ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<wgotol>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <write_margin@plt>
  mov    %rbp,%rdi
  lea    D7(%rip),%rsi
  call   <write_string@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <write_int@plt>
  mov    %rbp,%rdi
  lea    D6(%rip),%rsi
  call   <write_string@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <write_newline@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_infered: string
name: extract_symbolic_data
arguments: tag: D6
type_infered: string

tool: D7: break lab

tool: D6: ;

assistant: ```c
void wgotol(struct generator *g, int n) {
  write_margin(g);
  write_string(g, "break lab");
  write_int(g, n);
  write_string(g, ";");
  write_newline(g);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<wgotol>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <write_margin@plt>
  mov    %rbp,%rdi
  lea    D12(%rip),%rsi
  call   <write_string@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <write_int@plt>
  mov    %rbp,%rdi
  lea    D2(%rip),%rsi
  call   <write_string@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <write_newline@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D12
type: string
name: disassemble_data
arguments: data_label: D2
type: string

tool: D12: break lab

tool: D2: ;

assistant: ```c
void wgotol(struct generator *g, int n) {
  write_margin(g);
  write_string(g, "break lab");
  write_int(g, n);
  write_string(g, ";");
  write_newline(g);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <getint@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
data_type: string

tool: D2: %i %i


assistant: ```c
int main() {
  int a;
  int i = getint(&a);
  printf("%i %i\n", i, a);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <getint@plt>
  mov    %eax,%esi
  mov    0x4(%rsp),%edx
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D5
data_type: string

tool: D5: %i %i


assistant: ```c
int main() {
  int a;
  int i = getint(&a);
  printf("%i %i\n", i, a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <getint@plt>
  mov    0x4(%rsp),%edx
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D6
datatype: string

tool: D6: %i %i


assistant: ```c
int main() {
  int a;
  int i = getint(&a);
  printf("%i %i\n", i, a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <getint@plt>
  mov    0x4(%rsp),%edx
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type: string

tool: D4: %i %i


assistant: ```c
int main() {
  int a;
  int i = getint(&a);
  printf("%i %i\n", i, a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ece391_strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L17
L32:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    $0x0,%eax
  jmp    L6
L10:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L17:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L32
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbq %al,%rdx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbq %al,%rcx
  mov    %rdx,%rax
  sub    %rcx,%rax
L6:
  pop    %rbp
  ret
```

assistant: ```c
int32_t ece391_strcmp(const uint8_t *s1, const uint8_t *s2) {
  while (*s1 == *s2) {
    if (*s1 == '\0')
      return 0;
    s1++;
    s2++;
  }
  return ((int32_t)*s1) - ((int32_t)*s2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ece391_strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  cmp    %cl,%dl
  jne    L15
  mov    $0x1,%eax
L8:
  test   %dl,%dl
  je     L18
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  je     L8
L15:
  movsbq %dl,%rax
  movsbq %cl,%rcx
  sub    %rcx,%rax
  ret
L18:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int32_t ece391_strcmp(const uint8_t *s1, const uint8_t *s2) {
  while (*s1 == *s2) {
    if (*s1 == '\0')
      return 0;
    s1++;
    s2++;
  }
  return ((int32_t)*s1) - ((int32_t)*s2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ece391_strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movsbq (%rsi),%rcx
  cmp    %dl,%cl
  jne    L17
  mov    $0x1,%eax
  jmp    L5
  cs nopw 0x0(%rax,%rax,1)
L35:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movsbq -0x1(%rsi,%rax,1),%rcx
  cmp    %cl,%dl
  jne    L17
L5:
  test   %dl,%dl
  jne    L35
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L17:
  movsbq %dl,%rax
  sub    %rcx,%rax
  ret
```

assistant: ```c
int32_t ece391_strcmp(const uint8_t *s1, const uint8_t *s2) {
  while (*s1 == *s2) {
    if (*s1 == '\0')
      return 0;
    s1++;
    s2++;
  }
  return ((int32_t)*s1) - ((int32_t)*s2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ece391_strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movsbq (%rsi),%rcx
  cmp    %dl,%cl
  jne    L23
  mov    $0x1,%eax
  jmp    L10
  cs nopw 0x0(%rax,%rax,1)
L22:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movsbq -0x1(%rsi,%rax,1),%rcx
  cmp    %cl,%dl
  jne    L23
L10:
  test   %dl,%dl
  jne    L22
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L23:
  movsbq %dl,%rax
  sub    %rcx,%rax
  ret
```

assistant: ```c
int32_t ece391_strcmp(const uint8_t *s1, const uint8_t *s2) {
  while (*s1 == *s2) {
    if (*s1 == '\0')
      return 0;
    s1++;
    s2++;
  }
  return ((int32_t)*s1) - ((int32_t)*s2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<SortArray_copy>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%r14d
  mov    %rdx,%r12
  mov    %rcx,%rbx
  mov    %r8d,%ebp
  lea    0x1(%r8),%edi
  mov    $0x4,%esi
  call   <calloc@plt>
  mov    %rax,%r15
  test   %ebp,%ebp
  js     L63
  lea    0x1(%rbp),%eax
  lea    (%r15,%rax,4),%rsi
  mov    %r15,%rax
  mov    %rbx,%rcx
  sub    %r15,%rcx
L2:
  mov    (%rcx,%rax,1),%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L2
L63:
  movslq %ebp,%rax
  cmp    %r14d,(%rbx,%rax,4)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    %rax,%r8
  test   %r14d,%r14d
  jle    L44
  mov    %r14d,%r14d
  lea    (%rax,%r14,4),%rsi
  mov    %rax,%rdx
  mov    %rax,%r9
  neg    %r9
  mov    %r13,%rdi
  sub    %rax,%rdi
L48:
  mov    (%rdi,%rdx,1),%ecx
  mov    %ecx,(%rdx)
  add    $0x4,%rdx
  cmp    %rsi,%rdx
  jne    L48
  add    %r9,%r12
L10:
  movslq (%r12,%rax,1),%rdx
  lea    (%rbx,%rdx,4),%rcx
  mov    (%rcx),%edx
  lea    0x1(%rdx),%edi
  mov    %edi,(%rcx)
  mov    (%rax),%ecx
  movslq %edx,%rdx
  mov    %ecx,0x0(%r13,%rdx,4)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L10
L44:
  mov    %r8,%rdi
  call   <free@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <EndIndex2BeginIndex@plt>
  test   %ebp,%ebp
  js     L29
  mov    %r15,%r12
  mov    $0x0,%r13d
  sub    %r15,%rbx
L59:
  mov    (%r12),%eax
  cmp    %eax,(%rbx,%r12,1)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x1,%r13d
  add    $0x4,%r12
  cmp    %r13d,%ebp
  jge    L59
L29:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void SortArray_copy(int *a, int na, int *key, int *index, int ni) {
  int i;
  int *newa;
  int *aux;
  aux = (int *)calloc(ni + 1, sizeof(int));
  for (i = 0; i <= ni; i++)
    aux[i] = index[i];
  assert(index[ni] == na);
  newa = (int *)calloc(na, sizeof(int));
  for (i = 0; i < na; i++)
    newa[i] = a[i];
  for (i = 0; i < na; i++)
    a[index[key[i]]++] = newa[i];
  free(newa);
  EndIndex2BeginIndex(index, ni);
  for (i = 0; i <= ni; i++)
    assert(aux[i] == index[i]);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<SortArray_copy>:
  endbr64
  push   %r15
  mov    %rcx,%r15
  push   %r14
  mov    %esi,%r14d
  mov    $0x4,%esi
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  lea    0x1(%r8),%edi
  push   %rbp
  push   %rbx
  mov    %r8d,%ebx
  sub    $0x8,%rsp
  call   <calloc@plt>
  movslq %ebx,%rdi
  mov    %rax,%rbp
  test   %ebx,%ebx
  js     L16
  mov    %r15,%rcx
  lea    0x4(%rax,%rdi,4),%rsi
  sub    %rax,%rcx
  nopl   0x0(%rax)
L18:
  mov    (%rcx,%rax,1),%edx
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  cmp    %rsi,%rax
  jne    L18
L16:
  cmp    %r14d,(%r15,%rdi,4)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x4,%esi
  mov    %r14d,%edi
  call   <calloc@plt>
  mov    %rax,%r8
  test   %r14d,%r14d
  jle    L3
  mov    %rax,%r9
  mov    %r12,%rdi
  lea    (%rax,%r14,4),%rsi
  mov    %rax,%rdx
  neg    %r9
  sub    %rax,%rdi
L2:
  mov    (%rdi,%rdx,1),%ecx
  add    $0x4,%rdx
  mov    %ecx,-0x4(%rdx)
  cmp    %rdx,%rsi
  jne    L2
  add    %r13,%r9
  nopw   0x0(%rax,%rax,1)
L20:
  movslq (%r9,%rax,1),%rdx
  add    $0x4,%rax
  lea    (%r15,%rdx,4),%rcx
  movslq (%rcx),%rdx
  lea    0x1(%rdx),%edi
  mov    %edi,(%rcx)
  mov    -0x4(%rax),%ecx
  mov    %ecx,(%r12,%rdx,4)
  cmp    %rax,%rsi
  jne    L20
L3:
  mov    %r8,%rdi
  call   <free@plt>
  mov    %ebx,%esi
  mov    %r15,%rdi
  call   <EndIndex2BeginIndex@plt>
  test   %ebx,%ebx
  js     L69
  xor    %r12d,%r12d
  sub    %rbp,%r15
  nopl   0x0(%rax)
L24:
  mov    0x0(%rbp),%eax
  xor    %edi,%edi
  cmp    %eax,(%r15,%rbp,1)
  sete   %dil
  add    $0x1,%r12d
  add    $0x4,%rbp
  call   <assert@plt>
  cmp    %r12d,%ebx
  jge    L24
L69:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void SortArray_copy(int *a, int na, int *key, int *index, int ni) {
  int i;
  int *newa;
  int *aux;
  aux = (int *)calloc(ni + 1, sizeof(int));
  for (i = 0; i <= ni; i++)
    aux[i] = index[i];
  assert(index[ni] == na);
  newa = (int *)calloc(na, sizeof(int));
  for (i = 0; i < na; i++)
    newa[i] = a[i];
  for (i = 0; i < na; i++)
    a[index[key[i]]++] = newa[i];
  free(newa);
  EndIndex2BeginIndex(index, ni);
  for (i = 0; i <= ni; i++)
    assert(aux[i] == index[i]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_filename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0xffffffff,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L21:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L11
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L11:
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  jl     L21
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *get_filename(char *path) {
  int last_delim = -1;
  for (int i = 0; i < strlen(path); ++i) {
    if (path[i] == '/')
      last_delim = i;
  }
  char *filename = path + last_delim + 1;
  return filename;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_filename>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%ebx
  mov    $0xffffffff,%r12d
  jmp    L17
L11:
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  cmove  %r13d,%r12d
  add    $0x1,%rbx
L17:
  mov    %ebx,%r13d
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L11
  movslq %r12d,%r12
  lea    0x1(%rbp,%r12,1),%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *get_filename(char *path) {
  int last_delim = -1;
  for (int i = 0; i < strlen(path); ++i) {
    if (path[i] == '/')
      last_delim = i;
  }
  char *filename = path + last_delim + 1;
  return filename;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_filename>:
  endbr64
  push   %r12
  mov    $0xffffffff,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L13
  nopl   (%rax)
L8:
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  cmove  %ebx,%r12d
  add    $0x1,%rbx
L13:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L8
  movslq %r12d,%r12
  pop    %rbx
  lea    0x1(%rbp,%r12,1),%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *get_filename(char *path) {
  int last_delim = -1;
  for (int i = 0; i < strlen(path); ++i) {
    if (path[i] == '/')
      last_delim = i;
  }
  char *filename = path + last_delim + 1;
  return filename;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_filename>:
  endbr64
  push   %r12
  mov    $0xffffffff,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L3
  nopl   (%rax)
L6:
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  cmove  %ebx,%r12d
  add    $0x1,%rbx
L3:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L6
  movslq %r12d,%r12
  pop    %rbx
  lea    0x1(%rbp,%r12,1),%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *get_filename(char *path) {
  int last_delim = -1;
  for (int i = 0; i < strlen(path); ++i) {
    if (path[i] == '/')
      last_delim = i;
  }
  char *filename = path + last_delim + 1;
  return filename;
}
```
user: What is the c source code of the assembly code below:

```asm
<MergeSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jne    L11
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x24(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  jmp    L12
L11:
  mov    -0x28(%rbp),%eax
  sub    -0x24(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x8(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <MergeSort>
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edi
  mov    -0x28(%rbp),%edx
  mov    -0x8(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %edx,%ecx
  mov    %edi,%edx
  mov    %rax,%rdi
  call   <MergeSort>
  mov    -0x28(%rbp),%edi
  mov    -0xc(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <Merge@plt>
L12:
  nop
  leave
  ret
```

assistant: ```c
void MergeSort(int *source, int *target, int s, int t) {
  int m, *temp;
  if (s == t) {
    target[s] = source[s];
  } else {
    temp = (int *)malloc(sizeof(int) * (t - s + 1));
    m = (s + t) / 2;
    MergeSort(source, temp, s, m);
    MergeSort(source, temp, m + 1, t);
    Merge(temp, target, s, m, t);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<MergeSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %edx,%ebx
  cmp    %ecx,%edx
  jne    L3
  movslq %edx,%rbx
  mov    (%rdi,%rbx,4),%eax
  mov    %eax,(%rsi,%rbx,4)
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  mov    %ecx,%ebp
  mov    %ecx,%eax
  sub    %edx,%eax
  lea    0x4(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r15
  lea    (%rbx,%rbp,1),%eax
  mov    %eax,%r14d
  shr    $0x1f,%r14d
  add    %eax,%r14d
  sar    %r14d
  mov    %r14d,%ecx
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <MergeSort>
  lea    0x1(%r14),%edx
  mov    %ebp,%ecx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <MergeSort>
  mov    %ebp,%r8d
  mov    %r14d,%ecx
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <Merge@plt>
  jmp    L18
```

assistant: ```c
void MergeSort(int *source, int *target, int s, int t) {
  int m, *temp;
  if (s == t) {
    target[s] = source[s];
  } else {
    temp = (int *)malloc(sizeof(int) * (t - s + 1));
    m = (s + t) / 2;
    MergeSort(source, temp, s, m);
    MergeSort(source, temp, m + 1, t);
    Merge(temp, target, s, m, t);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<MergeSort>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %ecx,%r12d
  jne    L1
  mov    (%rdi,%r12,4),%eax
  mov    %eax,(%rsi,%r12,4)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %ecx,%eax
  mov    %ecx,%ebx
  sub    %r12d,%eax
  lea    0x4(,%rax,4),%edi
  call   <malloc@plt>
  mov    %r12d,%edx
  mov    %r15,%rdi
  mov    %rax,%rbp
  lea    (%r12,%rbx,1),%eax
  mov    %eax,%r13d
  mov    %rbp,%rsi
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  mov    %r13d,%ecx
  call   <MergeSort>
  lea    0x1(%r13),%edx
  mov    %ebx,%ecx
  mov    %rbp,%rsi
  mov    %r15,%rdi
  call   <MergeSort>
  add    $0x8,%rsp
  mov    %ebx,%r8d
  mov    %r13d,%ecx
  pop    %rbx
  mov    %r12d,%edx
  mov    %r14,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <Merge@plt>
```

assistant: ```c
void MergeSort(int *source, int *target, int s, int t) {
  int m, *temp;
  if (s == t) {
    target[s] = source[s];
  } else {
    temp = (int *)malloc(sizeof(int) * (t - s + 1));
    m = (s + t) / 2;
    MergeSort(source, temp, s, m);
    MergeSort(source, temp, m + 1, t);
    Merge(temp, target, s, m, t);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L7
  mov    $0x0,%eax
  jmp    L9
L7:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    %rax,%rdi
  call   <check_name@plt>
L9:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc < 2)
    return 0;
  return check_name(&argv[1]);
}
```
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
  mov    $0x0,%eax
  cmp    $0x1,%edi
  jle    L6
  sub    $0x8,%rsp
  lea    0x8(%rsi),%rdi
  call   <check_name@plt>
  add    $0x8,%rsp
  ret
L6:
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc < 2)
    return 0;
  return check_name(&argv[1]);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L6
  lea    0x8(%rsi),%rdi
  jmp    1050 <check_name@plt>
L6:
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc < 2)
    return 0;
  return check_name(&argv[1]);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L8
  lea    0x8(%rsi),%rdi
  jmp    1050 <check_name@plt>
L8:
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc < 2)
    return 0;
  return check_name(&argv[1]);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x10000,%edi
  call   <f@plt>
  mov    %rax,%rbx
  mov    $0x10000,%edi
  call   <g@plt>
  mov    %eax,%edi
  call   <h@plt>
  cmp    %rax,%rbx
  je     L2
  mov    $0xffffffff,%edi
  call   <_exit@plt>
L2:
  mov    $0x0,%edi
  call   <_exit@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main() {
  if (f(65536) != h(g(65536)))
    _exit(-1);
  _exit(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    $0x10000,%edi
  call   <f@plt>
  mov    %rax,%rbx
  mov    $0x10000,%edi
  call   <g@plt>
  mov    %eax,%edi
  call   <h@plt>
  cmp    %rax,%rbx
  jne    L19
L4:
  mov    $0x0,%edi
  call   <_exit@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
L19:
  mov    $0xffffffff,%edi
  call   <_exit@plt>
  jmp    L4
```

assistant: ```c
int main() {
  if (f(65536) != h(g(65536)))
    _exit(-1);
  _exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x10000,%edi
  call   <f@plt>
  mov    $0x10000,%edi
  mov    %rax,%rbx
  call   <g@plt>
  mov    %eax,%edi
  call   <h@plt>
  cmp    %rax,%rbx
  je     L2
  or     $0xffffffff,%edi
  call   <_exit@plt>
L2:
  xor    %edi,%edi
  call   <_exit@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  if (f(65536) != h(g(65536)))
    _exit(-1);
  _exit(0);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x10000,%edi
  call   <f@plt>
  mov    $0x10000,%edi
  mov    %rax,%rbx
  call   <g@plt>
  mov    %eax,%edi
  call   <h@plt>
  cmp    %rax,%rbx
  je     L7
  or     $0xffffffff,%edi
  call   <_exit@plt>
L7:
  xor    %edi,%edi
  call   <_exit@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  if (f(65536) != h(g(65536)))
    _exit(-1);
  _exit(0);
}
```