declaration

What happens when I declare say multiple variables on a single line? e.g. int x, y, z; All are ints. The question is what are y and z in the following statement? int* x, y, z; Are they all int pointers? Only x is a pointer to int; y and z are regular ints. This is one aspect of C declaration syntax that trips some people up. C uses the concept of a declarator , which introduces the name of the thing being declared along with additional type information not provided by the type specifier . In the declaration int* x, y, z; the declarators are *x , y , and z (it's an accident of C syntax that you

Given that the name of an array is actually a pointer to the first element of an array, the following code: #include <stdio.h> int main(void) { int a[3] = {0, 1, 2}; int *p; p = a; printf("%d\n", p[1]); return 0; } prints 1 , as expected. Now, given that I can create a pointer that points to a pointer, I wrote the following: #include <stdio.h> int main(void) { int *p0; int **p1; int (*p2)[3]; int a[3] = {0, 1, 2}; p0 = a; p1 = &a; p2 = &a; printf("p0[1] = %d\n(*p1)[1] = %d\n(*p2)[1] = %d\n", p0[1], (*p1)[1], (*p2)[1]); return 0; } I expected it to compile and print p0[1] = 1 (*p1)[1] = 1 (*p2)