When i execute this code
#include<stdio.h>
int main() {
int (*x)[5];
printf("\nx = %u\nx+1 = %u\n&x = %u\n&x + 1 = %u",x,x+1,&x,&x+1);
}
This is the output in C or C++:
x = 134513520
x+1 = 134513540
&x = 3221191940
&x + 1 = 3221191944
Please explain. Also what is the difference between:
int x[5] and int (*x)[5] ?
int x[5] is an array of 5 integersint (*x)[5] is a pointer to an array of 5 integersWhen you increment a pointer, you increment by the size of the pointed to type. x+1 is therefore 5*sizeof(int) bytes larger than just x - giving the 8048370 and 8048384 hex values with a difference of 0x14, or 20.
&x is a pointer to a pointer - so when you increment it you add sizeof(a pointer) bytes - this gives the bf9b08b4 and bf9b08b8 hex values, with a difference of 4.
int x[5] is an array of 5 intsint (*x)[5] is a pointer to an array of 5 intsint* x[5] is an array of 5 pointers to ints
int (*x)[5];
declares a pointer to an array.
From the question title, you probably want
int* x[5];
instead, which declares an array of pointers.
int x[5];
declares a plain old array of ints.
