ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<execute>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x10,%rsp
  mov    %rdi,%rdx
  mov    %rsi,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  lea    0x800(%rsp),%rbx
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  lea    D23(%rip),%rsi
  mov    %rbx,%rdi
  call   <strtok@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <basename@plt>
  mov    %rax,(%rsp)
  lea    0x8(%rsp),%rbx
  lea    D23(%rip),%rbp
L14:
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,(%rbx)
  add    $0x8,%rbx
  test   %rax,%rax
  jne    L14
  call   <fork@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L32
L10:
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    %ebx,%eax
  add    $0x1010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L32:
  mov    %rsp,%rsi
  mov    %r12,%rdi
  call   <execv@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L10
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type: string
name: retrieve_data
arguments: data_tag: D23
type: string

tool: D0: %s/%s

tool: D23:  

assistant: ```c
int execute(char *path, char *prog) {
  int pid, i;
  char command[2 * 1024 + 1];
  int s = 0;
  char *name;
  char *argv[256];
  sprintf(command, "%s/%s", path, prog);
  name = strtok(command, " ");
  argv[0] = basename(name);
  for (i = 1; (argv[i] = strtok(NULL, " ")) != NULL; i++)
    ;
  pid = fork();
  if (pid == 0) {
    s = execv(name, argv);
    exit(0);
  }
  return pid;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<execute>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  lea    D16(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  lea    0x800(%rsp),%r12
  mov    %rdi,%rdx
  mov    %rsi,%rcx
  mov    %r12,%rdi
  lea    D11(%rip),%rsi
  lea    0x8(%rsp),%rbx
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strtok@plt>
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <basename@plt>
  mov    %rax,(%rsp)
L12:
  mov    %rbp,%rsi
  xor    %edi,%edi
  add    $0x8,%rbx
  call   <strtok@plt>
  mov    %rax,-0x8(%rbx)
  test   %rax,%rax
  jne    L12
  call   <fork@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  je     L5
L11:
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x1018,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %r13,%rdi
  mov    %rsp,%rsi
  call   <execv@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L11
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D16
type_infered: string
name: read_data
arguments: data_tag: D11
type_infered: string

tool: D16:  

tool: D11: %s/%s

assistant: ```c
int execute(char *path, char *prog) {
  int pid, i;
  char command[2 * 1024 + 1];
  int s = 0;
  char *name;
  char *argv[256];
  sprintf(command, "%s/%s", path, prog);
  name = strtok(command, " ");
  argv[0] = basename(name);
  for (i = 1; (argv[i] = strtok(NULL, " ")) != NULL; i++)
    ;
  pid = fork();
  if (pid == 0) {
    s = execv(name, argv);
    exit(0);
  }
  return pid;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<execute>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  lea    D2(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  lea    0x800(%rsp),%r12
  mov    %rdi,%rdx
  mov    %rsi,%rcx
  mov    %r12,%rdi
  lea    D8(%rip),%rsi
  lea    0x8(%rsp),%rbx
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strtok@plt>
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <basename@plt>
  mov    %rax,(%rsp)
L3:
  mov    %rbp,%rsi
  xor    %edi,%edi
  add    $0x8,%rbx
  call   <strtok@plt>
  mov    %rax,-0x8(%rbx)
  test   %rax,%rax
  jne    L3
  call   <fork@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  je     L20
L26:
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x1018,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %r13,%rdi
  mov    %rsp,%rsi
  call   <execv@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L26
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D2
datatype: string
name: retrieve_data
arguments: label: D8
datatype: string

tool: D2:  

tool: D8: %s/%s

assistant: ```c
int execute(char *path, char *prog) {
  int pid, i;
  char command[2 * 1024 + 1];
  int s = 0;
  char *name;
  char *argv[256];
  sprintf(command, "%s/%s", path, prog);
  name = strtok(command, " ");
  argv[0] = basename(name);
  for (i = 1; (argv[i] = strtok(NULL, " ")) != NULL; i++)
    ;
  pid = fork();
  if (pid == 0) {
    s = execv(name, argv);
    exit(0);
  }
  return pid;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pe>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L10:
  mov    -0x4(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x4(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  shl    $0x7,%eax
  mov    %eax,%esi
  mov    -0x20(%rbp),%rax
  add    %rax,%rsi
  mov    -0x4(%rbp),%eax
  shl    $0x7,%eax
  mov    %eax,%edi
  mov    -0x18(%rbp),%rax
  add    %rdi,%rax
  mov    %rax,%rdi
  call   <needwun@plt>
  addl   $0x1,-0x4(%rbp)
L13:
  cmpl   $0x7f,-0x4(%rbp)
  jbe    L10
  nop
  nop
  leave
  ret
```

assistant: ```c
void pe(char *seqA, char *seqB, char *alignedA, char *alignedB) {
#pragma HLS inline off
  for (unsigned i = 0; i < 256 / 2; i++) {
    needwun(seqA + i * 128, seqB + i * 128, alignedA + i * (128 + 128), alignedB + i * (128 + 128));
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<pe>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    %rdx,%r12
  mov    %rcx,%rbp
  lea    0x4000(%rdi),%r14
L3:
  mov    %rbp,%rcx
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <needwun@plt>
  sub    $0xffffffffffffff80,%rbx
  sub    $0xffffffffffffff80,%r13
  add    $0x100,%r12
  add    $0x100,%rbp
  cmp    %r14,%rbx
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void pe(char *seqA, char *seqB, char *alignedA, char *alignedB) {
#pragma HLS inline off
  for (unsigned i = 0; i < 256 / 2; i++) {
    needwun(seqA + i * 128, seqB + i * 128, alignedA + i * (128 + 128), alignedB + i * (128 + 128));
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<pe>:
  endbr64
  push   %r14
  lea    0x4000(%rdi),%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  nop
L2:
  mov    %rbp,%rcx
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  sub    $0xffffffffffffff80,%rbx
  sub    $0xffffffffffffff80,%r13
  add    $0x100,%r12
  add    $0x100,%rbp
  call   <needwun@plt>
  cmp    %r14,%rbx
  jne    L2
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void pe(char *seqA, char *seqB, char *alignedA, char *alignedB) {
#pragma HLS inline off
  for (unsigned i = 0; i < 256 / 2; i++) {
    needwun(seqA + i * 128, seqB + i * 128, alignedA + i * (128 + 128), alignedB + i * (128 + 128));
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<pe>:
  endbr64
  push   %r14
  lea    0x4000(%rdi),%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  nop
L5:
  mov    %rbp,%rcx
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  sub    $0xffffffffffffff80,%rbx
  sub    $0xffffffffffffff80,%r13
  add    $0x100,%r12
  add    $0x100,%rbp
  call   <needwun@plt>
  cmp    %r14,%rbx
  jne    L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void pe(char *seqA, char *seqB, char *alignedA, char *alignedB) {
#pragma HLS inline off
  for (unsigned i = 0; i < 256 / 2; i++) {
    needwun(seqA + i * 128, seqB + i * 128, alignedA + i * (128 + 128), alignedB + i * (128 + 128));
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<myrand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    D8(%rip),%eax
  imul   $0x19660d,%eax,%eax
  add    $0x3c6ef35f,%eax
  mov    %eax,D29(%rip)
  mov    D29(%rip),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %edx,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L5
  mov    -0x4(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  jmp    L14
L5:
  mov    -0x4(%rbp),%eax
L14:
  pop    %rbp
  ret
```

assistant: ```c
int myrand(int n) {
  static int lcg = 0;
  int x;
  lcg = 1664525 * lcg + 1013904223;
  x = lcg % n;
  return x < 0 ? x + n : x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<myrand>:
  endbr64
  imul   $0x19660d,0x2f1d(%rip),%eax        # 4024 <lcg.0>
  add    $0x3c6ef35f,%eax
  mov    %eax,D6(%rip)
  cltd
  idiv   %edi
  add    %edx,%edi
  test   %edx,%edx
  mov    %edx,%eax
  cmovs  %edi,%eax
  ret
```

assistant: ```c
int myrand(int n) {
  static int lcg = 0;
  int x;
  lcg = 1664525 * lcg + 1013904223;
  x = lcg % n;
  return x < 0 ? x + n : x;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<myrand>:
  endbr64
  imul   $0x19660d,0x2f16(%rip),%eax        # 4024 <lcg.0>
  add    $0x3c6ef35f,%eax
  cltd
  mov    %eax,D6(%rip)
  idiv   %edi
  add    %edx,%edi
  mov    %edx,%eax
  test   %edx,%edx
  cmovs  %edi,%eax
  ret
```

assistant: ```c
int myrand(int n) {
  static int lcg = 0;
  int x;
  lcg = 1664525 * lcg + 1013904223;
  x = lcg % n;
  return x < 0 ? x + n : x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<myrand>:
  endbr64
  imul   $0x19660d,0x2f16(%rip),%eax        # 4024 <lcg.0>
  add    $0x3c6ef35f,%eax
  cltd
  mov    %eax,D9(%rip)
  idiv   %edi
  add    %edx,%edi
  mov    %edx,%eax
  test   %edx,%edx
  cmovs  %edi,%eax
  ret
```

assistant: ```c
int myrand(int n) {
  static int lcg = 0;
  int x;
  lcg = 1664525 * lcg + 1013904223;
  x = lcg % n;
  return x < 0 ? x + n : x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    $0x0,%eax
  call   *%rdx
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int (*func)();
  func = (int (*)())argv[1];
  (int)(*func)();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   *0x8(%rsi)
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int (*func)();
  func = (int (*)())argv[1];
  (int)(*func)();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   *0x8(%rsi)
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int (*func)();
  func = (int (*)())argv[1];
  (int)(*func)();
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   *0x8(%rsi)
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int (*func)();
  func = (int (*)())argv[1];
  (int)(*func)();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<lb_discr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x30(%rbp)
  mov    %r8,-0x40(%rbp)
  mov    %r9d,-0x44(%rbp)
  cmpq   $0x0,-0x40(%rbp)
  jne    L101
  movq   $0x0,-0x10(%rbp)
  jmp    L17
L101:
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L17
  mov    $0x1,%eax
  jmp    L39
L17:
  cmpq   $0x0,0x10(%rbp)
  jne    L69
  movq   $0x0,-0x8(%rbp)
  jmp    L81
L69:
  mov    0x10(%rbp),%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L81
  cmpq   $0x0,-0x10(%rbp)
  je     L11
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L11:
  mov    $0x1,%eax
  jmp    L39
L81:
  mov    -0x10(%rbp),%r8
  mov    -0x30(%rbp),%edi
  mov    -0x38(%rbp),%rcx
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%rax
  sub    $0x8,%rsp
  push   0x18(%rbp)
  push   -0x8(%rbp)
  mov    -0x44(%rbp),%esi
  push   %rsi
  mov    %r8,%r9
  mov    %edi,%r8d
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L61
  cmpq   $0x0,-0x10(%rbp)
  je     L92
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L92:
  cmpq   $0x0,-0x8(%rbp)
  je     L61
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L61:
  mov    -0x14(%rbp),%eax
L39:
  leave
  ret
```

assistant: ```c
int lb_discr(struct lb *p, int width_word, const char *word, int width_pre, const char *pre, int width_post, const char *post, const void *info) {
  char *p1, *p2;
  int rc;
  if (pre == NULL)
    p1 = NULL;
  else {
    p1 = strdup(pre);
    if (p1 == NULL)
      return 1;
  }
  if (post == NULL)
    p2 = NULL;
  else {
    p2 = strdup(post);
    if (p2 == NULL) {
      if (p1 != NULL)
        free(p1);
      return 1;
    }
  }
  rc = lb_add(p, 4, width_word, word, width_pre, p1, width_post, p2, info);
  if (rc != 0) {
    if (p1 != NULL)
      free(p1);
    if (p2 != NULL)
      free(p2);
  }
  return rc;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<lb_discr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,%r12d
  mov    %rdx,%r13
  mov    %ecx,%r14d
  mov    %r8,%rbx
  mov    %r9d,%r15d
  mov    0x50(%rsp),%rbp
  test   %r8,%r8
  je     L1
  mov    %r8,%rdi
  call   <strdup@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L84
L1:
  test   %rbp,%rbp
  je     L78
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L23
L78:
  sub    $0x8,%rsp
  push   0x60(%rsp)
  push   %rbp
  push   %r15
  mov    %rbx,%r9
  mov    %r14d,%r8d
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    $0x4,%esi
  mov    0x28(%rsp),%rdi
  call   <lb_add@plt>
  mov    %eax,%r12d
  add    $0x20,%rsp
  test   %eax,%eax
  je     L35
  test   %rbx,%rbx
  je     L60
  mov    %rbx,%rdi
  call   <free@plt>
L60:
  test   %rbp,%rbp
  je     L35
  mov    %rbp,%rdi
  call   <free@plt>
L35:
  mov    %r12d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  mov    $0x1,%r12d
  test   %rbx,%rbx
  je     L35
  mov    %rbx,%rdi
  call   <free@plt>
  jmp    L35
L84:
  mov    $0x1,%r12d
  jmp    L35
```

assistant: ```c
int lb_discr(struct lb *p, int width_word, const char *word, int width_pre, const char *pre, int width_post, const char *post, const void *info) {
  char *p1, *p2;
  int rc;
  if (pre == NULL)
    p1 = NULL;
  else {
    p1 = strdup(pre);
    if (p1 == NULL)
      return 1;
  }
  if (post == NULL)
    p2 = NULL;
  else {
    p2 = strdup(post);
    if (p2 == NULL) {
      if (p1 != NULL)
        free(p1);
      return 1;
    }
  }
  rc = lb_add(p, 4, width_word, word, width_pre, p1, width_post, p2, info);
  if (rc != 0) {
    if (p1 != NULL)
      free(p1);
    if (p2 != NULL)
      free(p2);
  }
  return rc;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<lb_discr>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x18,%rsp
  mov    %r9d,0xc(%rsp)
  mov    0x50(%rsp),%rbp
  test   %r8,%r8
  je     L103
  mov    %r8,%rdi
  call   <strdup@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L50
  test   %rbp,%rbp
  je     L97
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L88
  sub    $0x8,%rsp
  mov    %r12,%rdi
  mov    %r15,%r9
  mov    %ebx,%r8d
  push   0x60(%rsp)
  mov    %r14,%rcx
  mov    %r13d,%edx
  mov    $0x4,%esi
  push   %rax
  mov    0x24(%rsp),%eax
  push   %rax
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L63
L21:
  xor    %r12d,%r12d
L54:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L103:
  test   %rbp,%rbp
  je     L95
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L50
  sub    $0x8,%rsp
  mov    %r12,%rdi
  xor    %r9d,%r9d
  mov    %ebx,%r8d
  push   0x60(%rsp)
  mov    %r14,%rcx
  mov    %r13d,%edx
  mov    $0x4,%esi
  push   %rax
  mov    0x24(%rsp),%eax
  push   %rax
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %eax,%eax
  je     L21
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L54
  nopl   0x0(%rax)
L97:
  sub    $0x8,%rsp
  mov    %r12,%rdi
  mov    %r15,%r9
  mov    %ebx,%r8d
  push   0x60(%rsp)
  mov    %r14,%rcx
  mov    %r13d,%edx
  mov    $0x4,%esi
  push   $0x0
  mov    0x24(%rsp),%eax
  push   %rax
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %eax,%eax
  je     L21
  mov    %r15,%rdi
  call   <free@plt>
  jmp    L54
  nopl   0x0(%rax)
L95:
  sub    $0x8,%rsp
  mov    %ecx,%r8d
  xor    %r9d,%r9d
  mov    %rdx,%rcx
  push   0x60(%rsp)
  mov    %esi,%edx
  mov    $0x4,%esi
  push   $0x0
  mov    0x24(%rsp),%eax
  push   %rax
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %eax,%eax
  je     L21
  jmp    L54
L63:
  mov    %r15,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L54
L88:
  mov    %r15,%rdi
  call   <free@plt>
L50:
  mov    $0x1,%r12d
  jmp    L54
```

assistant: ```c
int lb_discr(struct lb *p, int width_word, const char *word, int width_pre, const char *pre, int width_post, const char *post, const void *info) {
  char *p1, *p2;
  int rc;
  if (pre == NULL)
    p1 = NULL;
  else {
    p1 = strdup(pre);
    if (p1 == NULL)
      return 1;
  }
  if (post == NULL)
    p2 = NULL;
  else {
    p2 = strdup(post);
    if (p2 == NULL) {
      if (p1 != NULL)
        free(p1);
      return 1;
    }
  }
  rc = lb_add(p, 4, width_word, word, width_pre, p1, width_post, p2, info);
  if (rc != 0) {
    if (p1 != NULL)
      free(p1);
    if (p2 != NULL)
      free(p2);
  }
  return rc;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<lb_discr>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x18,%rsp
  mov    %r9d,0xc(%rsp)
  mov    0x50(%rsp),%rbp
  test   %r8,%r8
  je     L98
  mov    %r8,%rdi
  call   <strdup@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L60
  test   %rbp,%rbp
  je     L142
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L47
  sub    $0x8,%rsp
  mov    %r12,%rdi
  mov    %r15,%r9
  mov    %ebx,%r8d
  push   0x60(%rsp)
  mov    %r14,%rcx
  mov    %r13d,%edx
  mov    $0x4,%esi
  push   %rax
  mov    0x24(%rsp),%eax
  push   %rax
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L12
L32:
  xor    %r12d,%r12d
L86:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L98:
  test   %rbp,%rbp
  je     L133
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L60
  sub    $0x8,%rsp
  mov    %r12,%rdi
  xor    %r9d,%r9d
  mov    %ebx,%r8d
  push   0x60(%rsp)
  mov    %r14,%rcx
  mov    %r13d,%edx
  mov    $0x4,%esi
  push   %rax
  mov    0x24(%rsp),%eax
  push   %rax
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %eax,%eax
  je     L32
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L86
  nopl   0x0(%rax)
L142:
  sub    $0x8,%rsp
  mov    %r12,%rdi
  mov    %r15,%r9
  mov    %ebx,%r8d
  push   0x60(%rsp)
  mov    %r14,%rcx
  mov    %r13d,%edx
  mov    $0x4,%esi
  push   $0x0
  mov    0x24(%rsp),%eax
  push   %rax
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %eax,%eax
  je     L32
  mov    %r15,%rdi
  call   <free@plt>
  jmp    L86
  nopl   0x0(%rax)
L133:
  sub    $0x8,%rsp
  mov    %ecx,%r8d
  xor    %r9d,%r9d
  mov    %rdx,%rcx
  push   0x60(%rsp)
  mov    %esi,%edx
  mov    $0x4,%esi
  push   $0x0
  mov    0x24(%rsp),%eax
  push   %rax
  call   <lb_add@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %eax,%eax
  je     L32
  jmp    L86
L12:
  mov    %r15,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L86
L47:
  mov    %r15,%rdi
  call   <free@plt>
L60:
  mov    $0x1,%r12d
  jmp    L86
```

assistant: ```c
int lb_discr(struct lb *p, int width_word, const char *word, int width_pre, const char *pre, int width_post, const char *post, const void *info) {
  char *p1, *p2;
  int rc;
  if (pre == NULL)
    p1 = NULL;
  else {
    p1 = strdup(pre);
    if (p1 == NULL)
      return 1;
  }
  if (post == NULL)
    p2 = NULL;
  else {
    p2 = strdup(post);
    if (p2 == NULL) {
      if (p1 != NULL)
        free(p1);
      return 1;
    }
  }
  rc = lb_add(p, 4, width_word, word, width_pre, p1, width_post, p2, info);
  if (rc != 0) {
    if (p1 != NULL)
      free(p1);
    if (p2 != NULL)
      free(p2);
  }
  return rc;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_dictStringKeyValCopyHTValDup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <_dictAlloc@plt>
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *_dictStringKeyValCopyHTValDup(void *privdata, const void *val) {
  int len = strlen(val);
  char *copy = _dictAlloc(len + 1);
  ((void)privdata);
  memcpy(copy, val, len);
  copy[len] = '\0';
  return copy;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_dictStringKeyValCopyHTValDup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r12
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  lea    0x1(%rax),%edi
  mov    $0x0,%eax
  call   <_dictAlloc@plt>
  movslq %eax,%rbp
  movslq %ebx,%rbx
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  movb   $0x0,0x0(%rbp,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void *_dictStringKeyValCopyHTValDup(void *privdata, const void *val) {
  int len = strlen(val);
  char *copy = _dictAlloc(len + 1);
  ((void)privdata);
  memcpy(copy, val, len);
  copy[len] = '\0';
  return copy;
}
```