CODEWITHSUNDEEP
c64-bit
Samboy786
Samboy786

Reputation: 101

A win32 vs x64 code comparison

I was playing around with some code for a while :

    int n;
    char *ptr;
    scanf("%d",&n); 
    ptr = (char *) &n; // pointer to the integer
    printf("\na[0]=%p",*ptr); // print the values at the next 4 memory locations
    printf("\na[1]=%p",*(ptr+1));
    printf("\na[2]=%p",*(ptr+2));
    printf("\na[3]=%p",*(ptr+3));

this code gives me a clean output of 

FFFFFFFF
FFFFFFFF
FFFFFFFF
FFFFFFFF

for a input of -1 on win32 settings in VisualC++ '12.

However when I compile this code for a x64 architecture i get -

00000000FFFFFFFF
00000000FFFFFFFF
00000000FFFFFFFF
00000000FFFFFFFF

for the same input value of -1. The size of integer is same across both the architectures. My question is why the memory layout is not all F's throughout the 4 locations ?

Upvotes: 1

Views: 195

Answers (2)

chux
chux

Reputation: 153507

%p and *ptr not behaving as OP expects.

printf("\na[0]=%p",*ptr); // print the values at the next 4 memory locations

ptr has the value of address of int n. Note: The type of ptr is still a pointer to a char. *ptr says to dereference ptr getting a char. This char has the 8-bit value of 0xFF as it is one of the bytes of n, which is certainly 0xFFFFFFFF (-1). This char is promoted to an int as it is a varidac parameter to printf(). Thus 0xFF becomes 0xFFFFFFFF as int is likely a 4-byte integer on your machine. printf() sees the %p format specifier, thus expecting to see a 4-byte pointer in "VisualC++ '12". So that is you output. On "x64", an 8-byte pointer is expected, but only 4 bytes are defined (the 0xFFFFFFFF). The next 4 byte it grabbed are undefined but turn out to be 0x00000000. Thus you get "00000000FFFFFFFF".

The next line printf("\na[1]=%p",*(ptr+1)) simple gets the next char of n, which is also 0xFF. And then the same promotion to int with a value of 0xFFFFFFFF occurs.

Suspect the OP wants

printf("\na[0] = 0x%08X",*  ((int *)ptr)  ); // print the values at the next 4 memory locations

Upvotes: 1

Michael F
Michael F

Reputation: 40829

The p specifier of printf expects a void *. Instead, you're passing an int (exercise: why is it an int?), therefore the behaviour is undefined. Use a correct specifier for what you want to print (c, x or u, perhaps?).

Upvotes: 3

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