ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x3,%edi
  call   <SumDivisibleBy@plt>
  mov    %eax,%ebx
  mov    $0x5,%edi
  call   <SumDivisibleBy@plt>
  add    %eax,%ebx
  mov    $0xf,%edi
  call   <SumDivisibleBy@plt>
  mov    %ebx,%esi
  sub    %eax,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_guessed: string

tool: D2: %d


assistant: ```c
int main(void) {
  printf("%d\n", SumDivisibleBy(3) + SumDivisibleBy(5) - SumDivisibleBy(15));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x3,%edi
  call   <SumDivisibleBy@plt>
  mov    $0x5,%edi
  mov    %eax,%ebx
  call   <SumDivisibleBy@plt>
  mov    $0xf,%edi
  add    %eax,%ebx
  call   <SumDivisibleBy@plt>
  mov    %ebx,%esi
  lea    D2(%rip),%rdi
  sub    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type_infered: string

tool: D2: %d


assistant: ```c
int main(void) {
  printf("%d\n", SumDivisibleBy(3) + SumDivisibleBy(5) - SumDivisibleBy(15));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x3,%edi
  call   <SumDivisibleBy@plt>
  mov    $0x5,%edi
  mov    %eax,%ebx
  call   <SumDivisibleBy@plt>
  mov    $0xf,%edi
  add    %eax,%ebx
  call   <SumDivisibleBy@plt>
  mov    %ebx,%esi
  lea    D6(%rip),%rdi
  sub    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type: string

tool: D6: %d


assistant: ```c
int main(void) {
  printf("%d\n", SumDivisibleBy(3) + SumDivisibleBy(5) - SumDivisibleBy(15));
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strncat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strncat(char *dst, char *src, unsigned int n) {
  memcpy((dst + strlen(dst)), src, n);
  return dst;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strncat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  call   <strlen@plt>
  movslq %eax,%rdi
  add    %rbx,%rdi
  mov    %r12d,%edx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strncat(char *dst, char *src, unsigned int n) {
  memcpy((dst + strlen(dst)), src, n);
  return dst;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strncat>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  movslq %eax,%rdi
  add    %r12,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strncat(char *dst, char *src, unsigned int n) {
  memcpy((dst + strlen(dst)), src, n);
  return dst;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strncat>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  movslq %eax,%rdi
  add    %r12,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strncat(char *dst, char *src, unsigned int n) {
  memcpy((dst + strlen(dst)), src, n);
  return dst;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memset0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L11
L5:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  addq   $0x1,-0x18(%rbp)
L11:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L5
  nop
  pop    %rbp
  ret
```

assistant: ```c
void memset0(char *m, long len) {
  char *t;
  t = m + len;
  while (m < t) {
    *m = (char)0;
    m = m + 1;
  }
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<memset0>:
  endbr64
  add    %rdi,%rsi
  cmp    %rsi,%rdi
  jae    L1
L4:
  movb   $0x0,(%rdi)
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L4
L1:
  ret
```

assistant: ```c
void memset0(char *m, long len) {
  char *t;
  t = m + len;
  while (m < t) {
    *m = (char)0;
    m = m + 1;
  }
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memset0>:
  endbr64
  lea    (%rdi,%rsi,1),%rax
  mov    %rsi,%rdx
  cmp    %rax,%rdi
  jae    L6
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  ret
```

assistant: ```c
void memset0(char *m, long len) {
  char *t;
  t = m + len;
  while (m < t) {
    *m = (char)0;
    m = m + 1;
  }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memset0>:
  endbr64
  lea    (%rdi,%rsi,1),%rax
  mov    %rsi,%rdx
  cmp    %rax,%rdi
  jae    L9
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L9:
  ret
```

assistant: ```c
void memset0(char *m, long len) {
  char *t;
  t = m + len;
  while (m < t) {
    *m = (char)0;
    m = m + 1;
  }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<base_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L17
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  jmp    L0
L17:
  mov    -0x18(%rbp),%rax
L0:
  leave
  ret
```

assistant: ```c
char *base_name(char *path) {
  char *base = strrchr(path, '/');
  return base ? (base + 1) : path;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<base_name>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x2f,%esi
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  cmovne %rdx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *base_name(char *path) {
  char *base = strrchr(path, '/');
  return base ? (base + 1) : path;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<base_name>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *base_name(char *path) {
  char *base = strrchr(path, '/');
  return base ? (base + 1) : path;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<base_name>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *base_name(char *path) {
  char *base = strrchr(path, '/');
  return base ? (base + 1) : path;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x660,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x658(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x654(%rbp)
  jmp    L21
L11:
  lea    -0x650(%rbp),%rdx
  mov    -0x654(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x654(%rbp),%eax
  cltq
  mov    -0x650(%rbp,%rax,4),%edx
  mov    -0x654(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  mov    -0x654(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x654(%rbp),%eax
  cltq
  mov    %edx,-0x330(%rbp,%rax,4)
  mov    -0x654(%rbp),%eax
  cltq
  mov    -0x330(%rbp,%rax,4),%edx
  mov    -0x654(%rbp),%eax
  cltq
  mov    %edx,-0x4c0(%rbp,%rax,4)
  addl   $0x1,-0x654(%rbp)
L21:
  mov    -0x658(%rbp),%eax
  cmp    %eax,-0x654(%rbp)
  jl     L11
  mov    -0x658(%rbp),%edx
  lea    -0x650(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sort_1@plt>
  mov    -0x658(%rbp),%edx
  lea    -0x4c0(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sort_2@plt>
  mov    -0x658(%rbp),%edx
  lea    -0x330(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sort_3@plt>
  mov    -0x658(%rbp),%edx
  lea    -0x1a0(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sort_4@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D19
type_guessed: string
name: read_data
arguments: label: D21
type_guessed: string

tool: D19: Enter the length of the array : 

tool: D21: %d

assistant: ```c
int main() {
  int a[100], a2[100], a3[100], a4[100];
  int len;
  int i;
  printf("Enter the length of the array : ");
  scanf("%d", &len);
  for (i = 0; i < len; i++) {
    scanf("%d", &a[i]);
    a2[i] = a3[i] = a4[i] = a[i];
  }
  sort_1(&a[0], len);
  sort_2(&a2[0], len);
  sort_3(&a3[0], len);
  sort_4(&a4[0], len);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x660,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x658(%rsp)
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L25
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D3(%rip),%r12
L1:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x0(%rbp),%eax
  mov    %eax,0x4c0(%rsp,%rbx,4)
  mov    %eax,0x330(%rsp,%rbx,4)
  mov    %eax,0x1a0(%rsp,%rbx,4)
  mov    0xc(%rsp),%esi
  add    $0x1,%rbx
  add    $0x4,%rbp
  cmp    %ebx,%esi
  jg     L1
L25:
  lea    0x10(%rsp),%rdi
  call   <sort_1@plt>
  lea    0x1a0(%rsp),%rdi
  mov    0xc(%rsp),%esi
  call   <sort_2@plt>
  lea    0x330(%rsp),%rdi
  mov    0xc(%rsp),%esi
  call   <sort_3@plt>
  lea    0x4c0(%rsp),%rdi
  mov    0xc(%rsp),%esi
  call   <sort_4@plt>
  mov    0x658(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    $0x0,%eax
  add    $0x660,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D24
type_guessed: string
name: parse_data
arguments: data_tag: D3
type_guessed: string

tool: D24: Enter the length of the array : 

tool: D3: %d

assistant: ```c
int main() {
  int a[100], a2[100], a3[100], a4[100];
  int len;
  int i;
  printf("Enter the length of the array : ");
  scanf("%d", &len);
  for (i = 0; i < len; i++) {
    scanf("%d", &a[i]);
    a2[i] = a3[i] = a4[i] = a[i];
  }
  sort_1(&a[0], len);
  sort_2(&a2[0], len);
  sort_3(&a3[0], len);
  sort_4(&a4[0], len);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D17(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D18(%rip),%rbp
  push   %rbx
  sub    $0x678,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x668(%rsp)
  xor    %eax,%eax
  lea    0x4d0(%rsp),%r12
  lea    0x340(%rsp),%r13
  call   <printf@plt>
  lea    0x1c(%rsp),%rsi
  mov    %rbp,%rdi
  lea    0x1b0(%rsp),%r14
  call   <scanf@plt>
  mov    0x1c(%rsp),%esi
  lea    0x20(%rsp),%rax
  mov    %rax,0x8(%rsp)
  test   %esi,%esi
  jle    L19
  mov    %rax,%r15
  xor    %ebx,%ebx
  lea    0x4d0(%rsp),%r12
  lea    0x340(%rsp),%r13
  lea    0x1b0(%rsp),%r14
  nop
L12:
  mov    %r15,%rsi
  mov    %rbp,%rdi
  add    $0x4,%r15
  call   <scanf@plt>
  mov    -0x4(%r15),%eax
  mov    0x1c(%rsp),%esi
  mov    %eax,(%r12,%rbx,4)
  mov    %eax,0x0(%r13,%rbx,4)
  mov    %eax,(%r14,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,%esi
  jg     L12
L19:
  mov    0x8(%rsp),%rdi
  call   <sort_1@plt>
  mov    0x1c(%rsp),%esi
  mov    %r14,%rdi
  call   <sort_2@plt>
  mov    0x1c(%rsp),%esi
  mov    %r13,%rdi
  call   <sort_3@plt>
  mov    0x1c(%rsp),%esi
  mov    %r12,%rdi
  call   <sort_4@plt>
  mov    0x668(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x678,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D17
datatype: string
name: unpack_data
arguments: data_tag: D18
datatype: string

tool: D17: Enter the length of the array : 

tool: D18: %d

assistant: ```c
int main() {
  int a[100], a2[100], a3[100], a4[100];
  int len;
  int i;
  printf("Enter the length of the array : ");
  scanf("%d", &len);
  for (i = 0; i < len; i++) {
    scanf("%d", &a[i]);
    a2[i] = a3[i] = a4[i] = a[i];
  }
  sort_1(&a[0], len);
  sort_2(&a2[0], len);
  sort_3(&a3[0], len);
  sort_4(&a4[0], len);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  lea    D5(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D18(%rip),%rbp
  push   %rbx
  sub    $0x678,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x668(%rsp)
  xor    %eax,%eax
  lea    0x4d0(%rsp),%r12
  lea    0x340(%rsp),%r13
  call   <printf@plt>
  lea    0x1c(%rsp),%rsi
  mov    %rbp,%rdi
  lea    0x1b0(%rsp),%r14
  call   <scanf@plt>
  mov    0x1c(%rsp),%esi
  lea    0x20(%rsp),%rax
  mov    %rax,0x8(%rsp)
  test   %esi,%esi
  jle    L5
  mov    %rax,%r15
  xor    %ebx,%ebx
  lea    0x4d0(%rsp),%r12
  lea    0x340(%rsp),%r13
  lea    0x1b0(%rsp),%r14
  nop
L7:
  mov    %r15,%rsi
  mov    %rbp,%rdi
  add    $0x4,%r15
  call   <scanf@plt>
  mov    -0x4(%r15),%eax
  mov    0x1c(%rsp),%esi
  mov    %eax,(%r12,%rbx,4)
  mov    %eax,0x0(%r13,%rbx,4)
  mov    %eax,(%r14,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,%esi
  jg     L7
L5:
  mov    0x8(%rsp),%rdi
  call   <sort_1@plt>
  mov    0x1c(%rsp),%esi
  mov    %r14,%rdi
  call   <sort_2@plt>
  mov    0x1c(%rsp),%esi
  mov    %r13,%rdi
  call   <sort_3@plt>
  mov    0x1c(%rsp),%esi
  mov    %r12,%rdi
  call   <sort_4@plt>
  mov    0x668(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x678,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D5
type: string
name: parse_data
arguments: data_tag: D18
type: string

tool: D5: Enter the length of the array : 

tool: D18: %d

assistant: ```c
int main() {
  int a[100], a2[100], a3[100], a4[100];
  int len;
  int i;
  printf("Enter the length of the array : ");
  scanf("%d", &len);
  for (i = 0; i < len; i++) {
    scanf("%d", &a[i]);
    a2[i] = a3[i] = a4[i] = a[i];
  }
  sort_1(&a[0], len);
  sort_2(&a2[0], len);
  sort_3(&a3[0], len);
  sort_4(&a4[0], len);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<egcd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  jmp    L14
L10:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L14:
  cmpl   $0x0,-0x18(%rbp)
  jne    L10
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int egcd(int a, int b) {
  int rem;
  while (b != 0) {
    rem = a % b;
    a = b;
    b = rem;
  }
  return a;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<egcd>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  je     L12
L0:
  mov    %esi,%ecx
  cltd
  idiv   %esi
  mov    %edx,%esi
  mov    %ecx,%eax
  test   %edx,%edx
  jne    L0
L25:
  mov    %ecx,%eax
  ret
L12:
  mov    %edi,%ecx
  jmp    L25
```

assistant: ```c
int egcd(int a, int b) {
  int rem;
  while (b != 0) {
    rem = a % b;
    a = b;
    b = rem;
  }
  return a;
}
```
user: Translate the below assembly code into C programming language:

```asm
<egcd>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edx
  test   %esi,%esi
  je     L0
  nopl   0x0(%rax)
L11:
  mov    %edx,%r8d
  cltd
  idiv   %r8d
  mov    %r8d,%eax
  test   %edx,%edx
  jne    L11
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %edi,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int egcd(int a, int b) {
  int rem;
  while (b != 0) {
    rem = a % b;
    a = b;
    b = rem;
  }
  return a;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<egcd>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edx
  test   %esi,%esi
  je     L9
  nopl   0x0(%rax)
L1:
  mov    %edx,%r8d
  cltd
  idiv   %r8d
  mov    %r8d,%eax
  test   %edx,%edx
  jne    L1
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %edi,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int egcd(int a, int b) {
  int rem;
  while (b != 0) {
    rem = a % b;
    a = b;
    b = rem;
  }
  return a;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<path_concat>:
  endbr64
  mov    %rdi,%r8
  test   %rdi,%rdi
  je     L93
  test   %rsi,%rsi
  je     L93
  sub    $0x1,%rsi
  lea    (%rdi,%rsi,1),%r9
  test   %rdx,%rdx
  je     L275
  cmpb   $0x0,(%rdx)
  je     L217
  mov    %rdx,%rax
L182:
  mov    %rax,%rdi
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L182
  cmp    %rdi,%rdx
  jae    L278
L187:
  cmpb   $0x2f,(%rdi)
  jne    L278
  sub    $0x1,%rdi
  cmp    %rdi,%rdx
  jne    L187
  mov    %rdx,%rdi
L243:
  cmp    %rdi,%rdx
  ja     L130
  mov    %r8,%rax
  cmp    %r9,%r8
  jae    L130
L60:
  add    $0x1,%rdx
  add    $0x1,%rax
  movzbl -0x1(%rdx),%esi
  mov    %sil,-0x1(%rax)
  cmp    %rdi,%rdx
  ja     L267
  cmp    %rax,%r9
  ja     L60
  jmp    L267
L93:
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  ret
L278:
  mov    %rdi,%rax
  sub    %rdx,%rax
  cmp    %rax,%rsi
  jae    L243
  mov    D24(%rip),%rax
  mov    (%rax),%eax
L229:
  movb   $0x0,(%r8)
  ret
L130:
  mov    %r8,%rax
  jmp    L267
L275:
  mov    %rdi,%rax
  jmp    L267
L217:
  mov    %rdi,%rax
L267:
  test   %rcx,%rcx
  je     L137
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L137
  cmp    $0x2f,%dl
  jne    L91
L210:
  add    $0x1,%rcx
  cmpb   $0x2f,(%rcx)
  je     L210
L91:
  cmp    %r9,%rax
  jae    L185
  cmp    %r8,%rax
  ja     L20
L185:
  movzbl (%rcx),%edx
  cmp    %r9,%rax
  jae    L163
  test   %dl,%dl
  je     L163
L71:
  add    $0x1,%rcx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  movzbl (%rcx),%edx
  cmp    %rax,%r9
  jbe    L163
  test   %dl,%dl
  jne    L71
L163:
  test   %dl,%dl
  jne    L174
L137:
  movb   $0x0,(%rax)
  mov    $0x0,%eax
  ret
L20:
  cmpb   $0x2f,-0x1(%rax)
  je     L185
  movb   $0x2f,(%rax)
  lea    0x1(%rax),%rax
  jmp    L185
L174:
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  jmp    L229
```

assistant: ```c
int path_concat(char *path, size_t path_size, const char *head, const char *tail) {
  int ret;
  const char *p, *src;
  char *dst, *dst_end;
  if (!path || path_size < 1)
    return ENOBUFS;
  dst = path;
  dst_end = path + path_size - 1;
  if (head && *head) {
    for (p = head; *p; p++)
      ;
    for (p--; p > head && *p == '/'; p--)
      ;
    if ((p - head) > path_size - 1) {
      ret = ENOBUFS;
      goto fail;
    }
    for (src = head; src <= p && dst < dst_end;) {
      *dst++ = *src++;
    }
  }
  if (tail && *tail) {
    for (p = tail; *p && *p == '/'; p++)
      ;
    if (dst > path)
      if (dst < dst_end && *(dst - 1) != '/') {
        *dst++ = '/';
      }
    for (src = p; *src && dst < dst_end;) {
      *dst++ = *src++;
    }
    if (*src) {
      ret = ENOBUFS;
      goto fail;
    }
  }
  *dst = 0;
  return 0;
fail:
  *path = '\0';
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<path_concat>:
  endbr64
  test   %rdi,%rdi
  je     L177
  test   %rsi,%rsi
  je     L177
  sub    $0x1,%rsi
  lea    (%rdi,%rsi,1),%r9
  test   %rdx,%rdx
  je     L108
  movzbl (%rdx),%r10d
  test   %r10b,%r10b
  je     L108
  mov    %rdx,%r8
  nop
L273:
  mov    %r8,%rax
  lea    0x1(%r8),%r8
  cmpb   $0x0,0x1(%rax)
  jne    L273
  cmp    %rax,%rdx
  jb     L161
  jmp    L185
  nopl   0x0(%rax)
L18:
  sub    $0x1,%rax
  cmp    %rax,%rdx
  je     L45
L161:
  cmpb   $0x2f,(%rax)
  je     L18
L185:
  mov    %rax,%r8
  sub    %rdx,%r8
  cmp    %r8,%rsi
  jae    L45
L267:
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  movb   $0x0,(%rdi)
  ret
  cs nopw 0x0(%rax,%rax,1)
L108:
  mov    %rdi,%rsi
L283:
  test   %rcx,%rcx
  je     L69
  movzbl (%rcx),%eax
  test   %al,%al
  je     L69
  cmp    %rdi,%rsi
  seta   %dl
  cmp    %r9,%rsi
  setb   %r8b
  and    %r8d,%edx
  cmp    $0x2f,%al
  jne    L289
  nopw   0x0(%rax,%rax,1)
L132:
  movzbl 0x1(%rcx),%eax
  add    $0x1,%rcx
  cmp    $0x2f,%al
  je     L132
  test   %dl,%dl
  jne    L235
L308:
  test   %al,%al
  jne    L213
  jmp    L69
  cs nopw 0x0(%rax,%rax,1)
L116:
  add    $0x1,%rcx
  mov    %al,(%rsi)
  add    $0x1,%rsi
  movzbl (%rcx),%eax
  test   %al,%al
  je     L69
L213:
  cmp    %rsi,%r9
  ja     L116
  jmp    L267
  nopl   0x0(%rax,%rax,1)
L235:
  cmpb   $0x2f,-0x1(%rsi)
  jne    L89
  test   %al,%al
  jne    L116
L69:
  movb   $0x0,(%rsi)
  xor    %eax,%eax
  ret
L177:
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  ret
  nopw   0x0(%rax,%rax,1)
L45:
  cmp    %rax,%rdx
  ja     L108
  mov    %rdi,%rsi
  cmp    %r9,%rdi
  jb     L254
  jmp    L108
  cs nopw 0x0(%rax,%rax,1)
L228:
  movzbl (%rdx),%r10d
L254:
  add    $0x1,%rdx
  add    $0x1,%rsi
  mov    %r10b,-0x1(%rsi)
  cmp    %rax,%rdx
  ja     L283
  cmp    %rsi,%r9
  ja     L228
  jmp    L283
  nopl   0x0(%rax,%rax,1)
L289:
  test   %dl,%dl
  je     L213
  cmpb   $0x2f,-0x1(%rsi)
  je     L116
  cs nopw 0x0(%rax,%rax,1)
L89:
  movb   $0x2f,(%rsi)
  add    $0x1,%rsi
  movzbl (%rcx),%eax
  jmp    L308
```

assistant: ```c
int path_concat(char *path, size_t path_size, const char *head, const char *tail) {
  int ret;
  const char *p, *src;
  char *dst, *dst_end;
  if (!path || path_size < 1)
    return ENOBUFS;
  dst = path;
  dst_end = path + path_size - 1;
  if (head && *head) {
    for (p = head; *p; p++)
      ;
    for (p--; p > head && *p == '/'; p--)
      ;
    if ((p - head) > path_size - 1) {
      ret = ENOBUFS;
      goto fail;
    }
    for (src = head; src <= p && dst < dst_end;) {
      *dst++ = *src++;
    }
  }
  if (tail && *tail) {
    for (p = tail; *p && *p == '/'; p++)
      ;
    if (dst > path)
      if (dst < dst_end && *(dst - 1) != '/') {
        *dst++ = '/';
      }
    for (src = p; *src && dst < dst_end;) {
      *dst++ = *src++;
    }
    if (*src) {
      ret = ENOBUFS;
      goto fail;
    }
  }
  *dst = 0;
  return 0;
fail:
  *path = '\0';
  return ret;
}
```