To provide more of a visualization of Matt's answer, here's a good diagram of how MIPS architecture handles their operations. This is a 1-bit ALU:

Picture can be found in the textbook Computer Organization and Design MIPS Edition: The Hardware/Software Interface by Patterson and Hennessy.

So you can see the "Operation" bit(s) that "selects" the operation that multiplexes a particular function that can be performed between "a" and "b": add, AND, OR, less than, subtract, etc. If we invert "a" or "b", as shown with "Ainvert" and "Binvert", we can perform a subtraction. There are other operations like overflow or zero detection. More operations can be accomplished, which has been proven with more modern architectures but this is simply just an example of a 32-bit MIPS ALU:

Picture from the same source as mentioned above.

So basically in order to get a result from two numbers, you are first choose an operation, e.g. 0x02 for addition shown here, then the binary numbers for "a" and "b", e.g. a[0:31] = 111010...001.

To provide more of a visualization of Matt's answer, here's a good diagram of how MIPS architecture handles their operations. This is a 1-bit ALU:

Picture can be found in the textbook Computer Organization and Design MIPS Edition: The Hardware/Software Interface by Patterson and Hennessy.

So you can see the "Operation" bit(s) that "selects" the operation that multiplexes a particular function that can be performed between "a" and "b": add, AND, OR, less than, subtract, etc. If we invert "a" or "b", as shown with "Ainvert" and "Binvert", we can perform a subtraction. There are other operations like overflow or zero detection. More operations can be accomplished, which has been proven with more modern architectures but this is simply just an example of a 32-bit MIPS ALU:

Picture from the same source as mentioned above.

So basically in order to get a result from two numbers, you are first choose an operation, e.g. 0x02 for addition shown here, then the binary numbers for "a" and "b", e.g. a[0:31] = 111010...001 and b[0:31] = 101010...010, and then it will output the result, e.g. 010 (for add) 111010...001 (a), 101010...010 (b) --> (1)010100...011 stored in memory but oh wait, there's an overflow with the extra (1)... Raise that flag This is pretty much how your Assembly works.

To provide more of a visualization of Matt's answer, here's a good diagram of how MIPS architecture handles their operations. This is a 1-bit ALU:

Picture can be found in the textbook Computer Organization and Design MIPS Edition: The Hardware/Software Interface by Patterson and Hennessy.

So you can see the "Operation" bit(s) that "selects" the operation that multiplexes a particular function that can be performed between "a" and "b": add, AND, OR, less than, subtract, etc. If we invert "a" or "b", as shown with "Ainvert" and "Binvert", we can perform a subtraction. There are other operations like overflow or zero detection. More operations can be accomplished, which has been proven with more modern architectures but this is simply just an example of a 32-bit MIPS ALU:

Picture from the same source as mentioned above.

To provide more of a visualization of Matt's answer, here's a good diagram of how MIPS architecture handles their operations. This is a 1-bit ALU:

Picture can be found in the textbook Computer Organization and Design MIPS Edition: The Hardware/Software Interface by Patterson and Hennessy.

So you can see the "Operation" bit(s) that "selects" the operation that multiplexes a particular function that can be performed between "a" and "b": add, AND, OR, less than, subtract, etc. If we invert "a" or "b", as shown with "Ainvert" and "Binvert", we can perform a subtraction. There are other operations like overflow or zero detection. More operations can be accomplished, which has been proven with more modern architectures but this is simply just an example of a 32-bit MIPS ALU:

Picture from the same source as mentioned above.

So basically in order to get a result from two numbers, you are first choose an operation, e.g. 0x02 for addition shown here, then the binary numbers for "a" and "b", e.g. a[0:31] = 111010...001.