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What happens when you use an incorrect load/store instruction

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• Important warning
  • Warning:

    You must use the correct load and store instructions to tranfer data in the (memory) variables to the registers inside the CPU !!! Use When the data type of the variable is ------------ ------------------------------------------ ldrsb/strb byte ldrsh/strh short ldr /str int Your program will contain a bug (= program error) if you use an incorrect load/store instruction

    In this webpage, I will show you what will happen when you use an incorrect load/store instruction (and quiz you if you can understand the cause of the error)

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• Prelude to "understanding what happens when you use an incorrect load instruction"
  • Fact:

    A computer is a machine that does exactly what it is told to do In other words: A computer cannot think !! (Well, you may have head about Artificial Intelligence, that's not the computer. AI is a very large computer program that examines a lot of data using a computer - the computer itself is not intelligent) A computer is just as dumb as a door nail !!!

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• Review of the load instruction
  We have learned what the load instructions do:

  Syntax and meaning of the basic form of the load register instruction is:

  Syntax Meaning of the instruction ---------------- -------------------------------------------------- ldr rN, [rM] Load 4 bytes from memory at the address given in rM into the (destination) register rN The 4 bytes are stored in the register as follows: ldrsh rN, [rM] Load 2 bytes from memory at the address given in rM into the (destination) register rN AND fill the left half of register with the sign bit The 2 bytes are stored at the right and then sign extended to 32 bits: ldrsb rN, [rM] Load 1 byte from memory at the address given in rM into the (destination) register rN AND fill the left 3/4 of register with the sign bit The byte is stored at the right and then sign extended to 32 bits:

  Let's see how well you can apply your understanding...

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• What happens when you use an incorrect load instruction
  • A program contains 2 short variables x and y stored as follows:

    In the following program, I transfer the value stored at the memory location x: into the registers using 3 different load instruction:

    main: // short x contains 254 = 0000000011111110 movw r0, #:lower16:x // Moves the address of memory movt r0, #:upper16:x // variable x into register r0 // Move short x into r1 using ldrsh (correct !) ldrsh r1,[r0] // r1 = 254, correct // Move short x into r2 using ldrsb (wrong value !) ldrsb r2,[r0] // r2 = -1, WRONG !! // Can you explain why r2 contains -1 ? // Move short x into r3 using ldr (wrong value !) ldr r3,[r0] // r2 = 65790 = 216 + 254, WRONG !! // Can you explain why r2 contains 216 + 254 ? /* -------------------------------------------------- Begin of the permanent program variables -------------------------------------------------- */ .data x: .2byte 254 // short typed variable y: .2byte 1 // short typed variable .end

    The ldrsh instruction will transfer the correct value (254) into the register r1

    But::

    The ldrsb instruction stores a wrong value −2 The ldr instruction stores a wrong value 65790

    Can you explain why using what you know about these instructions (see above) ???

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  • Example Program: (Demo above code)                                                
    • Prog file: /home/cs255001/demo/asm/2-mov/wrong-ldr.s

    How to run the program:

    To compile:   as255 wrong-ldr          To run: use EGTAPI

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  • Why does the ldrsb instruction store −2 into a register:

    The ldrsb instruction will load a byte from memory location x: and sign-extend the byte into 32 bits: So the register will contain the 2s complement code 11111111111111111111111111111110 which represents the value −2 !!!

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  • Why does the ldr instruction store 65790 = 2¹⁶ + 254 into a register:

    The ldr instruction will load a int (= 4 bytes) from memory location x: which will read too many bytes: The ldr instructions has also read the bytes from the variable y into the register !!! So the register will contain the 2s complement code 00000000000000010000000011111110 which represents the value 216+254 = 65790 !!!

   

   

   

   

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• Review of the store instruction
  We have learned what the store instructions do:

  Syntax and meaning of the basic form of the store register instruction

  Syntax Meaning of the instruction ---------------- -------------------------------------------------- str rN, [rM] Store 4 bytes from register rN to memory (consecutively) at the address given in register rM The 4 bytes are taken from the register as follows: strh rN, [rM] Store 2 bytes from register rN to memory (consecutively) at the address given in register rM The 2 bytes are taken from the register as follows: strb rN, [rM] Store 1 byte from register rN to memory (consecutively) at the address given in register rM The byte is taken from the register as follows:

  Let's see how well you can apply your understanding...

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• What happens when you use an incorrect store instruction
  • The next program contains the following 2 short variables x and y:

    In the main( ) method, I will update the memory variable at the memory location x: with the value -2 (stored in the register r1) using 3 different store instructions:

    main: movw r0, #:lower16:x // Moves the address of memory movt r0, #:upper16:x // variable x into register r0 mov r1, #-2 // Move -2 into short var x (correct !) strh r1,[r0] // x = -2, correct // Change strh to strb and you will get x = 254 !! // Can you explain why x contains 254 ? // Change strh to str and you will update y = -1 !! // Can you explain why y gets updated to -1 ? /* -------------------------------------------------- Begin of the permanent program variables -------------------------------------------------- */ .data x: .2byte 4 // a short typed variable y: .2byte 7 // a short typed variable .end

    You need to edit the program and recompile and run again to see the following results:

    The strh instruction will update the variable x with the correct value -2 and does not mistakenly update other variable But:: The strb instruction stores a wrong value 254 in the variable x The str instruction will also update another variable y (sets the variable y = -1 !!!)

    Can you explain why using what you know about these store instructions (see above) ???

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  • Example Program: (Demo above code)                                                
    • Prog file: /home/cs255001/demo/asm/2-mov/wrong-str.s

    How to run the program:

    To compile:   as255 wrong-str          To run: use EGTAPI

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  • Why does the strb instruction store 254 into the variable x:

    The strb instruction will update a byte value in register r1 into the memory location x:: So the short typed memory variable x will contain the 2s complement code 0000000011111110 which represents the value 254 (decimal) !!!

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  • Why does the str instruction also update the variable y = -1:

    The str instruction will transfer a int (= 4 bytes) from a register to the memory location x:. The 4 bytes are stored in memory as follows: So the str instructions has also updated the bytes of the variable y in the memory !!! The short variable y will contain the 2s complement code 1111111111111111 which represents the value −1 !!!