GCC asm內聯約束，衝突的寄存器分配

我製作了一些ARM內聯彙編代碼。
在Semaphore.s中，我看到gcc使用寄存器r3來表示兩個變量：「success」和「change」。我想知道我的限制是否有問題？
首先最相關的行代碼：GCC asm內聯約束，衝突的寄存器分配

ASM直列：

"1: MVN %[success], #0 @ success=TRUE=~FALSE\n\t" 
"LDREX %[value], %[signal] @ try to get exclusive access\n\t" 
"ADDS %[newValue], %[value], %[change] @ new value = value + change\n\t"

約束：

: [signal] "+m" (signal), [success] "=r" (success), [locked] "=r" (locked), [newValue] "=r" (newValue), [value] "=r" (value) 
: [borderValue] "r" (borderValue), [change] "r" (change) 
: "cc"

符號文件：

1: MVN r3, #0 @ success=TRUE=~FALSE 
LDREX r0, [r7, #12] @ try to get exclusive access 
ADDS r1, r0, r3 @ new value = value + change

更多源和生成的符號如下。從符號文件

BOOLEAN Semaphore_exclusiveChange (INT32U * signal, INT32S change, INT32U borderValue) 
{ 
BOOLEAN success; 
INT32U locked;// exclusive status 
INT32U newValue; 
INT32U value; 

asm (
"1: MVN %[success], #0 @ success=TRUE=~FALSE\n\t" 
"LDREX %[value], %[signal] @ new to get exclusive access\n\t" 
"ADDS %[newValue], %[value], %[change] @ new value = value + change\n\t" 
"ITE MI @ if (new value<0) \n\t" 
" SUBSMI %[newValue], %[newValue] @ (new value<0): new value=0, set zero flag \n\t" 
"@ else\n\t" 
" CMPPL %[newValue], %[borderValue] @ (new value>=0): if new value > border value \n\t" 
"\n\[email protected] zero flag is either: new value=0 or =bordervalue\n\t" 
"ITE HI @ if new signal level > border value \n\t" // 
" MOVHI  %[success], #0 @ fail to raise signal, success=FALSE \n\t" 
"\[email protected] else\n\t" 
" MOVLS  %[value], %[newValue] @ use new value \n\t" // ok 
"STREX %[locked], %[value], %[signal] @ new exclusive store of value\n\t" 
"TST %[locked],%[locked] @ is locked? \n\t" 
"IT NE @ if locked \n\t" 
"BNE 1b @ try again\n\t" 
"DMB @ memory barrier\n\t" // 

: [signal] "+m" (signal), [success] "=r" (success), [locked] "=r" (locked), [newValue] "=r" (newValue), [value] "=r" (value) 
: [borderValue] "r" (borderValue), [change] "r" (change) 
: "cc" ); 
return success; 
}

相關的文字：

Semaphore_exclusiveChange: 
.LFB2: 
    .loc 1 10 0 
    @ args = 0, pretend = 0, frame = 32 
    @ frame_needed = 1, uses_anonymous_args = 0 
    @ link register save eliminated. 
    push {r7} 
.LCFI0: 
    sub sp, sp, #36 
.LCFI1: 
    add r7, sp, #0 
.LCFI2: 
    str r0, [r7, #12] 
    str r1, [r7, #8] 
    str r2, [r7, #4] 
    .loc 1 16 0 
    ldr r2, [r7, #4] 
    ldr r3, [r7, #8] 
@ 16 "../drivers/Semaphore.c" 1 
    1: MVN r3, #0 @ success=TRUE=~FALSE 
    LDREX r0, [r7, #12] @ new to get exclusive access 
    ADDS r1, r0, r3 @ new value = value + change 
    ITE MI @ if (new value<0) 
    SUBSMI r1, r1 @ (new value<0): new value=0, set zero flag 
    @ else 
    CMPPL r1, r2 @ (new value>=0): if new value > border value 

    @ zero flag is either: new value=0 or =bordervalue 
    ITE HI @ if new signal level > border value 
    MOVHI  r3, #0 @ fail to raise signal, success=FALSE 
     @ else 
    MOVLS  r0, r1 @ use new value 
    STREX r2, r0, [r7, #12] @ new exclusive store of value 
    TST r2,r2 @ is locked? 
    IT NE @ if locked 
    BNE 1b @ try again 
    DMB @ memory barrier 

@ 0 "" 2 
    .thumb 
    strb r3, [r7, #19] 
    str r2, [r7, #20] 
    str r1, [r7, #24] 
    str r0, [r7, #28] 
    .loc 1 38 0 
    ldrb r3, [r7, #19] @ zero_extendqisi2 
    .loc 1 39 0 
    mov r0, r3 
    add r7, r7, #36 
    mov sp, r7 
    pop {r7} 
    bx lr

來源

2011-04-11 Morten Nedertoft

我編輯了你的問題來解決一個非常小的代碼格式問題。 – 2011-04-11 12:45:44

需要約束「成功」進一步 '&'：

: [signal] "+m" (signal), [success] "=&r" (success), [locked] "=r" (locked), [newValue] "=r" (newValue), [value] "=r" (value)

這標誌着它作爲一個 '早期撞'。否則，編譯器將假定所有輸出都是在所有輸入被消耗後產生的，並且可以自由地將相同的寄存器用於不同的輸出和輸入。如果您有「輸入/輸出」值，則需要使用「重複值」約束。

來源

2011-04-11 12:45:08 hirschhornsalz

GCC asm內聯約束，衝突的寄存器分配

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