C 语言和 ARM 汇编 - 6 分支跳转

if .. else ..

一个使用 if 进行条件判断的例子

int a = 4;
int b = 8;
int d = 0;
if (a > b)
    d = 1;
else
    d = 2;

局部变量对应位置

a -> sp[12]
b -> sp[8]
d -> sp[4]

汇编代码

# 准备堆栈
    sub sp, sp, #16
# sp[12] = 4
    movs    r3, #4
    str r3, [sp, #12]
# sp[8] = 8
    movs    r3, #8
    str r3, [sp, #8]
# sp[4] = 0
    movs    r3, #0
    str r3, [sp, #4]
# 比较 sp[8] sp[12]
    ldr r2, [sp, #12]
    ldr r3, [sp, #8]
    cmp r2, r3
# 如果 sp[12] > sp[8]，跳转到 .L22
    ble .L2
    movs    r3, #1
    str r3, [sp, #4]
    b   .L1
.L2:
    movs    r3, #2
    str r3, [sp, #4]
.L1:
# 恢复堆栈指针
    add sp, sp, #16
    @ sp needed
    bx  lr

switch .. case

例子

int a = 4;
int b = 8;
int d = 0;
switch(b)
{
    case 2:
        a++;
        break;
    case 8:
        b++;
        break;
    default:
        d++;
        break;
}

对应的汇编代码

    sub sp, sp, #16
    movs    r3, #4
    str r3, [sp, #12]
    movs    r3, #8
    str r3, [sp, #8]
    movs    r3, #0
    str r3, [sp, #4]
    ldr r3, [sp, #8]
# if sp[8] == 2，跳转到 .L3
    cmp r3, #2
    beq .L3
# if sp[8] == 8，跳转到 .L4
    cmp r3, #8
    beq .L4
    b   .L6
.L3:
    ldr r3, [sp, #12]
    adds    r3, r3, #1
    str r3, [sp, #12]
    b   .L1
.L4:
    ldr r3, [sp, #8]
    adds    r3, r3, #1
    str r3, [sp, #8]
    b   .L1
.L6:
    ldr r3, [sp, #4]
    adds    r3, r3, #1
    str r3, [sp, #4]
    nop
.L1:
    add sp, sp, #16
    @ sp needed
    bx  lr

每个 case 结束，都有一个跳转指令 b，对应了代码里面的 break，如果没有写 break，也就是 fall though，会继续执行下面的 case。