在使用跨平台GCC for ARM CPU在Ubuntu上编译C代码时，对于每个汇编指令，源数据将从堆栈加载，结果将写入堆栈。为什么GCC不使用更多寄存器来减少加载/存储数据的时间？ 是因为ARM是加载和存储结构，GCC会对每个汇编指令的数据使用加载/存储吗？
以下是我的冒泡排序功能：
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;
}
以下是使用Ubuntu上的跨平台GCC工具从上面的C代码编译的汇编代码：
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/存储的时间：
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 "
}
答案 0 :(得分：2)
使用优化-O3，gcc将使用更多寄存器。
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