How to show the starting address of some variables in C?

问题

#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <string.h>

extern char **environ;

int global_x = 10;                  // initialised global variable
int global_y;                       // un-initialised global variable
char global_array1[] = "Hello, world!";     // initialised global array and a string literal
char global_array2[10];             // un-initialised global array
char *global_pointer1 = "bye!";         // global pointer to a string literal 
char *global_pointer2;              // un-initialised global pointer 
float global_float = 100.1;         // initialised global variable
double global_double;               // un-initialised global variable

#define ONEGB  1073741824
#define ONEMB  1048576
#define ONEKB  1024

char *addr(unsigned long a)
{
    unsigned long r; // remainder

    r = (unsigned long) a;
    int gb = (int) ( r / ONEGB );

    r -=  gb * ONEGB;
    int mb = (int) ( r  / ONEMB );

    r -=  mb * ONEMB;
    int kb = (int) ( r  / ONEKB );

    r -=  kb * ONEKB;
    int b  = (int) ( r );

    char *p = malloc(64);

    sprintf(p, "%4dGB, %4dMB, %4dKB, %4d", gb, mb, kb, b);
    return p;
}

int f2(int x)
{
    char * f2_p;
    int f2_x = 21;

    f2_p = malloc(1000);         // dynamically allocated memory

    // print out the address of x
    // print out the addresses of f2_p, and f2_x
    // print out the starting address of the dynamically allocated memory
    .....

    L: f2_x = 10;
    return f2_x;
}

void f1(int x1, int x2, float x3, char x4, double x5, int x6)
{
    int f1_x = 10;
    int f1_y;
    char *f1_p1 = "This is inside f1";   // pointer to another string literal 
    char *f1_p2;

    f1_p2 = malloc(100);         // dynamically allocated memory

    // print out the addresses of x1, x2, x3, x4, x5, x6
    // print out the addresses of f1_x, f1_y, f1_p1, f1_p2
    // print out the address of the string literal "This is inside f1"
    .....

    f1_y = f2(10);
    return;
}

int main(int argc, char *argv[])
{

    // print out the addresses of argc, argv
    // print out the starting address and end address of the command line arguments of this process
    // print out the starting address and end address of the environment of this process
    // print out the starting addresses of function main, f1, and f2
    // print out the addresses of global_x, global_y, global_array1, global_array2, global_pointer1,
    //           global_pointer2, global_float, global_double
    // print out the addresses of string literals 10, "Hello, world!", "bye", 100.1

    ..... 

    // call function f1 with suitable arguments such as 12, -5, 33.7, 'A', 1.896e-10, 100 
    f1( .... );

    exit(0);
}

I tried to search on google, but cannot find sth useful, and in this case I just want to figure out how to print out the starting address of the dynamically allocated memory; print out the starting address and end address of the command line arguments of this process;print out the starting address and end address of the environment of this process;print out the starting addresses of function main, f1, and f2. anybody can help me?..thank you!

回答1:

In main :

print out the addresses of argc, argv - printf ("%d, %d", &argc, argv);

print out the starting address and end address of the command line arguments of this process - printf ("%d", (void *)argv);

print out the starting address and end address of the environment of this process - printf ("%d", (void *)environ);

print out the starting addresses of function main, f1, and f2 - printf ("%d %d %d", &main, &f1, &f2);

print out the addresses of global_x, global_y, global_array1, global_array2, global_pointer1, global_pointer2, global_float, global_double - just use the & operator in front of each variable whose address you want to print.

print out the addresses of string literals 10, "Hello, world!", "bye", 100.1 - printing addresses of string literals is not allowed.

In f1:

print out the addresses of x1, x2, x3, x4, x5, x6 - printf ("%d %d %d %d %d %d", &x1, &x2, &x3, &x4, &x5);

print out the addresses of f1_x, f1_y, f1_p1, f1_p2 - printf ("%d %d %d %d", &f1_x, &f1_y, f1_p1, f2_p2);

print out the address of the string literal "This is inside f1" - Taking address of a string literal is not allowed

In f2:

print out the address of x - printf ("%d", &x);

print out the addresses of f2_p, and f2_x - printf("%d", f2_p, &f2_x);

print out the starting address of the dynamically allocated memory - printf ("%d", f2_p);

回答2:

f1 and f2 (without the (parameter) block) is the starting address of the function For better clarification. f2(x); calls the function int f2(int x) passing the parameter x. "f2" without the parenthesis is the address of function int f2(int x)

回答3:

To declare a string literal as per my lecturer's guidance -

char *p = "this is a string literal";

printf("address of variable p starts at %p\n",  &p));
printf("address of the string literal starts at %p\n", p);

回答4:

For that you need the & operator. It takes the address of a variable

Some more info on pointer operators can be found here

to the starting address of a regular variable:

printf("%p", &variable);

These variables don't realy have an end address, but i think what you want is something like:

printf("%p", (&variable + 1));

which prints the next available address

回答5:

If you want to see the address of a variable you need to use the & operator or the straight pointer (in the case of dynamically allocated memory).

int main(int argc, char *argv[]){  
   int * arr;
   int i = 0;
   arr = malloc(100 * sizeof(int));
   printf("dynamic array starts at: %#x and ends at: %#x\n",arr,arr+100);
   printf("static i is at: %#x\n",&i);
}

Output:

mike@linux-4puc:~> ./a.out 
dynamic array starts at: 0x804b008 and ends at: 0x804b198
static i is at: 0xbfc1f6d8

来源：https://stackoverflow.com/questions/12601140/how-to-show-the-starting-address-of-some-variables-in-c

标签

memory-address