ARMv8浮点输出内联汇编 [英] ARMv8 floating point output inline assembly

问题描述

要加两个整数，我写:

int sum;
asm volatile("add %0, x3, x4" : "=r"(sum) : :);

如何用两个浮子来做到这一点? 我试过了:

How can I do this with two floats? I tried:

float sum;
asm volatile("fadd %0, s3, s4" : "=r"(sum) : :);

但这给我一个错误:

错误:操作数1应该是SIMD向量寄存器-'fadd x0，s3，s4'

Error: operand 1 should be a SIMD vector register -- `fadd x0,s3,s4'

有什么想法吗?

推荐答案

ARMv7 double:%P修饰符

ARMv7 double: %P modifier

GCC开发人员通过 https通知我正确的ARMv7修饰符加倍. ://gcc.gnu.org/bugzilla/show_bug.cgi?id = 89482#c4 也许有一天我应该不再懒惰并grep GCC:

GCC devs informed me the correct undocumented modifier for ARMv7 doubles at https://gcc.gnu.org/bugzilla/show_bug.cgi?id=89482#c4 Maybe I should stop being lazy and grep GCC some day:

main.c

#include <assert.h>

int main(void) {
    double my_double = 1.5;
    __asm__ (
        "vmov.f64 d0, 1.0;"
        "vadd.f64 %P[my_double], %P[my_double], d0;"
        : [my_double] "+w" (my_double)
        :
        : "d0"
    );
    assert(my_double == 2.5);
}

编译并运行:

sudo apt-get install qemu-user gcc-arm-linux-gnueabihf
arm-linux-gnueabihf-gcc -O3 -std=c99 -ggdb3 -march=armv7-a -marm \
  -pedantic -Wall -Wextra -o main.out main.c
qemu-arm -L /usr/arm-linux-gnueabihf main.out

反汇编包含:

   0x00010320 <+4>:     08 7b b7 ee     vmov.f64        d7, #120        ; 0x3fc00000  1.5
   0x00010324 <+8>:     00 0b b7 ee     vmov.f64        d0, #112        ; 0x3f800000  1.0
   0x00010328 <+12>:    00 7b 37 ee     vadd.f64        d7, d7, d0

在Ubuntu 16.04，GCC 5.4.0，QEMU 2.5.0中进行了测试.

Tested in Ubuntu 16.04, GCC 5.4.0, QEMU 2.5.0.

源代码定义点

• ARM: https://github.com/gcc-mirror/gcc/blob/gcc-8_2_0-release/gcc/config/arm/arm.c#L22466
• aarch64: https://github.com/gcc-mirror/gcc/blob/gcc-8_2_0-release/gcc/config/aarch64/aarch64.c#L6743

• ARM: https://github.com/gcc-mirror/gcc/blob/gcc-8_2_0-release/gcc/config/arm/arm.c#L22466
• aarch64: https://github.com/gcc-mirror/gcc/blob/gcc-8_2_0-release/gcc/config/aarch64/aarch64.c#L6743

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