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

```asm
<printFreqs>:
  endbr64
  test   %edi,%edi
  jle    L6
  push   %r14
  lea    D7(%rip),%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L3:
  movsd  (%r12,%rbx,8),%xmm0
  mov    0x0(%r13,%rbx,8),%rsi
  mov    %r14,%rdi
  add    $0x1,%rbx
  call   <printBothOpen@plt>
  cmp    %rbp,%rbx
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L6:
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
datatype: string

tool: D7: freq pi(%s): %f


assistant: ```c
void printFreqs(int n, double *f, char **names) {
  int k;
  for (k = 0; k < n; k++)
    printBothOpen("freq pi(%s): %f\n", names[k], f[k]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<printFreqs>:
  endbr64
  test   %edi,%edi
  jle    L16
  push   %r14
  lea    D1(%rip),%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L15:
  movsd  (%r12,%rbx,8),%xmm0
  mov    0x0(%r13,%rbx,8),%rsi
  mov    %r14,%rdi
  add    $0x1,%rbx
  call   <printBothOpen@plt>
  cmp    %rbx,%rbp
  jne    L15
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L16:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D1
type_infered: string

tool: D1: freq pi(%s): %f


assistant: ```c
void printFreqs(int n, double *f, char **names) {
  int k;
  for (k = 0; k < n; k++)
    printBothOpen("freq pi(%s): %f\n", names[k], f[k]);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <parse_args@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L11
  mov    -0x20(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <start_server@plt>
  jmp    L16
L11:
  cmpl   $0x2,-0x4(%rbp)
  jne    L12
  mov    -0x20(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <start_client@plt>
  jmp    L16
L12:
  mov    $0x0,%eax
  jmp    L7
L16:
  mov    $0x0,%eax
L7:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int result;
  result = parse_args(argc, argv);
  if (result == 1)
    start_server(argc, argv);
  else if (result == 2)
    start_client(argc, argv);
  else
    return 0;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  call   <parse_args@plt>
  cmp    $0x1,%eax
  je     L10
  cmp    $0x2,%eax
  je     L7
L3:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <start_server@plt>
  jmp    L3
L7:
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <start_client@plt>
  jmp    L3
```

assistant: ```c
int main(int argc, char **argv) {
  int result;
  result = parse_args(argc, argv);
  if (result == 1)
    start_server(argc, argv);
  else if (result == 2)
    start_client(argc, argv);
  else
    return 0;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <parse_args@plt>
  cmp    $0x1,%eax
  je     L15
  cmp    $0x2,%eax
  je     L19
L5:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L19:
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <start_client@plt>
  jmp    L5
L15:
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <start_server@plt>
  jmp    L5
```

assistant: ```c
int main(int argc, char **argv) {
  int result;
  result = parse_args(argc, argv);
  if (result == 1)
    start_server(argc, argv);
  else if (result == 2)
    start_client(argc, argv);
  else
    return 0;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <parse_args@plt>
  cmp    $0x1,%eax
  je     L34
  cmp    $0x2,%eax
  je     L17
L27:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L17:
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <start_client@plt>
  jmp    L27
L34:
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <start_server@plt>
  jmp    L27
```

assistant: ```c
int main(int argc, char **argv) {
  int result;
  result = parse_args(argc, argv);
  if (result == 1)
    start_server(argc, argv);
  else if (result == 2)
    start_client(argc, argv);
  else
    return 0;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<lua_internaldofile>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <luaL_opt_string@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <lua_dofile@plt>
  test   %rax,%rax
  jne    L16
  mov    $0x0,%eax
  call   <passresults@plt>
  test   %rax,%rax
  jne    L16
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <lua_pushusertag@plt>
L16:
  nop
  leave
  ret
```

assistant: ```c
void lua_internaldofile(void) {
  char *fname = luaL_opt_string(1, 0);
  if (lua_dofile(fname) == 0)
    if (passresults() == 0)
      lua_pushusertag(0, 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<lua_internaldofile>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <luaL_opt_string@plt>
  mov    %rax,%rdi
  call   <lua_dofile@plt>
  test   %rax,%rax
  je     L4
L17:
  add    $0x8,%rsp
  ret
L4:
  call   <passresults@plt>
  test   %rax,%rax
  jne    L17
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <lua_pushusertag@plt>
  jmp    L17
```

assistant: ```c
void lua_internaldofile(void) {
  char *fname = luaL_opt_string(1, 0);
  if (lua_dofile(fname) == 0)
    if (passresults() == 0)
      lua_pushusertag(0, 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<lua_internaldofile>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x1,%edi
  call   <luaL_opt_string@plt>
  mov    %rax,%rdi
  call   <lua_dofile@plt>
  test   %rax,%rax
  je     L3
L19:
  add    $0x8,%rsp
  ret
  cs nopw 0x0(%rax,%rax,1)
L3:
  call   <passresults@plt>
  test   %rax,%rax
  jne    L19
  xor    %esi,%esi
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    10a0 <lua_pushusertag@plt>
```

assistant: ```c
void lua_internaldofile(void) {
  char *fname = luaL_opt_string(1, 0);
  if (lua_dofile(fname) == 0)
    if (passresults() == 0)
      lua_pushusertag(0, 0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<lua_internaldofile>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x1,%edi
  call   <luaL_opt_string@plt>
  mov    %rax,%rdi
  call   <lua_dofile@plt>
  test   %rax,%rax
  je     L0
L11:
  add    $0x8,%rsp
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  call   <passresults@plt>
  test   %rax,%rax
  jne    L11
  xor    %esi,%esi
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    10a0 <lua_pushusertag@plt>
```

assistant: ```c
void lua_internaldofile(void) {
  char *fname = luaL_opt_string(1, 0);
  if (lua_dofile(fname) == 0)
    if (passresults() == 0)
      lua_pushusertag(0, 0);
}
```
user: What is the c source code of the assembly code below:

```asm
<strcoll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L27
L24:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L27:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L13
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L13
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L24
L13:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int strcoll(const char *s1, const char *s2) {
  while (*s1 && *s2 && *s1 == *s2)
    ++s1, ++s2;
  return *(unsigned char *)s1 - *(unsigned char *)s2;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcoll>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L32
L11:
  movzbl (%rsi),%edx
  cmp    %al,%dl
  jne    L32
  test   %dl,%dl
  je     L32
  add    $0x1,%rdi
  add    $0x1,%rsi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L11
L32:
  movzbl %al,%eax
  movzbl (%rsi),%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int strcoll(const char *s1, const char *s2) {
  while (*s1 && *s2 && *s1 == *s2)
    ++s1, ++s2;
  return *(unsigned char *)s1 - *(unsigned char *)s2;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcoll>:
  endbr64
  jmp    L25
  cs nopw 0x0(%rax,%rax,1)
L34:
  movzbl (%rsi),%edx
  cmp    %al,%dl
  jne    L1
  test   %dl,%dl
  je     L1
  add    $0x1,%rdi
  add    $0x1,%rsi
L25:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L34
  xor    %eax,%eax
L1:
  movzbl (%rsi),%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int strcoll(const char *s1, const char *s2) {
  while (*s1 && *s2 && *s1 == *s2)
    ++s1, ++s2;
  return *(unsigned char *)s1 - *(unsigned char *)s2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcoll>:
  endbr64
  jmp    L30
  cs nopw 0x0(%rax,%rax,1)
L39:
  movzbl (%rsi),%edx
  cmp    %al,%dl
  jne    L0
  test   %dl,%dl
  je     L0
  add    $0x1,%rdi
  add    $0x1,%rsi
L30:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L39
  xor    %eax,%eax
L0:
  movzbl (%rsi),%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int strcoll(const char *s1, const char *s2) {
  while (*s1 && *s2 && *s1 == *s2)
    ++s1, ++s2;
  return *(unsigned char *)s1 - *(unsigned char *)s2;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<calMod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x1,-0x8(%rbp)
  movq   $0x1,-0x10(%rbp)
  jmp    L9
L17:
  mov    -0x8(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x20(%rbp)
  mov    %rdx,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    $0x0,%edx
  divq   -0x20(%rbp)
  mov    %rdx,%rax
  imul   -0x8(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x20(%rbp)
  mov    %rdx,%rcx
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    $0x0,%edx
  divq   -0x20(%rbp)
  mov    %rdx,%rax
  add    %rcx,%rax
  add    $0x3,%rax
  mov    $0x0,%edx
  divq   -0x20(%rbp)
  mov    %rdx,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L9:
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jb     L17
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long long calMod(unsigned long long n, unsigned long int prime) {
  unsigned long long i;
  unsigned long long modulo = 1;
  for (i = 1; i < n; i++) {
    modulo = modulo % prime;
    modulo = (((((6 * modulo) % prime) * modulo) % prime) + ((10 * modulo) % prime) + 3) % prime;
  }
  return modulo;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<calMod>:
  endbr64
  mov    %rsi,%rcx
  cmp    $0x1,%rdi
  jbe    L29
  mov    $0x1,%edx
  mov    $0x1,%r8d
L17:
  mov    %rdx,%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rsi
  lea    (%rdx,%rdx,2),%rax
  add    %rax,%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rax
  imul   %rsi,%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%r9
  lea    (%rsi,%rsi,4),%rax
  add    %rax,%rax
  mov    $0x0,%edx
  div    %rcx
  lea    0x3(%r9,%rdx,1),%rax
  mov    $0x0,%edx
  div    %rcx
  add    $0x1,%r8
  cmp    %r8,%rdi
  jne    L17
L13:
  mov    %rdx,%rax
  ret
L29:
  mov    $0x1,%edx
  jmp    L13
```

assistant: ```c
unsigned long long calMod(unsigned long long n, unsigned long int prime) {
  unsigned long long i;
  unsigned long long modulo = 1;
  for (i = 1; i < n; i++) {
    modulo = modulo % prime;
    modulo = (((((6 * modulo) % prime) * modulo) % prime) + ((10 * modulo) % prime) + 3) % prime;
  }
  return modulo;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<calMod>:
  endbr64
  mov    %rsi,%rcx
  mov    $0x1,%edx
  cmp    $0x1,%rdi
  jbe    L13
  mov    $0x1,%r8d
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %rdx,%rax
  xor    %edx,%edx
  add    $0x1,%r8
  div    %rcx
  lea    (%rdx,%rdx,2),%rax
  mov    %rdx,%rsi
  xor    %edx,%edx
  add    %rax,%rax
  div    %rcx
  mov    %rdx,%rax
  xor    %edx,%edx
  imul   %rsi,%rax
  div    %rcx
  lea    (%rsi,%rsi,4),%rax
  add    %rax,%rax
  mov    %rdx,%r9
  xor    %edx,%edx
  div    %rcx
  lea    0x3(%r9,%rdx,1),%rax
  xor    %edx,%edx
  div    %rcx
  cmp    %r8,%rdi
  jne    L12
L13:
  mov    %rdx,%rax
  ret
```

assistant: ```c
unsigned long long calMod(unsigned long long n, unsigned long int prime) {
  unsigned long long i;
  unsigned long long modulo = 1;
  for (i = 1; i < n; i++) {
    modulo = modulo % prime;
    modulo = (((((6 * modulo) % prime) * modulo) % prime) + ((10 * modulo) % prime) + 3) % prime;
  }
  return modulo;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<calMod>:
  endbr64
  mov    %rsi,%rcx
  mov    $0x1,%edx
  cmp    $0x1,%rdi
  jbe    L4
  mov    $0x1,%r8d
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %rdx,%rax
  xor    %edx,%edx
  add    $0x1,%r8
  div    %rcx
  lea    (%rdx,%rdx,2),%rax
  mov    %rdx,%rsi
  xor    %edx,%edx
  add    %rax,%rax
  div    %rcx
  mov    %rdx,%rax
  xor    %edx,%edx
  imul   %rsi,%rax
  div    %rcx
  lea    (%rsi,%rsi,4),%rax
  add    %rax,%rax
  mov    %rdx,%r9
  xor    %edx,%edx
  div    %rcx
  lea    0x3(%r9,%rdx,1),%rax
  xor    %edx,%edx
  div    %rcx
  cmp    %r8,%rdi
  jne    L17
L4:
  mov    %rdx,%rax
  ret
```

assistant: ```c
unsigned long long calMod(unsigned long long n, unsigned long int prime) {
  unsigned long long i;
  unsigned long long modulo = 1;
  for (i = 1; i < n; i++) {
    modulo = modulo % prime;
    modulo = (((((6 * modulo) % prime) * modulo) % prime) + ((10 * modulo) % prime) + 3) % prime;
  }
  return modulo;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<try_mfocrf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2a,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int try_mfocrf(int x) {
  int res;
  res = 42;
  return res;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<try_mfocrf>:
  endbr64
  mov    $0x2a,%eax
  ret
```

assistant: ```c
int try_mfocrf(int x) {
  int res;
  res = 42;
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<try_mfocrf>:
  endbr64
  mov    $0x2a,%eax
  ret
```

assistant: ```c
int try_mfocrf(int x) {
  int res;
  res = 42;
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<try_mfocrf>:
  endbr64
  mov    $0x2a,%eax
  ret
```

assistant: ```c
int try_mfocrf(int x) {
  int res;
  res = 42;
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<removeBackListTest>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    D31(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <addFrontList@plt>
  lea    D48(%rip),%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <addFrontList@plt>
  lea    D43(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <addFrontList@plt>
  lea    D46(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printList@plt>
  lea    D26(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <removeBackList@plt>
  lea    D76(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printList@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <removeBackList@plt>
  lea    D78(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printList@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <removeBackList@plt>
  lea    D58(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printList@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D25
type_guessed: string
name: disassemble_data
arguments: data_tag: D31
type_guessed: string
name: disassemble_data
arguments: data_tag: D48
type_guessed: string
name: disassemble_data
arguments: data_tag: D43
type_guessed: string
name: disassemble_data
arguments: data_tag: D46
type_guessed: string
name: disassemble_data
arguments: data_tag: D26
type_guessed: string
name: disassemble_data
arguments: data_tag: D76
type_guessed: string
name: disassemble_data
arguments: data_tag: D78
type_guessed: string
name: disassemble_data
arguments: data_tag: D58
type_guessed: string

tool: D25: 
Testing removeBackList...


tool: D31: Add 3 to the front of the list...


tool: D48: Add 4 to the front of the list...


tool: D43: Add 10 to the front of the list...


tool: D46: The list content should be: [10->4->3]


tool: D26: 

Calling removeBackList(lst)


tool: D76: The list content should be: [10->4]


tool: D78: The list content should be: [10]


tool: D58: The list content should be empty!


assistant: ```c
void removeBackListTest(struct linkedList *lst) {
  printf("\nTesting removeBackList...\n");
  printf("Add 3 to the front of the list...\n");
  addFrontList(lst, 3);
  printf("Add 4 to the front of the list...\n");
  addFrontList(lst, 4);
  printf("Add 10 to the front of the list...\n");
  addFrontList(lst, 10);
  printf("The list content should be: [10->4->3]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be: [10->4]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be: [10]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be empty!\n");
  printList(lst);
}
```
user: Translate the below assembly code into C programming language:

```asm
<removeBackListTest>:
  endbr64
  push   %r12
  lea    D82(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D16(%rip),%rdi
  sub    $0x8,%rsp
  call   <printf@plt>
  lea    D71(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <addFrontList@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <addFrontList@plt>
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <addFrontList@plt>
  lea    D49(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printList@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <removeBackList@plt>
  lea    D56(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printList@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <removeBackList@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printList@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <removeBackList@plt>
  lea    D88(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    10b0 <printList@plt>
```

tool_calls: name: unpack_data
arguments: tag: D82
type_guessed: string
name: unpack_data
arguments: tag: D16
type_guessed: string
name: unpack_data
arguments: tag: D71
type_guessed: string
name: unpack_data
arguments: tag: D30
type_guessed: string
name: unpack_data
arguments: tag: D38
type_guessed: string
name: unpack_data
arguments: tag: D49
type_guessed: string
name: unpack_data
arguments: tag: D56
type_guessed: string
name: unpack_data
arguments: tag: D4
type_guessed: string
name: unpack_data
arguments: tag: D88
type_guessed: string

tool: D82: 

Calling removeBackList(lst)


tool: D16: 
Testing removeBackList...


tool: D71: Add 3 to the front of the list...


tool: D30: Add 4 to the front of the list...


tool: D38: Add 10 to the front of the list...


tool: D49: The list content should be: [10->4->3]


tool: D56: The list content should be: [10->4]


tool: D4: The list content should be: [10]


tool: D88: The list content should be empty!


assistant: ```c
void removeBackListTest(struct linkedList *lst) {
  printf("\nTesting removeBackList...\n");
  printf("Add 3 to the front of the list...\n");
  addFrontList(lst, 3);
  printf("Add 4 to the front of the list...\n");
  addFrontList(lst, 4);
  printf("Add 10 to the front of the list...\n");
  addFrontList(lst, 10);
  printf("The list content should be: [10->4->3]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be: [10->4]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be: [10]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be empty!\n");
  printList(lst);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<removeBackListTest>:
  endbr64
  push   %r12
  lea    D31(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D72(%rip),%rdi
  sub    $0x8,%rsp
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <addFrontList@plt>
  lea    D62(%rip),%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <addFrontList@plt>
  lea    D85(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <addFrontList@plt>
  lea    D44(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printList@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <removeBackList@plt>
  lea    D80(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printList@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <removeBackList@plt>
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printList@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <removeBackList@plt>
  lea    D55(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    10b0 <printList@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D31
type: string
name: unpack_data
arguments: data_tag: D72
type: string
name: unpack_data
arguments: data_tag: D15
type: string
name: unpack_data
arguments: data_tag: D62
type: string
name: unpack_data
arguments: data_tag: D85
type: string
name: unpack_data
arguments: data_tag: D44
type: string
name: unpack_data
arguments: data_tag: D80
type: string
name: unpack_data
arguments: data_tag: D26
type: string
name: unpack_data
arguments: data_tag: D55
type: string

tool: D31: 

Calling removeBackList(lst)


tool: D72: 
Testing removeBackList...


tool: D15: Add 3 to the front of the list...


tool: D62: Add 4 to the front of the list...


tool: D85: Add 10 to the front of the list...


tool: D44: The list content should be: [10->4->3]


tool: D80: The list content should be: [10->4]


tool: D26: The list content should be: [10]


tool: D55: The list content should be empty!


assistant: ```c
void removeBackListTest(struct linkedList *lst) {
  printf("\nTesting removeBackList...\n");
  printf("Add 3 to the front of the list...\n");
  addFrontList(lst, 3);
  printf("Add 4 to the front of the list...\n");
  addFrontList(lst, 4);
  printf("Add 10 to the front of the list...\n");
  addFrontList(lst, 10);
  printf("The list content should be: [10->4->3]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be: [10->4]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be: [10]\n");
  printList(lst);
  printf("\n\nCalling removeBackList(lst)\n");
  removeBackList(lst);
  printf("The list content should be empty!\n");
  printList(lst);
}
```