c - 为什么GCC在编译C代码时不使用更多寄存器

Question

在Ubuntu上使用ARM CPU的跨平台GCC编译C代码时，对于每条汇编指令，将从堆栈加载源数据并将结果写入堆栈。为什么GCC不使用更多的寄存器来减少加载/存储数据的时间？是不是因为ARM是load and store结构，GCC对每一条汇编指令的数据都会使用load/store的方式？

以下是我的冒泡排序函数:

bool bubble_sort(int* data, int dataSize) {
    int i;
    int j;
    int dataTmp;

    for(i=0; i<dataSize-1; i++) {
        for(j=0; j<dataSize-1; j++) {
            if(data[j] > data[j+1]) {
                dataTmp = data[j+1];
                data[j+1] = data[j];
                data[j] = dataTmp;
            }
        }
    }
    return true;
}

以下是上述C代码在Ubuntu上使用跨平台GCC工具编译的汇编代码:

00008d0c <bubble_sort>:
    8d0c:   e52db004    push    {fp}        ; (str fp, [sp, #-4]!)
    8d10:   e28db000    add fp, sp, #0
    8d14:   e24dd01c    sub sp, sp, #28
    8d18:   e50b0018    str r0, [fp, #-24]
    8d1c:   e50b101c    str r1, [fp, #-28]
    8d20:   e3a03000    mov r3, #0
    8d24:   e50b3010    str r3, [fp, #-16]
    8d28:   ea000032    b   8df8 <bubble_sort+0xec>
    8d2c:   e3a03000    mov r3, #0
    8d30:   e50b300c    str r3, [fp, #-12]
    8d34:   ea000027    b   8dd8 <bubble_sort+0xcc>
    8d38:   e51b300c    ldr r3, [fp, #-12]
    8d3c:   e1a03103    lsl r3, r3, #2
    8d40:   e51b2018    ldr r2, [fp, #-24]
    8d44:   e0823003    add r3, r2, r3
    8d48:   e5932000    ldr r2, [r3]
    8d4c:   e51b300c    ldr r3, [fp, #-12]
    8d50:   e2833001    add r3, r3, #1
    8d54:   e1a03103    lsl r3, r3, #2
    8d58:   e51b1018    ldr r1, [fp, #-24]
    8d5c:   e0813003    add r3, r1, r3
    8d60:   e5933000    ldr r3, [r3]
    8d64:   e1520003    cmp r2, r3
    8d68:   da000017    ble 8dcc <bubble_sort+0xc0>
    8d6c:   e51b300c    ldr r3, [fp, #-12]
    8d70:   e2833001    add r3, r3, #1
    8d74:   e1a03103    lsl r3, r3, #2
    8d78:   e51b2018    ldr r2, [fp, #-24]
    8d7c:   e0823003    add r3, r2, r3
    8d80:   e5933000    ldr r3, [r3]
    8d84:   e50b3008    str r3, [fp, #-8]
    8d88:   e51b300c    ldr r3, [fp, #-12]
    8d8c:   e2833001    add r3, r3, #1
    8d90:   e1a03103    lsl r3, r3, #2
    8d94:   e51b2018    ldr r2, [fp, #-24]
    8d98:   e0823003    add r3, r2, r3
    8d9c:   e51b200c    ldr r2, [fp, #-12]
    8da0:   e1a02102    lsl r2, r2, #2
    8da4:   e51b1018    ldr r1, [fp, #-24]
    8da8:   e0812002    add r2, r1, r2
    8dac:   e5922000    ldr r2, [r2]
    8db0:   e5832000    str r2, [r3]
    8db4:   e51b300c    ldr r3, [fp, #-12]
    8db8:   e1a03103    lsl r3, r3, #2
    8dbc:   e51b2018    ldr r2, [fp, #-24]
    8dc0:   e0823003    add r3, r2, r3
    8dc4:   e51b2008    ldr r2, [fp, #-8]
    8dc8:   e5832000    str r2, [r3]
    8dcc:   e51b300c    ldr r3, [fp, #-12]
    8dd0:   e2833001    add r3, r3, #1
    8dd4:   e50b300c    str r3, [fp, #-12]
    8dd8:   e51b301c    ldr r3, [fp, #-28]
    8ddc:   e2432001    sub r2, r3, #1
    8de0:   e51b300c    ldr r3, [fp, #-12]
    8de4:   e1520003    cmp r2, r3
    8de8:   caffffd2    bgt 8d38 <bubble_sort+0x2c>
    8dec:   e51b3010    ldr r3, [fp, #-16]
    8df0:   e2833001    add r3, r3, #1
    8df4:   e50b3010    str r3, [fp, #-16]
    8df8:   e51b301c    ldr r3, [fp, #-28]
    8dfc:   e2432001    sub r2, r3, #1
    8e00:   e51b3010    ldr r3, [fp, #-16]
    8e04:   e1520003    cmp r2, r3
    8e08:   caffffc7    bgt 8d2c <bubble_sort+0x20>
    8e0c:   e3a03001    mov r3, #1
    8e10:   e1a00003    mov r0, r3
    8e14:   e28bd000    add sp, fp, #0
    8e18:   e8bd0800    ldmfd   sp!, {fp}
    8e1c:   e12fff1e    bx  lr

以下是我使用内联汇编编写的代码，以使用更多寄存器来减少加载/存储时间:

void bubble_sort(int *data, int size) {
    //r2 = i
    //r3 = j
    asm volatile(
"       STMDB SP!, {r0-r7}  \n\t"
"                               "
"       SUB r1, r1, #1      \n\t" // r3 = size-1
"       LSL r1, r1, #2      \n\t" // int
"                               "
"       MOV r2, #0          \n\t" // r2 = i = 0
"                               "
"loop1:                     \n\t"
"       CMP r2, r1          \n\t" // r2 == r3: i == size-1
"       BEQ loop_end        \n\t"  
"                               "
"       MOV r3, #0          \n\t" // r3 = j = 1
"       MOV r4, r0          \n\t" // r4 = data
"       ADD r5, r4, #4      \n\t" // r5 = data + 4
"                               "
"loop2:                     \n\t"
"       CMP r3, r1          \n\t" // r3 == r1: j==size
"       BEQ loop1_end       \n\t"  
"                               "
"       LDR r6, [r4], #0    \n\t" //data[j-1]
"       LDR r7, [r5], #0    \n\t" //data[j]
"       CMP r6, r7          \n\t"
"       BLE incr_addr       \n\t"
"                               "
"       STR r6, [r5], #4    \n\t" //swp data, address + 4
"       STR r7, [r4], #4    \n\t"
"       B   loop2_end       \n\t"
"                               "
"incr_addr:                 \n\t"
"       ADD r4, r4, #4      \n\t" // r4 = r4 + 4
"       ADD r5, r5, #4      \n\t" // r5 = r5 + 4
"                               "
"loop2_end:                 \n\t"
"       ADD r3, r3, #4      \n\t"
"       B   loop2           \n\t"
"                               "
"loop1_end:                 \n\t"
"       ADD r2, r2, #4      \n\t"
"       B   loop1           \n\t"
"                               "
"loop_end:                  \n\t"
"                               "
"       LDMIA SP!,{r0-r7}       "
    );
    //back to caller
    //"      BX   lr              "
}

Answer 1

使用优化-O3，gcc将使用更多寄存器。

00008e10 <bubble_sort>:
    8e10:   e92d0030    push    {r4, r5}
    8e14:   e2414001    sub r4, r1, #1
    8e18:   e3540000    cmp r4, #0
    8e1c:   da00000d    ble 8e58 <bubble_sort+0x48>
    8e20:   e080c101    add ip, r0, r1, lsl #2
    8e24:   e2805004    add r5, r0, #4
    8e28:   e3a00000    mov r0, #0
    8e2c:   e1a03005    mov r3, r5
    8e30:   e5131004    ldr r1, [r3, #-4]
    8e34:   e4932004    ldr r2, [r3], #4
    8e38:   e1510002    cmp r1, r2
    8e3c:   c5031004    strgt   r1, [r3, #-4]
    8e40:   c5032008    strgt   r2, [r3, #-8]
    8e44:   e153000c    cmp r3, ip
    8e48:   1afffff8    bne 8e30 <bubble_sort+0x20>
    8e4c:   e2800001    add r0, r0, #1
    8e50:   e1500004    cmp r0, r4
    8e54:   1afffff4    bne 8e2c <bubble_sort+0x1c>
    8e58:   e3a00001    mov r0, #1
    8e5c:   e8bd0030    pop {r4, r5}
    8e60:   e12fff1e    bx  lr