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Help understanding part of this generated assembly code


By : ivester
Date : October 15 2020, 01:20 AM
I wish this help you cmpl $0x0,-0x4(%esp); setg %al means compare -0x4(%esp) (floop_id in your code) against 0, and set %al to 1 if it's greater, or 0 otherwise.
test %al, %al here isn't doing anything. I don't know why it's in the assembly. (Normally, testing a value with itself is used to get the signum of the value (i.e., zero, positive, or negative), but the result of this isn't being used here. Chances are, it was going to do a conditional branch (to implement the loop), but seeing as your loop is empty, it got removed.)
code :


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Understanding optimized assembly code generated by gcc


By : Riber
Date : March 29 2020, 07:55 AM
This might help you That instruction is used to fill space for alignment purposes. Loops can be faster when they start on aligned addresses, because the processor loads memory into the decoder in chunks. By aligning the beginnings of loops and functions, it becomes more likely that they will be at the beginning of one of these chunks. This prevents previous instructions which will not be used from being loaded, maximizes the number of future instructions that will, and, possibly most importantly, ensures that the first instruction is entirely in the first chunk, so it does not take two loads to execute it.
The compiler knows that it is best to align the loop, and has two options to do so. It can either place a jump to the beginning of the loop, or fill the gap with no-ops and let the processor flow through them. Jump instructions break the flow of instructions and often cause wasted cycles on modern processors, so adding them unnecessarily is inadvisable. For a short distance like this no-ops are better.

Understanding assembly generated by C function call


By : Ashutosh Dwivedi
Date : March 29 2020, 07:55 AM
should help you out , Ok, now it does something, I'll annotate it for you
code :
calc:
    pushl   %ebp           ; \
    movl    %esp, %ebp     ; /  set up basic stack frame
    movl    8(%ebp), %edx  ; load x
    movl    16(%ebp), %ecx ; load z
    leal    (%edx,%edx,2), %edx ; calculate x + 2 * x
    movl    12(%ebp), %eax ; load y
    leal    (%edx,%eax,2), %eax ; calculate (x + 2 * x) + (2 * y)
    movl    %ecx, %edx     ; make a temp copy of z
    sall    $4, %edx       ; calculate z * 16
    subl    %ecx, %edx     ; calculate (z * 16) - z
    addl    %edx, %eax     ; calculate final sum
    popl    %ebp
    ret

Understanding this part arm assembly code


By : Hetal Mehta
Date : March 29 2020, 07:55 AM
Does that help The instructions starting with a . are really assembler directives. You can look them up in GAS: ARM machine directives .syntax unified signals the use of unified ARM / Thumb assembly syntax. The concept is explained here and here. .thumb_func signals the start of a Thumb mode function for ARM-Thumb interwork. The concept is explained here and here. raise_privilege looks exactly like a void raise_privilege(void) leaf function (i.e. it doesn't call other functions) in C to me. Call it with:
code :
bl raise_privilege

Need help in understanding basic Assembly code which was generated from C code


By : Tok Uban
Date : March 29 2020, 07:55 AM
help you fix your problem You are compiling with optimizations enabled. GCC was smart enough to perform all of those calculations at compile-time, and replace all of that useless code with a simple constant.
First of all, x and y will get replaced with their constant expressions:
code :
    int x = 10;
    int y = 10;
    int firstArg = 20;
    int secondArg = 10;
    value = firstArg + secondArg;
int main(int argc, char *argv[]) {
    return 30;
}
main:
    movl    $30, %eax
    ret
add:
    movl    8(%esp), %eax
    addl    4(%esp), %eax
    ret

Compilers: Understanding assembly code generated from small programs


By : Nguyen Thanh Tung
Date : March 29 2020, 07:55 AM
wish of those help Regarding the second question, since the C99 standard it's allowed to not have an explicit return 0 in the main function, the compiler will add it implicitly. Note that this is only for the main function, no other function.
As for the third question, the rbp register acts as the frame pointer.
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