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C++ to Assembly

  1. Jun 28, 2008 #1
    Hi all,

    I have this small function in C++

    void myfunc(int a, int b) {
    int c = 1;
    int d = 2;
    c += b;
    d += a;


    the assembly code and my comments follow:

    subl $8, %esp ;; subtract 8 from %esp, what we are doing here is decrementing the stack pointer by 8 and then writing the value at the new top of stack address, used for allocating space for local variables.

    movl $1, -4(%ebp) ;; copy 1 (which is the values stored in y) at the location using %ebp as the base address with an offset -4, the register is a pointer, the displacement specified how far from the pointer

    movl $2, -8(%ebp) ;; copy 2 (which is the value stored in z) at the location using %ebp as the base address with an offset -8, the register is a pointer, the displacement specified how far from the pointer

    movl 12(%ebp), %edx ;; copy what is at %ebp + offset 12 into %edx, what this means that the last instruction pushed to the stack is copied to %edx

    leal -4(%ebp), %eax ;; this is a variant of movl and instead of copying the data at %ebp + offset -4, its storing the effective address into the destination

    addl %edx, (%eax)

    movl 8(%ebp), %edx

    leal -8(%ebp), %eax

    addl %edx, (%eax)

    Am I correct in the comments of the lines, if not can someone please help me, also I am confused on leal. Thanks a bunch
  2. jcsd
  3. Jun 28, 2008 #2


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    Homework Helper

    Your comment on the first leal is correct, it's a load effective address instruction.

    Some of the assembly code is missing. There shoud be a push ebp, then a mov esp,ebp, before the first instruction you have. After this, [epb+0] = original ebp, [epb+4] = return address, [ebp+8] = first function parameter, [ebp+12] = second function parameter. [ebp-4] = first local variable, [ebp-8] = second local variable.

    Note that this syntax is reversed from the Intel standard, where the operands are ordered as destination, source. Also the "l" such as movl, aren't used in the Intel standard, since operand size is determined by register name or specific declartion (dx or word ptr for 16 bit, edx or dword ptr for 32 bit, rdx or qword ptr for 64 bit).

    Depending on the level of optimization, some of the function parameters are in registers instead on the stack. Microsoft has _fastcall convention as an option for 32 bit code. For 64 bit code in Microsoft enviroment the variations were done away with and the convention is similar to the _fastcall convention of 32 bit code, were the first 4 parameters are located in registers. Even though the parameters are located in registers, rsp is subtracted as if the paramters were passed on the stack, as a default place to store the parameters if the called function wishes to use the space.
  4. Jun 29, 2008 #3


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    Staff: Mentor

    This function doesn't in fact do anything, what is the sense of translating it to assembly? Or is it just an "art for art's sake" exercise?
  5. Jul 1, 2008 #4


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    great answer, through I'm only familar with the intel standard.
  6. Jul 2, 2008 #5

    jim mcnamara

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    Staff: Mentor

    Most C++ compilers can generate assembly code as output. To see what Jeff means by "missing" look into your compiler's manual to find the options to create asm as the endpoint of compilation.
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