我有以下 C 程序：
__int64_t fib_tr_const (__int64_t n, __int64_t n_1, __int64_t n_2)
{
if (n == 0x42 || n == 0x43 ) {
return n_1 + n_2;
}
return 0x31 + fib_tr_const (n-1, n_2, n_1 + n_2);
}
void main (int argc, char ** argv)
{
__int64_t n = atoi (argv[1]);
// printf("fib(%ld) = %ld\n", n, fib(n));
// printf("fib_tr(%ld) = %ld\n", n, fib_tr(n, 0, 1));
__int64_t volatile result = 0x4567;
printf ("11111\n");
printf ("11111\n");
result = fib_tr_const(n, 0x123, 0x234);
printf ("22222\n");
printf ("22222\n");
printf("fib_tr_const(%ld) = %ld\n", n, result);
}
我用 建造了它-O2。
gcc tail_call.c -O2 -o tail_call_const_O2
我反汇编了二进制文件：
主要()：
00000000000010a0 <main>:
10a0: f3 0f 1e fa endbr64
10a4: 41 54 push %r12
10a6: ba 0a 00 00 00 mov $0xa,%edx
10ab: 48 83 ec 10 sub $0x10,%rsp
10af: 48 8b 7e 08 mov 0x8(%rsi),%rdi
10b3: 31 f6 xor %esi,%esi
10b5: e8 c6 ff ff ff callq 1080 <strtol@plt>
10ba: 48 8d 3d 43 0f 00 00 lea 0xf43(%rip),%rdi # 2004 <_IO_stdin_used+0x4>
10c1: 48 c7 44 24 08 67 45 movq $0x4567,0x8(%rsp)
10c8: 00 00
10ca: 4c 63 e0 movslq %eax,%r12
10cd: e8 9e ff ff ff callq 1070 <puts@plt>
10d2: 48 8d 3d 2b 0f 00 00 lea 0xf2b(%rip),%rdi # 2004 <_IO_stdin_used+0x4>
10d9: e8 92 ff ff ff callq 1070 <puts@plt>
------------------START-------------------
10de: 49 8d 44 24 ff lea -0x1(%r12),%rax
10e3: 48 83 f8 01 cmp $0x1,%rax
10e7: 76 75 jbe 115e <main+0xbe>
10e9: 4c 89 e0 mov %r12,%rax
10ec: 31 c9 xor %ecx,%ecx
10ee: ba 01 00 00 00 mov $0x1,%edx
10f3: 31 ff xor %edi,%edi
10f5: eb 0c jmp 1103 <main+0x63>
10f7: 66 0f 1f 84 00 00 00 nopw 0x0(%rax,%rax,1)
10fe: 00 00
1100: 48 89 f2 mov %rsi,%rdx
1103: 48 8d 34 17 lea (%rdi,%rdx,1),%rsi
1107: 48 89 c7 mov %rax,%rdi
110a: 48 83 e8 01 sub $0x1,%rax
110e: 48 01 f9 add %rdi,%rcx
1111: 48 89 d7 mov %rdx,%rdi
1114: 48 83 f8 02 cmp $0x2,%rax
1118: 75 e6 jne 1100 <main+0x60>
111a: 48 01 f2 add %rsi,%rdx
111d: 48 01 d1 add %rdx,%rcx
1120: 48 8d 3d e3 0e 00 00 lea 0xee3(%rip),%rdi # 200a <_IO_stdin_used+0xa>
1127: 48 89 4c 24 08 mov %rcx,0x8(%rsp)
------------------END-------------------
112c: e8 3f ff ff ff callq 1070 <puts@plt>
1131: 48 8d 3d d2 0e 00 00 lea 0xed2(%rip),%rdi # 200a <_IO_stdin_used+0xa>
1138: e8 33 ff ff ff callq 1070 <puts@plt>
113d: 48 8b 4c 24 08 mov 0x8(%rsp),%rcx
1142: 48 83 c4 10 add $0x10,%rsp
1146: 31 c0 xor %eax,%eax
1148: 4c 89 e2 mov %r12,%rdx
114b: 48 8d 35 be 0e 00 00 lea 0xebe(%rip),%rsi # 2010 <_IO_stdin_used+0x10>
1152: bf 01 00 00 00 mov $0x1,%edi
1157: 41 5c pop %r12
1159: e9 32 ff ff ff jmpq 1090 <__printf_chk@plt>
115e: ba 01 00 00 00 mov $0x1,%edx
1163: 31 c9 xor %ecx,%ecx
1165: eb b6 jmp 111d <main+0x7d>
1167: 66 0f 1f 84 00 00 00 nopw 0x0(%rax,%rax,1)
116e: 00 00
fib_tr_const()：
00000000000012e0 <fib_tr_const>:
12e0: f3 0f 1e fa endbr64
12e4: 48 8d 47 be lea -0x42(%rdi),%rax
12e8: 48 89 f9 mov %rdi,%rcx
12eb: 48 83 f8 01 cmp $0x1,%rax
12ef: 77 0a ja 12fb <fib_tr_const+0x1b>
12f1: eb 35 jmp 1328 <fib_tr_const+0x48>
12f3: 0f 1f 44 00 00 nopl 0x0(%rax,%rax,1)
12f8: 48 89 c2 mov %rax,%rdx
12fb: 48 83 ef 01 sub $0x1,%rdi
12ff: 48 8d 04 16 lea (%rsi,%rdx,1),%rax
1303: 48 89 d6 mov %rdx,%rsi
1306: 48 83 ff 43 cmp $0x43,%rdi
130a: 75 ec jne 12f8 <fib_tr_const+0x18>
130c: 48 8d 34 49 lea (%rcx,%rcx,2),%rsi
1310: 48 01 c2 add %rax,%rdx
1313: 48 c1 e6 04 shl $0x4,%rsi
1317: 48 8d 8c 31 2d f3 ff lea -0xcd3(%rcx,%rsi,1),%rcx
131e: ff
131f: 48 8d 04 0a lea (%rdx,%rcx,1),%rax
1323: c3 retq
1324: 0f 1f 40 00 nopl 0x0(%rax)
1328: 48 89 d0 mov %rdx,%rax
132b: 48 89 f2 mov %rsi,%rdx
132e: 31 c9 xor %ecx,%ecx
1330: 48 01 c2 add %rax,%rdx
1333: 48 8d 04 0a lea (%rdx,%rcx,1),%rax
1337: c3 retq
1338: 0f 1f 84 00 00 00 00 nopl 0x0(%rax,%rax,1)
133f: 00
有了地标printf ("11111\n");和printf ("22222\n");，因此在汇编中。我相信和puts之间的部分应该是调用的地方。STARTENDmain()fib_tr_const()
我没有看到任何call从jmpmain() 分支到 fib_tr_const() 的指令。那么这个函数是如何调用的呢？
我将标记点之间的汇编与 fib_tr_const() 的汇编进行了比较。asm 代码看起来很相似。看来 fib_tr_const() 函数是内联的。
另一个问题出现了——如果函数是内联的，为什么代码仍然存在于最终的二进制文件中？如果删除它，可以节省空间，那么为什么不发生这种情况呢？
