05. Addressing

(Do in on sugon server)

• do not forget to use compile script

  • …or static linking: cc -O0 -static -mno-abicalls -mno-gpopt your_file.c -o your_program_name

• printf("%p\n", &variable) — print pointer to variable, which in C strictly means address

  • write a program that prints addresses of a pair of local and a pair of global variables
  • compare the addresses by sight

  • interpret this

    • unlike .data segment in MARS, real world .rdata segment is not at 0x1001000, but just after .text segment ends!)

• int *P; — «pointer to integer» variable type (i. e. address of memory part that keeps an integer value)

• Array name is array address, it equals to the array zero element address

  • i. e. A == &A[0]

• Address arithmetic. Write a program, that

  • prints character (char) array zero and fifths element addresses

  • prints integer (int) array zero and fifths element addresses

  • interpret this

• Compare two global arrays A and B addresses by just subtracting them!

  •    1 #include <stdio.h>
       2 
       3 int A[5]={1,2,3,4,5};
       4 int B[5]={9,8,7,6,5};
       5 
       6 int main(int argc, char *argv[]) {
       7   printf("A, B, B-A: %p, %p, %d\n", A, B, B-A);
       8   return 0;
       9 }
    

  • Why B-A = 5?

  • How to calculate real distance between A and B? Try it.

    1. with sizeof() ?

    2. with (int) cast ?

• Unexpected: remove both «={…}» initializers and compile code dynamically (cc prog.c -o prog)

  • There's a chance that A address became larger than B! This is because the order of uninitialized elements in memory is not guaranteed

• Alignment. Compile this:

     1 #include <stdio.h>
     2 struct STRU {
     3   int a;
     4   int b;
     5   double f;
     6 };
     7 
     8 struct STRU Massive[10];
     9 void main(void) {
    10   int i=3, j=4;
    11   printf("Size of STRU: %d, STRU[10]: %d\n", sizeof(struct STRU), sizeof(Massive));
    12   printf("%d, %lf\n", Massive[i].a, Massive[j].f);
    13   printf("%p, %p, %p\n", &Massive[i].a, &Massive[i].b, &Massive[i].f);
    14  
    15 }
  

• Rewrite the structure as this:

     1 struct STRU {
     2   int a;
     3   double f;
     4   int b;
     5 };
  

  • compile and run

  • interptet changes!

  • (this is because alignment, see .align instructions in assrmber code)

H/W

If you have non-MIPS environment, try these programs there