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Accessing arrays using pointer arithmetic

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• Using pointer arithmetic to access array elements
  • Using pointer arithmetic to access array elements

    Because the expression: p + x evaluates to: the address of the xth array element following the array element that is pointed to by p !! Therefore: the expression: *( p + x ) evaluates to: the variable that is the xth array element following the array element that is pointed to by p !!

    Example:

    Explanation:

    If p = &a[0], then: *p is an alias for a[0] *(p+0) is also an alias for a[0] *(p+1) is also an alias for a[1] *(p+2) is also an alias for a[2]

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  • Example C program:

    #include <stdio.h> int main(int argc, char *argv[]) { int a[10] = { 11, 22, 33, 44, 55, 66, 77, 88, 99, 777 }; int* p; p = &a[0]; // p points to variable a[0] printf("*(p + 0) = %d, a[0] = %d\n", *(p+0), a[0] ); printf("*(p + 1) = %d, a[1] = %d\n", *(p+1), a[1] ); printf("*(p + 2) = %d, a[2] = %d\n", *(p+2), a[2] ); printf("*(p + 3) = %d, a[3] = %d\n", *(p+3), a[3] ); }

    Output:

    *(p + 0) = 11, a[0] = 11 *(p + 1) = 22, a[1] = 22 *(p + 2) = 33, a[2] = 33 *(p + 3) = 44, a[3] = 44

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  • Example Program: (Demo above code)                                                
    • Prog file: /home/cs255001/demo/C/set2/pointer-arithm2.c

    How to run the program:

    To compile:   gcc pointer-arithm2.c To run:          ./a.out