What does MOV EAX, DWORD PTR DS:[ESI] mean and what does it do?

Solution 1:

Registers in square brackets such as [ESI] are dereferenced pointers. The instruction you quote moves the DWORD (a 32-bit/4-byte value) in memory location specified by ESI into register EAX. In your case, memory location 00402050, read as a DWORD, contains 34333231.

Written in pseudo-C:

DWORD EAX;   /* Declaring the registers as we find them in silico */
DWORD ESI;

ESI = 0x00402050;  /* Set up your initial conditions for ESI */
EAX = *((DWORD *)ESI);   /* mov EAX, DWORD PTR [ESI] */
/*  ^ ^  ^^^^^^^    */
/*  | |     |       */
/*  | |     +-----------  From "DWORD PTR" we get "DWORD *" in C.          */
/*  | |             */ 
/*  | +-----------------  The C dereferencing operator * replaces [].      */
/*  |               */ 
/*  +-------------------  The C assignment operator = replaces mov opcode. */ 

In your case, it is not true that 0x00402050 "equals" the string "1234567890" -- rather it points to the memory which contains that string.

The value which you obtain, 0x34333231 is comprised from the ASCII values for the digits "1234", which are the first four bytes (i.e., the first DWORD) of the string. They appear in reversed order because the Intel architecture is "little endian" in the byte representation of a DWORD in memory.

In your example at this time, the mov instruction is loading ASCII characters as if they were the four bytes of an unsigned long value, when they are actually a string of single-byte characters.