CODEWITHSUNDEEP
cpointers
andypopa
andypopa

Reputation: 536

Define pointer to memory address and print contents C++

I want to print out things from given memory addresses.

Using this code, I can define a variable, define a pointer to it and print its contents using the pointer.

char buf[100];
void *p = buf;

strcpy(p, "Test string");

printf("Address: %X\n", &buf);
printf("Contents: %c\n", p);

What I want to do is to specify a certain memory address, and print out the contents of that block. I tried experimenting by incrementing and decremeting p, but it doesn't print out anything.

Upvotes: 1

Views: 5887

Answers (6)

david.pfx
david.pfx

Reputation: 10868

An interesting little problem. One that should occur to every programmer sooner or later, either out of curiosity or cussedness.

Essentially four parts.

  1. Express the address you want as an integer of the appropriate size.
  2. Cast the integer into a pointer to some type.
  3. Dereference the pointer to get a value of that type.
  4. Print the value.
  5. [if desired, increment address and repeat.]

For simplicity I'll use an address of 0x1000 and integer contents.

int address = 0x1000;
int* pcontent = (int*)address;
int content = *pcontent;
printf ("Address %p: content %08x\n", pcontent, content);

Two points should be made.

  1. This is undefined behaviour, but inevitable if you want to get a result.
  2. The address I chose could be valid memory or could trigger a trap error or it could do just about anything. You'll have to experiment with different addresses to find out which.

[Many years ago I used this strategy to print out the entire contents of memory on a machine that had just 96KB of memory, which led to some interesting hacking possibilities. But I digress.]

Upvotes: 5

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

You have to define pointer p as

char *p = buf;

or to use casting then incrementing p. For example

p = ( char *)p + n;

For example 

printf( "%s\n", ( char *)p + 5 );

You may not increment/decrement pointers of type void * because void in incomplete type.

Also the format specifiers in these calls

printf("Address: %X\n", &buf);
printf("Contents: %c\n", p);

are wrong.

If you want to print out a string you have to specify %s. if you wnat to print only one character then you should write

char *p = buf;
printf("Contents: %c\n", p[0]);

or 

void *p = buf;
printf("Contents: %c\n", ( ( char *)p )[0]);

Upvotes: 1

jspurim
jspurim

Reputation: 955

You could try something like this:

void printMemory(void* mem, int bytes){
    char* p = (char*) mem;
    for(int i=0;i<bytes;i++){
        printf("%x ",p[i] & 0xff);
        if(i%8 == 7)
            printf("\n");
    }
}

This will print 8 bytes per row.

Upvotes: 0

haccks
haccks

Reputation: 106012

You are using wrong specifier to print the address. Use %p and cast argument to void *. 

printf("Address: %p\n", (void*)buf);

and to print string 

printf("Contents: %s\n", p);

Upvotes: 3

Bill Lynch
Bill Lynch

Reputation: 81926

If we wanted to print random bytes from around memory, we would want to use:

char *p = buf;

printf("Address:  %p\n", p);
printf("Contents: 0x%02x\n", *p);

Let's note though, that given a random address, dereferencing it can cause the application to fail. You can not assume that all of the memory space is valid.

Upvotes: 0

Dipika
Dipika

Reputation: 584

You are using %c instead use %s to print string printf("Contents: %s\n", p);

Upvotes: 0

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