結論#
結論寫在前面
在 for 循環語句內定義局部循環變量，使用 AC6 編譯器或者 GCC 編譯器，都不會因此產生多次堆棧操作，而是會使用相同的兩個堆棧偏址。如果開優化，當二者邏輯功能沒有實際差異時，匯編將完全一樣。
事實上在 for 循環的同時定義循環變量是一個優秀的操作。對於將所有局部變量的定義全部提前到函數頭部，會產生事實上的負優化或無優化（依據優化等級和編譯器不同）。
由此引申，如果追求極致的性能，應當僅在使用到局部變量的分支中聲明該局部變量。
以下測試均以 stm32H7 為目標編譯
用如下的寫法討論局部變量的堆棧操作#
for(int i = 0; i < 50; i++)
{
for(int j = 0; j < 50; j++)
{
HAL_Delay(1);
}
}
感性來看，第一個循環每進行一次，都会聲明一個局部變量 j，那麼是否會由此產生多次的堆棧申請操作呢？
這部分的反匯編如下#
0x0000001e: LDR r0,[sp,#0]
0x00000020: STR r0,[sp,#8]
0x00000022: B {pc}+0x2 ; 0x24
0x00000024: LDR r0,[sp,#8]
0x00000026: CMP r0,#0x31
0x00000028: BGT {pc}+0x2c ; 0x54
0x0000002a: B {pc}+0x2 ; 0x2c
0x0000002c: MOVS r0,#0
0x0000002e: STR r0,[sp,#4]
0x00000030: B {pc}+0x2 ; 0x32
0x00000032: LDR r0,[sp,#4]
0x00000034: CMP r0,#0x31
0x00000036: BGT {pc}+0x14 ; 0x4a
0x00000038: B {pc}+0x2 ; 0x3a
0x0000003a: MOVS r0,#1
0x0000003c: BL HAL_Delay
0x00000040: B {pc}+0x2 ; 0x42
0x00000042: LDR r0,[sp,#4]
0x00000044: ADDS r0,#1
0x00000046: STR r0,[sp,#4]
0x00000048: B {pc}-0x16 ; 0x32
0x0000004a: B {pc}+0x2 ; 0x4c
0x0000004c: LDR r0,[sp,#8]
0x0000004e: ADDS r0,#1
0x00000050: STR r0,[sp,#8]
0x00000052: B {pc}-0x2e ; 0x24
外層循環#
不是我們的主要討論對象，反正就是會使用跳轉將內層循環執行 50 次
0x0000001e: LDR r0,[sp,#0]
0x00000020: STR r0,[sp,#8]
0x00000022: B {pc}+0x2 ; 0x24
0x00000024: LDR r0,[sp,#8]
0x00000026: CMP r0,#0x31
0x00000028: BGT {pc}+0x2c ; 0x54
0x0000002a: B {pc}+0x2 ; 0x2c
; .....內層循環
0x0000004a: B {pc}+0x2 ; 0x4c
0x0000004c: LDR r0,[sp,#8]
0x0000004e: ADDS r0,#1
0x00000050: STR r0,[sp,#8]
0x00000052: B {pc}-0x2e ; 0x24
內層循環#
0x0000002c: MOVS r0,#0
0x0000002e: STR r0,[sp,#4]
0x00000030: B {pc}+0x2 ; 0x32
0x00000032: LDR r0,[sp,#4]
0x00000034: CMP r0,#0x31
0x00000036: BGT {pc}+0x14 ; 0x4a
0x00000038: B {pc}+0x2 ; 0x3a
0x0000003a: MOVS r0,#1
0x0000003c: BL HAL_Delay
0x00000040: B {pc}+0x2 ; 0x42
0x00000042: LDR r0,[sp,#4]
0x00000044: ADDS r0,#1
0x00000046: STR r0,[sp,#4]
0x00000048: B {pc}-0x16 ; 0x32
2c、2e 兩句 將 sp+4 處堆棧的值置 0
然後利用增 1 和跳轉，執行 50 次循環
也就是每次執行外層循環都會有這套針對 sp+4 處堆棧的操作邏輯，外層循環每次都是針對 sp+8 處堆棧的操作邏輯
如果是預先定義局部變量呢？#
改為如下寫法
int i = 0;
int j = 0;
for(i = 0; i < 50; i++)
{
for(j = 0; j < 50; j++)
{
HAL_Delay(1);
}
}
這部分的反匯編如下#
0x0000001e: LDR r0,[sp,#0]
0x00000020: STR r0,[sp,#8]
0x00000022: STR r0,[sp,#4]
0x00000024: STR r0,[sp,#8]
0x00000026: B {pc}+0x2 ; 0x28
0x00000028: LDR r0,[sp,#8]
0x0000002a: CMP r0,#0x31
0x0000002c: BGT {pc}+0x2c ; 0x58
0x0000002e: B {pc}+0x2 ; 0x30
0x00000030: MOVS r0,#0
0x00000032: STR r0,[sp,#4]
0x00000034: B {pc}+0x2 ; 0x36
0x00000036: LDR r0,[sp,#4]
0x00000038: CMP r0,#0x31
0x0000003a: BGT {pc}+0x14 ; 0x4e
0x0000003c: B {pc}+0x2 ; 0x3e
0x0000003e: MOVS r0,#1
0x00000040: BL HAL_Delay
0x00000044: B {pc}+0x2 ; 0x46
0x00000046: LDR r0,[sp,#4]
0x00000048: ADDS r0,#1
0x0000004a: STR r0,[sp,#4]
0x0000004c: B {pc}-0x16 ; 0x36
0x0000004e: B {pc}+0x2 ; 0x50
0x00000050: LDR r0,[sp,#8]
0x00000052: ADDS r0,#1
0x00000054: STR r0,[sp,#8]
0x00000056: B {pc}-0x2e ; 0x28
可以看出，其循環部分（26-56）與此前的寫法（22-52）部分沒有差異，反而多將 (sp+4) 與 (sp+8) 置零的兩條語句，產生了負優化。
將循環複雜化是否會不一樣#
如下代碼，與反匯編，占用的 (sp+8) 與 (sp+12)，依然不會產生過多的堆棧操作
int test = 0;
for(int i = 0; i < 50; i++)
{
for(int j = 0; j < 50; j++)
{
if((test & 0x01) == 0)
HAL_Delay(1);
else
HAL_Delay(2);
}
test++;
}
0x0000001e: 9801 .. LDR r0,[sp,#4]
0x00000020: 9004 .. STR r0,[sp,#0x10]
0x00000022: 9003 .. STR r0,[sp,#0xc]
0x00000024: e7ff .. B {pc}+0x2 ; 0x26
0x00000026: 9803 .. LDR r0,[sp,#0xc]
0x00000028: 2831 1( CMP r0,#0x31
0x0000002a: dc21 !. BGT {pc}+0x46 ; 0x70
0x0000002c: e7ff .. B {pc}+0x2 ; 0x2e
0x0000002e: 2000 . MOVS r0,#0
0x00000030: 9002 .. STR r0,[sp,#8]
0x00000032: e7ff .. B {pc}+0x2 ; 0x34
0x00000034: 9802 .. LDR r0,[sp,#8]
0x00000036: 2831 1( CMP r0,#0x31
0x00000038: dc12 .. BGT {pc}+0x28 ; 0x60
0x0000003a: e7ff .. B {pc}+0x2 ; 0x3c
0x0000003c: f89d0010 .... LDRB r0,[sp,#0x10]
0x00000040: 07c0 .. LSLS r0,r0,#31
0x00000042: b920 . CBNZ r0,{pc}+0xc ; 0x4e
0x00000044: e7ff .. B {pc}+0x2 ; 0x46
0x00000046: 2001 . MOVS r0,#1
0x00000048: f7fffffe .... BL HAL_Delay
0x0000004c: e003 .. B {pc}+0xa ; 0x56
0x0000004e: 2002 . MOVS r0,#2
0x00000050: f7fffffe .... BL HAL_Delay
0x00000054: e7ff .. B {pc}+0x2 ; 0x56
0x00000056: e7ff .. B {pc}+0x2 ; 0x58
0x00000058: 9802 .. LDR r0,[sp,#8]
0x0000005a: 3001 .0 ADDS r0,#1
0x0000005c: 9002 .. STR r0,[sp,#8]
0x0000005e: e7e9 .. B {pc}-0x2a ; 0x34
如下代碼，將聲明提前，依然產生負優化了
int test = 0;
int i = 0;
int j = 0;
for(i = 0; i < 50; i++)
{
for(j = 0; j < 50; j++)
{
if((test & 0x01) == 0)
HAL_Delay(1);
else
HAL_Delay(2);
}
test++;
}
0x0000001e: 9801 .. LDR r0,[sp,#4]
0x00000020: 9004 .. STR r0,[sp,#0x10]
0x00000022: 9003 .. STR r0,[sp,#0xc]
0x00000024: 9002 .. STR r0,[sp,#8]
0x00000026: 9003 .. STR r0,[sp,#0xc]
0x00000028: e7ff .. B {pc}+0x2 ; 0x2a
0x0000002a: 9803 .. LDR r0,[sp,#0xc]
0x0000002c: 2831 1( CMP r0,#0x31
0x0000002e: dc21 !. BGT {pc}+0x46 ; 0x74
0x00000030: e7ff .. B {pc}+0x2 ; 0x32
0x00000032: 2000 . MOVS r0,#0
0x00000034: 9002 .. STR r0,[sp,#8]
0x00000036: e7ff .. B {pc}+0x2 ; 0x38
0x00000038: 9802 .. LDR r0,[sp,#8]
0x0000003a: 2831 1( CMP r0,#0x31
0x0000003c: dc12 .. BGT {pc}+0x28 ; 0x64
0x0000003e: e7ff .. B {pc}+0x2 ; 0x40
0x00000040: f89d0010 .... LDRB r0,[sp,#0x10]
0x00000044: 07c0 .. LSLS r0,r0,#31
0x00000046: b920 . CBNZ r0,{pc}+0xc ; 0x52
0x00000048: e7ff .. B {pc}+0x2 ; 0x4a
0x0000004a: 2001 . MOVS r0,#1
0x0000004c: f7fffffe .... BL HAL_Delay
0x00000050: e003 .. B {pc}+0xa ; 0x5a
0x00000052: 2002 . MOVS r0,#2
0x00000054: f7fffffe .... BL HAL_Delay
0x00000058: e7ff .. B {pc}+0x2 ; 0x5a
0x0000005a: e7ff .. B {pc}+0x2 ; 0x5c
0x0000005c: 9802 .. LDR r0,[sp,#8]
0x0000005e: 3001 .0 ADDS r0,#1
0x00000060: 9002 .. STR r0,[sp,#8]
0x00000062: e7e9 .. B {pc}-0x2a ; 0x38
0x00000064: 9804 .. LDR r0,[sp,#0x10]
0x00000066: 3001 .0 ADDS r0,#1
0x00000068: 9004 .. STR r0,[sp,#0x10]
0x0000006a: e7ff .. B {pc}+0x2 ; 0x6c
0x0000006c: 9803 .. LDR r0,[sp,#0xc]
0x0000006e: 3001 .0 ADDS r0,#1
0x00000070: 9003 .. STR r0,[sp,#0xc]
0x00000072: e7da .. B {pc}-0x48 ; 0x2a
使用優化#
O1#
仍然是上面的循環複雜化
for 內聲明
0x00000014: 2400 .$ MOVS r4,#0
0x00000016: bf00 .. NOP
0x00000018: f0040501 .... AND r5,r4,#1
0x0000001c: 2632 2& MOVS r6,#0x32
0x0000001e: bf00 .. NOP
0x00000020: 2002 . MOVS r0,#2
0x00000022: 2d00 .- CMP r5,#0
0x00000024: bf08 .. IT EQ
0x00000026: 2001 . MOVEQ r0,#1
0x00000028: f7fffffe .... BL HAL_Delay
0x0000002c: 3e01 .> SUBS r6,#1
0x0000002e: d1f7 .. BNE {pc}-0xe ; 0x20
0x00000030: 3401 .4 ADDS r4,#1
0x00000032: 2c32 2, CMP r4,#0x32
0x00000034: d1f0 .. BNE {pc}-0x1c ; 0x18
將聲明提前，二者完全一致
0x00000014: 2400 .$ MOVS r4,#0
0x00000016: bf00 .. NOP
0x00000018: f0040501 .... AND r5,r4,#1
0x0000001c: 2632 2& MOVS r6,#0x32
0x0000001e: bf00 .. NOP
0x00000020: 2002 . MOVS r0,#2
0x00000022: 2d00 .- CMP r5,#0
0x00000024: bf08 .. IT EQ
0x00000026: 2001 . MOVEQ r0,#1
0x00000028: f7fffffe .... BL HAL_Delay
0x0000002c: 3e01 .> SUBS r6,#1
0x0000002e: d1f7 .. BNE {pc}-0xe ; 0x20
0x00000030: 3401 .4 ADDS r4,#1
0x00000032: 2c32 2, CMP r4,#0x32
0x00000034: d1f0 .. BNE {pc}-0x1c ; 0x18
O2#
仍然是上面的循環複雜化
for 內聲明
0x00000014: 2500 .% MOVS r5,#0
0x00000016: bf00 .. NOP
0x00000018: 2402 .$ MOVS r4,#2
0x0000001a: 2632 2& MOVS r6,#0x32
0x0000001c: 07e8 .. LSLS r0,r5,#31
0x0000001e: bf08 .. IT EQ
0x00000020: 2401 .$ MOVEQ r4,#1
0x00000022: bf00 .. NOP
0x00000024: 4620 F MOV r0,r4
0x00000026: f7fffffe .... BL HAL_Delay
0x0000002a: 3e01 .> SUBS r6,#1
0x0000002c: d1fa .. BNE {pc}-0x8 ; 0x24
0x0000002e: 3501 .5 ADDS r5,#1
0x00000030: 2d32 2- CMP r5,#0x32
0x00000032: d1f1 .. BNE {pc}-0x1a ; 0x18
將聲明提前，二者完全一致
0x00000014: 2500 .% MOVS r5,#0
0x00000016: bf00 .. NOP
0x00000018: 2402 .$ MOVS r4,#2
0x0000001a: 2632 2& MOVS r6,#0x32
0x0000001c: 07e8 .. LSLS r0,r5,#31
0x0000001e: bf08 .. IT EQ
0x00000020: 2401 .$ MOVEQ r4,#1
0x00000022: bf00 .. NOP
0x00000024: 4620 F MOV r0,r4
0x00000026: f7fffffe .... BL HAL_Delay
0x0000002a: 3e01 .> SUBS r6,#1
0x0000002c: d1fa .. BNE {pc}-0x8 ; 0x24
0x0000002e: 3501 .5 ADDS r5,#1
0x00000030: 2d32 2- CMP r5,#0x32
0x00000032: d1f1 .. BNE {pc}-0x1a ; 0x18
O3#
O3 沒有討論價值，完全展開循環。
GCC 環境下的情況#
將局部變量提前定義，同樣是負優化
局部變量定義在 for 內，20 條
局部變量先定義，24 條
此文由 Mix Space 同步更新至 xLog 原始鏈接為 https://www.yono233.cn/posts/shoot/24_8_6_%E5%85%B3%E4%BA%8E%E5%B1%80%E9%83%A8%E5%8F%98%E9%87%8F%E7%9A%84%E6%A0%88%E8%A1%8C%E4%B8%BA%E2%80%94%E2%80%94%E7%94%B1%E5%BE%AA%E7%8E%AF%E8%AF%AD%E5%8F%A5%E5%86%85%E5%AE%9A%E4%B9%89%E5%BE%AA%E7%8E%AF%E5%8F%98%E9%87%8F%E5%BC%95%E7%94%B3