Reading (simple) assembly to find compile-time calculation resulting from using C++ templates

Question

I am in the process of comparing Fortran 90 vs C++ for a presentation. One of my comparisons relies on the assembly generated for simple programs by g++ and gfortran .

One example reads as follows:

#include<cstdio> // quick and dirty number formatting

template<int N>
double dot(double x[], double y[]){
  return x[N-1] * y[N-1] + dot<N-1>(x, y);
}

template<>
double dot<1>(double x[], double y[]){
  return x[0] * y[0];
}

int main(){
  double x[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
  double y[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
  printf("x.y = %23.16E\n", dot<10>(x, y));
}

The following assembler is generated by the command g++ -S -O3 myprogram.cpp using g++ 4.7.2 on OS X 10.7.4 x86_64-apple-darwin11.4.2:

    .text
    .align 4,0x90
    .globl __Z3dotILi1EEdPdS0_
__Z3dotILi1EEdPdS0_:
LFB2:
    movsd   (%rdi), %xmm0
    mulsd   (%rsi), %xmm0
    ret
LFE2:
    .cstring
LC1:
    .ascii "x.y = %23.16E\12\0"
    .section __TEXT,__text_startup,regular,pure_instructions
    .align 4
    .globl _main
_main:
LFB3:
    leaq    LC1(%rip), %rdi
    subq    $8, %rsp
LCFI0:
    movl    $1, %eax
    movsd   LC0(%rip), %xmm0
    call    _printf
    xorl    %eax, %eax
    addq    $8, %rsp
LCFI1:
    ret
LFE3:
    .literal8
    .align 3
LC0:
    .long   0
    .long   1081610240
    .section __TEXT,__eh_frame,coalesced,no_toc+strip_static_syms+live_support
EH_frame1:
    .set L$set$0,LECIE1-LSCIE1
    .long L$set$0
LSCIE1:
    .long   0
    .byte   0x1
    .ascii "zR\0"
    .byte   0x1
    .byte   0x78
    .byte   0x10
    .byte   0x1
    .byte   0x10
    .byte   0xc
    .byte   0x7
    .byte   0x8
    .byte   0x90
    .byte   0x1
    .align 3
LECIE1:
LSFDE1:
    .set L$set$1,LEFDE1-LASFDE1
    .long L$set$1
LASFDE1:
    .long   LASFDE1-EH_frame1
    .quad   LFB2-.
    .set L$set$2,LFE2-LFB2
    .quad L$set$2
    .byte   0
    .align 3
LEFDE1:
LSFDE3:
    .set L$set$3,LEFDE3-LASFDE3
    .long L$set$3
LASFDE3:
    .long   LASFDE3-EH_frame1
    .quad   LFB3-.
    .set L$set$4,LFE3-LFB3
    .quad L$set$4
    .byte   0
    .byte   0x4
    .set L$set$5,LCFI0-LFB3
    .long L$set$5
    .byte   0xe
    .byte   0x10
    .byte   0x4
    .set L$set$6,LCFI1-LCFI0
    .long L$set$6
    .byte   0xe
    .byte   0x8
    .align 3
LEFDE3:
    .constructor
    .destructor
    .align 1
    .subsections_via_symbols

The dot product is 385, and it seems that it was calculated at compile time, but I cannot seem to find exactly where. I suspect it is somewhere in the following assembler segment:

    movl    $1, %eax
    movsd   LC0(%rip), %xmm0
    call    _printf
    xorl    %eax, %eax
    addq    $8, %rsp
LCFI1:
    ret
LFE3:
    .literal8
    .align 3
LC0:
    .long   0
    .long   1081610240
    .section __TEXT,__eh_frame,coalesced,no_toc+strip_static_syms+live_support

My (very, very limited) understanding of assembly, would tell me that the dot product was calculated by the compiler and placed in a register (LC0). Then the instruction movsd LC0(%rip), %xmm0 places the value in a string, and calls printf on the resulting, formatted string.

Is this the case? Is the actual number 385 included somewhere in this output, or is it calculated elsewhwere?

Thank you!

EDIT:

In case anybody wonders how the assembly produced by gfortran looks like, I am attaching it below. Notice that even though is known at compile time, and I'm using Fortran's intrinsic dot_product operator, the generated assembly is substantially larger (130 lines vs. 90 lines in the C++ version), and it seems that the optimizer is not able to reduce the operation.

Program (notice that I am using the intrinsic, built-in dot_product operator):

PROGRAM MAIN
  REAL(8), DIMENSION(10):: X = (/1, 2, 3, 4, 5, 6, 7, 8, 9, 10/)
  REAL(8), DIMENSION(10):: Y = (/1, 2, 3, 4, 5, 6, 7, 8, 9, 10/)
  PRINT "(A, E23.16)", "x.y = ", DOT_PRODUCT(X, Y)
ENDPROGRAM MAIN

Assembly (gfortran -S -O3 myprogram.cpp using gcc 4.7.2 on OS X 10.7.4 x86_64-apple-darwin11.4.2)

    .cstring
LC0:
    .ascii "dotproduct-intrinsic.f90\0"
    .const
LC1:
    .ascii "(A, E23.16)"
LC2:
    .ascii "x.y = "
    .text
    .align 4,0x90
_MAIN__:
LFB0:
    leaq    LC0(%rip), %rax
    subq    $504, %rsp
LCFI0:
    movq    %rax, 24(%rsp)
    leaq    16(%rsp), %rdi
    leaq    LC1(%rip), %rax
    movl    $5, 32(%rsp)
    movq    %rax, 88(%rsp)
    movl    $11, 96(%rsp)
    movl    $4096, 16(%rsp)
    movl    $6, 20(%rsp)
    call    __gfortran_st_write
    leaq    16(%rsp), %rdi
    movl    $6, %edx
    leaq    LC2(%rip), %rsi
    call    __gfortran_transfer_character_write
    leaq    8(%rsp), %rsi
    movl    $8, %edx
    movabsq $4645480607818711040, %rax
    leaq    16(%rsp), %rdi
    movq    %rax, 8(%rsp)
    call    __gfortran_transfer_real_write
    leaq    16(%rsp), %rdi
    call    __gfortran_st_write_done
    addq    $504, %rsp
LCFI1:
    ret
LFE0:
    .section __TEXT,__text_startup,regular,pure_instructions
    .align 4
    .globl _main
_main:
LFB1:
    subq    $8, %rsp
LCFI2:
    call    __gfortran_set_args
    leaq    _options.3.1864(%rip), %rsi
    movl    $8, %edi
    call    __gfortran_set_options
    call    _MAIN__
    xorl    %eax, %eax
    addq    $8, %rsp
LCFI3:
    ret
LFE1:
    .const
    .align 5
_options.3.1864:
    .long   68
    .long   1023
    .long   0
    .long   0
    .long   1
    .long   1
    .long   0
    .long   1
    .section __TEXT,__eh_frame,coalesced,no_toc+strip_static_syms+live_support
EH_frame1:
    .set L$set$0,LECIE1-LSCIE1
    .long L$set$0
LSCIE1:
    .long   0
    .byte   0x1
    .ascii "zR\0"
    .byte   0x1
    .byte   0x78
    .byte   0x10
    .byte   0x1
    .byte   0x10
    .byte   0xc
    .byte   0x7
    .byte   0x8
    .byte   0x90
    .byte   0x1
    .align 3
LECIE1:
LSFDE1:
    .set L$set$1,LEFDE1-LASFDE1
    .long L$set$1
LASFDE1:
    .long   LASFDE1-EH_frame1
    .quad   LFB0-.
    .set L$set$2,LFE0-LFB0
    .quad L$set$2
    .byte   0
    .byte   0x4
    .set L$set$3,LCFI0-LFB0
    .long L$set$3
    .byte   0xe
    .byte   0x80,0x4
    .byte   0x4
    .set L$set$4,LCFI1-LCFI0
    .long L$set$4
    .byte   0xe
    .byte   0x8
    .align 3
LEFDE1:
LSFDE3:
    .set L$set$5,LEFDE3-LASFDE3
    .long L$set$5
LASFDE3:
    .long   LASFDE3-EH_frame1
    .quad   LFB1-.
    .set L$set$6,LFE1-LFB1
    .quad L$set$6
    .byte   0
    .byte   0x4
    .set L$set$7,LCFI2-LFB1
    .long L$set$7
    .byte   0xe
    .byte   0x10
    .byte   0x4
    .set L$set$8,LCFI3-LCFI2
    .long L$set$8
    .byte   0xe
    .byte   0x8
    .align 3
LEFDE3:
    .subsections_via_symbols

Edit 2

Thanks to @JerryCoffin's answer , I am able to verify that, indeed, the bit pattern

LC0:
    .long   0
    .long   1081610240

found in the assembly output corresponds to the number 385.

I used the exact same program provided by @JerryCoffin , namely:

#include <stdio.h>


    #pragma pack(1)
    struct x {
        long x, y;
    };    

    int main() { 
        x v = {0, 1081610240};

        printf("%f\n", *(double *)&v);
        return 0;
    }

with the only caveat that I had to compile it using 32 bits target: g++ verification.cpp -m32 .

Answer 1

Yes -- this:

LC0:
    .long   0
    .long   1081610240

Is the bit pattern for a double with the value 385 ¹ . So what's happening is that this:

LCFI0:
    movl    $1, %eax
    movsd   LC0(%rip), %xmm0
    call    _printf

...is loading that value into an XMM register, then calling printf to print it out. I can't say for sure, but if I had to guess, I would say the 1 is telling it that it's passing one parameter to be printed out.

---

¹ In case you care to verify that, try this:

#include <stdio.h>

#pragma pack(1)
struct x {
    long x, y;
};    

int main() { 
    x v = {0, 1081610240};

    printf("%f\n", *(double *)&v);
    return 0;
}

Officially, of course, this is non-portable, etc., but with the same compiler on the same machine, chances are about 99% that you'll get the same output I did -- 385 .