Solution for Exercise 19-6
ex1906
#include <stdio.h> int main() { int numbers[10]; int x; int *pn; pn = numbers; /* initialize pointer */ /* Fill array */ for(x=0;x<10;x++) { *pn=x+1; pn++; } pn = numbers; /* re-initialize pointer */ /* Display array */ for(x=0;x<10;x++) { printf("numbers[%d] = %d, address %p\n", x,*pn,pn); pn++; } return(0); }
Output
numbers[0] = 1, address 0x7ffeef1359f0
numbers[1] = 2, address 0x7ffeef1359f4
numbers[2] = 3, address 0x7ffeef1359f8
numbers[3] = 4, address 0x7ffeef1359fc
numbers[4] = 5, address 0x7ffeef135a00
numbers[5] = 6, address 0x7ffeef135a04
numbers[6] = 7, address 0x7ffeef135a08
numbers[7] = 8, address 0x7ffeef135a0c
numbers[8] = 9, address 0x7ffeef135a10
numbers[9] = 10, address 0x7ffeef135a14
Notes
* The only difference between this code and the solution for Exercise 19-5 is the change from array notation numbers[x]
to pointer notation *pn
in Line 24.
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