C / C++ / MFC

Notes:
· Do not use BITMAPCOREINFO or BIMAPCOREHEADER as these are for OS/2
compatibility only.
· Always follow the BITMAPINFOHEADER with the color table and only save 1, 4, or 8
bit DIBs
· Color table entries are RGBQUADs, which are stored in memory as BGR not RGB.
· For 24 bit color DIBs, the "pixels" are also stored in BGR order, not RGB.
· DIBs are stored ‘upside-down’ - the first pixel in the DIB is the lower-left corner of the
image, and the last pixel is the upper-right corner.
· DIBs can be larger than 64K, but be careful, a 24 bit pixel can straddle a 64K boundary!
· Pixels in 1, 4, and 8 bit DIBs are "always" color table indices, you must index through
the color table to determine the color value of a pixel.

jithAtran - ii wrote:

*p=2; // integer 2 is stored in 1000 and 1001 as(0000 0010)

jithAtran - ii wrote:

p=&j; // now doubt,here how 1000,1001 store the address of j

S. Senthil Kumar wrote:

jithAtran - ii wrote:
*p=2; // integer 2 is stored in 1000 and 1001 as(0000 0010)

I don't get what 1000 and 1001 are. If you consider them to be address of the pointer itself (&p), then your comment is incorrect. If you consider them to be the address where the pointer is pointing at, then your interpretation is correct. I'll assume you meant the latter.

S. Senthil Kumar wrote:

jithAtran - ii wrote:
p=&j; // now doubt,here how 1000,1001 store the address of j

All this does is change where p is pointing at. p will no longer hold 1000/1001, instead, it will hold whatever is the address of j (2000, 2001, for example). That explains why *p and j hold the same value

jithAtran - ii wrote:

the address of p is 1000/1001 and how it store the addresses 2000/2001 in bit format considering 1000 will hold 4 bits and 1001 will also hold 4 bits

S. Senthil Kumar wrote:

A byte is 8 bits

jithAtran - ii wrote:

Does the compiler holds anything like a stack to store the datatype?

char a = 'a';
char *pChar = &a;
void *pVoid = pChar;
int *pInt = (int *)pVoid;
*pInt = 12345;

int *i;
    static float *k;
    int j;
    i=&j;
    k=&j;//points to the same memory location
    printf("%d ,%f",*i,*k)

jithAtran - ii wrote:

I still have doubt in this part.How compiler knows i is an integer.where the command "You should read only 4 bytes for i since its an integer" is stored.

jithAtran - ii wrote:

Will it throw error if we performed like this.

float q;
i=&q;

S. Senthil Kumar wrote:

the compiler knows that i is a pointer to an integer

jithAtran - ii wrote:

That means i has been given an address space in the memory(1000/1001).
Now we are trying to store an address say

like
______________
__00__|___7D__ where 007D is the address of a
1000 1001 float variable.

Now here is my doubt.How compiler knows this assignment is wrong.

int *p = 0x007D;

float k = 10.0;
int *p = &k;

S. Senthil Kumar wrote:

the compiler allocates a certain number of bytes (depending on the platform) on the stack.

struct Data
{
    int number;
    int *pNumber;
};

// Code like this will allocate the memory for a "Data" struct on the stack.
// So the member "number" (not a pointer at all) and "pNumber" (is a pointer)
// will both be allocated on the stack.
void Function()
{
    Data IsOnTheStack;
    // Do stuff...
}

// Code like this will not use the stack.
// The memory will be in a data segment of the executable.
Data InDataSegment; // At global scope.

// Now code like this will place the struct (and its members) on the heap.
// The storage for the actual pointer "pData" may be on the heap, stack or
// a data segment in a module but the actual structure and it members are
// on the heap.
Data *pData = new Data;

Stephen Hewitt wrote:

This is not always the case: the pointer need not be allocated on the stack.