otherwise, this function has no way of knowing how big the array is.
+void printIntArray(int *arr, int size) {
+ int i;
+ for (i = 0; i < size; i++) {
+ printf("arr[%d] is: %d\n", i, arr[i]);
+ }
+}
+/*
+int my_arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
+int size = 10;
+printIntArray(my_arr, size);
+// will print "arr[0] is: 1" etc
+*/
+
// if referring to external variables outside function, must use extern keyword.
int i = 0;
void testFunc() {