Need help with this code on pointers
Hi All,
Please help me with the following piece of code. I do not understand why it prints the same value for a, *a and also **a. Could someone explain me how it works?
Thanks,Code:#include <conio.h> main() { int a[2][3][4] = { { 1,2,3,4, 5,6,7,8, 9,1,1,2 }, { 2,1,4,7, 6,7,8,9, 0,0,0,0 } }; clrscr(); printf("\n%u\n%u\n%u\n%d",a,*a,**a,***a); getch(); }
Babu
I took out conio.h and all dependencies upon it. When compiled with warnings turned up, the compiler gave me this:
Anyway, what I think the idea is behind what you want to know..... Array names, when taken by themselves (like for function calls), are generally pointers to the first element of the array. This means a will be a pointer to the first element of the entire array, which also happens to be what *a points to, which also happens to be what **a points to. The first element of the entire array also happens to be the first element of the first inner array, which also happens to be the pointer to the first innermost array.Code::4: warning: return type defaults to `int' : In function `main': :7: warning: missing braces around initializer :7: warning: (near initialization for `a[0][0]') :18: warning: implicit declaration of function `printf' :18: warning: unsigned int format, pointer arg (arg 2) :18: warning: unsigned int format, pointer arg (arg 3) :18: warning: unsigned int format, pointer arg (arg 4) :20: warning: control reaches end of non-void function
That may be confusing, so work it out on paper.
%u i have never seen this
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Check the man pages or google for the format flags of printf() and scanf() if you want to see a list of them. He's just printing everything as an unsigned int.
