Array pointer in C program

Question

Here is a C program in textbook, it asks a 3*5 2D array from users and prints the third line.

I am confused with int* p[5]. Why here needs to have [5], I think just int* p is OK. It can repeatedly add and point to the next memory space in the int array. And can anyone explain how pointer works in this program?

#include <stdio.h>
int main(void){
 int a[3][5];
 int i,j;
 int *p[5];
 p = &a[0];
 printf("Please input:\n");
 for(i = 0; i < 3; i++){
 for(j = 0; j<5;j++){
 scanf("%d\n",(*(p+i))+j);
 }
 }
 p = &a[2];
 printf("the third line is:\n");
 for(j = 0; j<5; j++){
 printf("%5d", *((*p)+j));
 }
 printf("\n");
}

• alk: In C a 2D array is just an array of a arrays.
• barak manos: You are correct. No need for p[5] here. Clearly, there is no assignment to any p[i].
• Stack Exchange Broke The Law: p = &a[0]; should not compile.
• too honest for this site: While it might work, it is still wrong and useless. Why not directly index into the 2D array? Anyway, int *p[5] is the wrong type. Just another author having trouble with arrays and pointers and the syntax in C. A pointer to the inner array would be int (*p)[5] - notice the parenthesis!
• Jennifer Q: But when I change it to int*p, it reports (*(p+i))+j) becomes to type int rather than int *. I just want to know what happens here?

Answer 1

int *p[5];

is an array of five pointers to int.

What you want is a pointer to an array of five ints

int (*p)[5];

because &a[0] is the address of the 1st element of a which is an int[5].

The compiler should have clearly issued at least a warning on this, if not an error, which would be expected.

More on this here: C pointer to array/array of pointers disambiguation

• Jennifer Q: yep, it reports a warning. There should be parenthesis. But most confused me is when I change it to int*p, (*(p+i))+j) becomes to type int rather than int *. Why that happens?
• M.M: @JenniferQ appying * to int * gives int
• alk: @JenniferQ: int * p defines p to point to an int. If you apply the de-referencing operator * to p you get what is points to, that is an int here.