I have argv[] defined as a char *. Using the following printf statements:
printf("%s\n",argv[1]); // prints out the entire string
printf("%p\n",&argv[1]); // & -> gets the address
printf("%c\n",argv[1][0]);// prints out the first char of second var
printf("%c\n",*argv[1]); //
It's this last one I don't understand. What does it mean to print *argv[1]? why isn't that the same as *argv[1][0] and how come you can't print out printf("%s\n",*argv[1]);. Also, why is &*argv[1] a different address then &argv[1]?
The array subscript operation a[i] is defined as *(a + i) - given the address a, offset i elements (not bytes) from that address and dereference the result. Thus, given a pointer p, *p is equivalent to *(p + 0), which is equivalent to p[0].
The type of argv is char **; given that, all of the following are true:
Expression Type Value
---------- ---- -----
argv char ** Pointer to a sequence of strings
*argv char * Equivalent to argv[0]
**argv char Equivalent to argv[0][0]
argv[i] char * Pointer to a single string
*argv[i] char Same as argv[i][0]
argv[i][j] char j'th character of i'th string
&argv[i] char ** Address of the pointer to the i'th string
Since the type of argv[i][j] is char, *argv[i][j] is not a valid expression.
Here's a bad visualization of the argv sequence:
+---+ +---+ +---+
argv | | ---> argv[0] | | ---------------------------> argv[0][0] | |
+---+ +---+ +---+ +---+
argv[1] | | -------> argv[1][0] | | argv[0][1] | |
+---+ +---+ +---+
... argv[1][1] | | ...
+---+ +---+ +---+
argv[argc] | | ---||| ... argv[0][n-1] | |
+---+ +---+ +---+
argv[1][m-1] | |
+---+
This may help explain the results of different expressions.