. What value will EAX contain after the following instructions execute? .data dVal DWORD 12345678h .code mov ax,3 mov WORD PTR dVal+2,ax mov eax,dVal

The given assembly code is doing the following:

1. It first moves the value 3 into the AX register. The AX register is the lower 16 bits of the EAX register.

2. Then it moves the value in AX (which is 3) into the third and fourth bytes of the variable dVal. The original value of dVal is 12345678h, which in binary is 0001 0010 0011 0100 0101 0110 0111 1000. After the move operation, the third and fourth bytes (0100 0101) are replaced by the binary representation of 3 (0000 0011), so dVal becomes 0001 0010 0000 0011 0110 0111 1000.

3. Finally, it moves the value of dVal into the EAX register.

So, after these instructions execute, EAX will contain the value 12000378h.

The instructions are written in Assembly language. Here's a step-by-step explanation of what happens:

1. mov ax,3: This instruction moves the value 3 into the AX register. The AX register is the lower 16 bits of the EAX register.

2. mov WORD PTR dVal+2,ax: This instruction moves the value in the AX register (which is 3) into the memory location dVal+2. Since dVal is a DWORD (32 bits or 4 bytes), dVal+2 refers to the third and fourth bytes of dVal. The original value of dVal is 12345678h (in hexadecimal), which in binary is 0001 0010 0011 0100 0101 0110 0111 1000. After this instruction, the third and fourth bytes (counting from the right) are replaced with the binary representation of 3 (0000 0011), so dVal becomes 0001 0010 0000 0011 0111 1000 in binary, or 1200378h in hexadecimal.

3. mov eax,dVal: This instruction moves the value in dVal into the EAX register. So, after this instruction, EAX will contain 1200378h (in hexadecimal).