ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bpf_update_elem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  lea    -0x10(%rbp),%rdx
  mov    $0x8,%ecx
  mov    %eax,%esi
  mov    $0x141,%edi
  call   <syscall@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int bpf_update_elem(int fd, void *key, void *value, unsigned long long flags) {
  union bpf_attr attr;
  memset(&attr, 0, sizeof(attr));
  attr.map_fd = fd;
  attr.key = ptr_to_u64(key);
  attr.value = ptr_to_u64(value);
  attr.flags = flags;
  return syscall(321, BPF_MAP_UPDATE_ELEM, &attr, sizeof(attr));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bpf_update_elem>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r14d
  mov    %rsi,%r13
  mov    %rdx,%r12
  mov    %rcx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    %r14d,(%rsp)
  mov    %r13,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,(%rsp)
  mov    %r12,%rdi
  call   <ptr_to_u64@plt>
  mov    %rbx,(%rsp)
  mov    $0x8,%ecx
  mov    %rbp,%rdx
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  mov    $0x141,%edi
  call   <syscall@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_update_elem(int fd, void *key, void *value, unsigned long long flags) {
  union bpf_attr attr;
  memset(&attr, 0, sizeof(attr));
  attr.map_fd = fd;
  attr.key = ptr_to_u64(key);
  attr.value = ptr_to_u64(value);
  attr.flags = flags;
  return syscall(321, BPF_MAP_UPDATE_ELEM, &attr, sizeof(attr));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<bpf_update_elem>:
  endbr64
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x8,%edx
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <memset@plt>
  mov    %r12,%rdi
  mov    %r13d,(%rsp)
  call   <ptr_to_u64@plt>
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  call   <ptr_to_u64@plt>
  mov    D8(%rip),%rax
  mov    %r14,%rdx
  mov    $0x8,%ecx
  mov    $0x141,%edi
  mov    %rbx,(%rsp)
  mov    (%rax),%esi
  call   <syscall@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_update_elem(int fd, void *key, void *value, unsigned long long flags) {
  union bpf_attr attr;
  memset(&attr, 0, sizeof(attr));
  attr.map_fd = fd;
  attr.key = ptr_to_u64(key);
  attr.value = ptr_to_u64(value);
  attr.flags = flags;
  return syscall(321, BPF_MAP_UPDATE_ELEM, &attr, sizeof(attr));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bpf_update_elem>:
  endbr64
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x8,%edx
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <memset@plt>
  mov    %r12,%rdi
  mov    %r13d,(%rsp)
  call   <ptr_to_u64@plt>
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  call   <ptr_to_u64@plt>
  mov    D7(%rip),%rax
  mov    %r14,%rdx
  mov    $0x8,%ecx
  mov    $0x141,%edi
  mov    %rbx,(%rsp)
  mov    (%rax),%esi
  call   <syscall@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_update_elem(int fd, void *key, void *value, unsigned long long flags) {
  union bpf_attr attr;
  memset(&attr, 0, sizeof(attr));
  attr.map_fd = fd;
  attr.key = ptr_to_u64(key);
  attr.value = ptr_to_u64(value);
  attr.flags = flags;
  return syscall(321, BPF_MAP_UPDATE_ELEM, &attr, sizeof(attr));
}
```
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)
  movl   $0x0,-0x8(%rbp)
  jmp    L9
L0:
  mov    -0x8(%rbp),%eax
  lea    -0x20(%rax),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0x38e38e39,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x14,-0x8(%rbp)
L9:
  cmpl   $0x12c,-0x8(%rbp)
  jle    L0
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type_guessed: string

tool: D3: %d	%d


assistant: ```c
int main(int argc, const char **argv) {
  for (int fahr = 0; fahr <= 300; fahr += 20) {
    int celsius = 5 * (fahr - 32) / 9;
    printf("%d\t%d\n", fahr, celsius);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0xffffff60,%ebp
  mov    $0x0,%ebx
  lea    D4(%rip),%r12
L7:
  movslq %ebp,%rdx
  imul   $0x38e38e39,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %ebp,%eax
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x14,%ebx
  add    $0x64,%ebp
  cmp    $0x140,%ebx
  jne    L7
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type_infered: string

tool: D4: %d	%d


assistant: ```c
int main(int argc, const char **argv) {
  for (int fahr = 0; fahr <= 300; fahr += 20) {
    int celsius = 5 * (fahr - 32) / 9;
    printf("%d\t%d\n", fahr, celsius);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xffffff60,%r12d
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L4:
  movslq %r12d,%rdx
  mov    %r12d,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  imul   $0x38e38e39,%rdx,%rdx
  sar    $0x1f,%eax
  add    $0x14,%ebx
  add    $0x64,%r12d
  sar    $0x21,%rdx
  sub    %eax,%edx
  call   <printf@plt>
  cmp    $0x140,%ebx
  jne    L4
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type_guessed: string

tool: D0: %d	%d


assistant: ```c
int main(int argc, const char **argv) {
  for (int fahr = 0; fahr <= 300; fahr += 20) {
    int celsius = 5 * (fahr - 32) / 9;
    printf("%d\t%d\n", fahr, celsius);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xffffff60,%r12d
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L9:
  movslq %r12d,%rdx
  mov    %r12d,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  imul   $0x38e38e39,%rdx,%rdx
  sar    $0x1f,%eax
  add    $0x14,%ebx
  add    $0x64,%r12d
  sar    $0x21,%rdx
  sub    %eax,%edx
  call   <printf@plt>
  cmp    $0x140,%ebx
  jne    L9
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D1
type: string

tool: D1: %d	%d


assistant: ```c
int main(int argc, const char **argv) {
  for (int fahr = 0; fahr <= 300; fahr += 20) {
    int celsius = 5 * (fahr - 32) / 9;
    printf("%d\t%d\n", fahr, celsius);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<readlink_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x64,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
L8:
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <xrealloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <readlink@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jns    L1
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L16
L1:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jge    L11
  mov    -0x8(%rbp),%rax
  jmp    L16
L11:
  shll   -0x10(%rbp)
  jmp    L8
L16:
  leave
  ret
```

assistant: ```c
char *readlink_malloc(const char *filename) {
  int size = 100;
  char *buffer = NULL;
  while (1) {
    buffer = (char *)xrealloc(buffer, size);
    int nchars = readlink(filename, buffer, size);
    if (nchars < 0) {
      free(buffer);
      return NULL;
    }
    if (nchars < size)
      return buffer;
    size *= 2;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<readlink_malloc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x0,%ebx
  mov    $0x64,%ebp
  jmp    L32
L16:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%ebx
L25:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  add    %ebp,%ebp
L32:
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <xrealloc@plt>
  mov    %rax,%rbx
  mov    %ebp,%edx
  mov    %rax,%rsi
  mov    %r12,%rdi
  call   <readlink@plt>
  test   %eax,%eax
  js     L16
  cmp    %eax,%ebp
  jle    L29
  jmp    L25
```

assistant: ```c
char *readlink_malloc(const char *filename) {
  int size = 100;
  char *buffer = NULL;
  while (1) {
    buffer = (char *)xrealloc(buffer, size);
    int nchars = readlink(filename, buffer, size);
    if (nchars < 0) {
      free(buffer);
      return NULL;
    }
    if (nchars < size)
      return buffer;
    size *= 2;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<readlink_malloc>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x64,%ebx
  jmp    L28
  nopl   0x0(%rax)
L20:
  cmp    %eax,%ebx
  jg     L7
  add    %ebx,%ebx
L28:
  mov    %rbp,%rdi
  mov    %ebx,%esi
  call   <xrealloc@plt>
  mov    %ebx,%edx
  mov    %r12,%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  call   <readlink@plt>
  test   %eax,%eax
  jns    L20
  mov    %rbp,%rdi
  xor    %ebp,%ebp
  call   <free@plt>
L7:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *readlink_malloc(const char *filename) {
  int size = 100;
  char *buffer = NULL;
  while (1) {
    buffer = (char *)xrealloc(buffer, size);
    int nchars = readlink(filename, buffer, size);
    if (nchars < 0) {
      free(buffer);
      return NULL;
    }
    if (nchars < size)
      return buffer;
    size *= 2;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<readlink_malloc>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x64,%ebx
  jmp    L27
  nopl   0x0(%rax)
L9:
  cmp    %eax,%ebx
  jg     L1
  add    %ebx,%ebx
L27:
  mov    %rbp,%rdi
  mov    %ebx,%esi
  call   <xrealloc@plt>
  mov    %ebx,%edx
  mov    %r12,%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  call   <readlink@plt>
  test   %eax,%eax
  jns    L9
  mov    %rbp,%rdi
  xor    %ebp,%ebp
  call   <free@plt>
L1:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *readlink_malloc(const char *filename) {
  int size = 100;
  char *buffer = NULL;
  while (1) {
    buffer = (char *)xrealloc(buffer, size);
    int nchars = readlink(filename, buffer, size);
    if (nchars < 0) {
      free(buffer);
      return NULL;
    }
    if (nchars < size)
      return buffer;
    size *= 2;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<matmult_tn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    %r8,-0x40(%rbp)
  mov    %r9,-0x48(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L28
L34:
  movl   $0x0,-0x14(%rbp)
  jmp    L6
L17:
  mov    -0x14(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x38(%rbp),%rax
  lea    (%rcx,%rax,1),%rsi
  mov    -0x14(%rbp),%eax
  imul   -0x24(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x48(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    -0x2c(%rbp),%eax
  mov    $0x1,%r8d
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    %eax,%edi
  call   <dot@plt>
  movq   %xmm0,%rax
  mov    %rax,(%rbx)
  addl   $0x1,-0x14(%rbp)
L6:
  mov    -0x14(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L17
  addl   $0x1,-0x18(%rbp)
L28:
  mov    -0x18(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L34
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void matmult_tn(int N, int M, int L, double *A, double *B, double *C) {
  int i, j;
  for (i = 0; i < N; i++) {
    for (j = 0; j < M; j++) {
      C[i + j * N] = dot(L, &A[i * L], 1, &B[j * L], 1);
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<matmult_tn>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rcx,0x28(%rsp)
  mov    %r8,0x30(%rsp)
  mov    %r9,0x38(%rsp)
  test   %edi,%edi
  jle    L31
  mov    %edi,%eax
  mov    %esi,%r14d
  mov    %edx,%r13d
  mov    %edi,%edi
  mov    %rdi,0x20(%rsp)
  movslq %edx,%rdx
  shl    $0x3,%rdx
  mov    %rdx,(%rsp)
  cltq
  shl    $0x3,%rax
  mov    %rax,0x8(%rsp)
  movl   $0x0,0x1c(%rsp)
  movq   $0x0,0x10(%rsp)
  jmp    L51
L18:
  movslq 0x1c(%rsp),%rax
  mov    0x28(%rsp),%rdi
  lea    (%rdi,%rax,8),%r15
  mov    0x30(%rsp),%r12
  mov    0x38(%rsp),%rax
  mov    0x10(%rsp),%rdx
  lea    (%rax,%rdx,8),%rbp
  mov    $0x0,%ebx
L37:
  mov    $0x1,%r8d
  mov    %r12,%rcx
  mov    $0x1,%edx
  mov    %r15,%rsi
  mov    %r13d,%edi
  call   <dot@plt>
  movsd  %xmm0,0x0(%rbp)
  add    $0x1,%ebx
  add    (%rsp),%r12
  add    0x8(%rsp),%rbp
  cmp    %ebx,%r14d
  jne    L37
L11:
  addq   $0x1,0x10(%rsp)
  mov    0x10(%rsp),%rax
  add    %r13d,0x1c(%rsp)
  cmp    0x20(%rsp),%rax
  je     L31
L51:
  test   %r14d,%r14d
  jg     L18
  jmp    L11
L31:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void matmult_tn(int N, int M, int L, double *A, double *B, double *C) {
  int i, j;
  for (i = 0; i < N; i++) {
    for (j = 0; j < M; j++) {
      C[i + j * N] = dot(L, &A[i * L], 1, &B[j * L], 1);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<matmult_tn>:
  endbr64
  push   %r15
  movslq %edi,%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %esi,0x18(%rsp)
  mov    %rcx,0x28(%rsp)
  mov    %r8,0x30(%rsp)
  mov    %r9,0x38(%rsp)
  test   %eax,%eax
  jle    L22
  movslq %edx,%r13
  mov    %rax,0x20(%rsp)
  mov    %edx,%ebx
  lea    0x0(,%rax,8),%r14
  lea    0x0(,%r13,8),%rsi
  movl   $0x0,0x1c(%rsp)
  mov    %rsi,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
L30:
  mov    0x18(%rsp),%eax
  test   %eax,%eax
  jle    L7
  movslq 0x1c(%rsp),%rax
  mov    0x28(%rsp),%rsi
  xor    %r15d,%r15d
  mov    0x30(%rsp),%r13
  mov    0x10(%rsp),%rdi
  lea    (%rsi,%rax,8),%r12
  mov    0x38(%rsp),%rax
  lea    (%rax,%rdi,8),%rbp
  mov    %r13,%rax
  mov    %rbp,%r13
  mov    %rax,%rbp
  nopl   (%rax)
L27:
  mov    %rbp,%rcx
  mov    $0x1,%r8d
  mov    $0x1,%edx
  mov    %ebx,%edi
  mov    %r12,%rsi
  add    $0x1,%r15d
  call   <dot@plt>
  add    0x8(%rsp),%rbp
  movsd  %xmm0,0x0(%r13)
  add    %r14,%r13
  cmp    %r15d,0x18(%rsp)
  jne    L27
L7:
  addq   $0x1,0x10(%rsp)
  mov    0x10(%rsp),%rax
  add    %ebx,0x1c(%rsp)
  cmp    0x20(%rsp),%rax
  jne    L30
L22:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void matmult_tn(int N, int M, int L, double *A, double *B, double *C) {
  int i, j;
  for (i = 0; i < N; i++) {
    for (j = 0; j < M; j++) {
      C[i + j * N] = dot(L, &A[i * L], 1, &B[j * L], 1);
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<matmult_tn>:
  endbr64
  push   %r15
  movslq %edi,%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %esi,0x4(%rsp)
  mov    %r8,0x18(%rsp)
  mov    %r9,0x20(%rsp)
  test   %eax,%eax
  jle    L25
  test   %esi,%esi
  jle    L25
  mov    %rax,0x28(%rsp)
  shl    $0x3,%rax
  movslq %edx,%rbp
  mov    %edx,%r13d
  mov    %rax,0x8(%rsp)
  shl    $0x3,%rbp
  mov    %rcx,%rbx
  movq   $0x0,0x10(%rsp)
  nopw   0x0(%rax,%rax,1)
L38:
  mov    0x20(%rsp),%rax
  mov    0x10(%rsp),%rsi
  xor    %r15d,%r15d
  mov    0x18(%rsp),%r12
  lea    (%rax,%rsi,8),%r14
  mov    %r14,%rax
  mov    %r12,%r14
  mov    %rax,%r12
  nop
L11:
  mov    %r14,%rcx
  mov    $0x1,%r8d
  mov    %rbx,%rsi
  mov    %r13d,%edi
  mov    $0x1,%edx
  add    $0x1,%r15d
  add    %rbp,%r14
  call   <dot@plt>
  movsd  %xmm0,(%r12)
  add    0x8(%rsp),%r12
  cmp    %r15d,0x4(%rsp)
  jne    L11
  addq   $0x1,0x10(%rsp)
  add    %rbp,%rbx
  mov    0x10(%rsp),%rax
  cmp    %rax,0x28(%rsp)
  jne    L38
L25:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void matmult_tn(int N, int M, int L, double *A, double *B, double *C) {
  int i, j;
  for (i = 0; i < N; i++) {
    for (j = 0; j < M; j++) {
      C[i + j * N] = dot(L, &A[i * L], 1, &B[j * L], 1);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xd,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <times_two@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D6
type_guessed: string

tool: D6: %d

assistant: ```c
int main() {
  int num = 13;
  times_two(&num);
  printf("%d", num);
}
```
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    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xd,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <times_two@plt>
  mov    0x4(%rsp),%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
data_type: string

tool: D0: %d

assistant: ```c
int main() {
  int num = 13;
  times_two(&num);
  printf("%d", num);
}
```
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
  movl   $0xd,0x4(%rsp)
  call   <times_two@plt>
  mov    0x4(%rsp),%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

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

tool: D6: %d

assistant: ```c
int main() {
  int num = 13;
  times_two(&num);
  printf("%d", num);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0xd,0x4(%rsp)
  call   <times_two@plt>
  mov    0x4(%rsp),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D7
data_type: string

tool: D7: %d

assistant: ```c
int main() {
  int num = 13;
  times_two(&num);
  printf("%d", num);
}
```
user: What is the c source code of the assembly code below?

```asm
<memcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  xor    %rdx,%rax
  and    $0x7,%eax
  test   %rax,%rax
  jne    L18
  jmp    L17
L59:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L17:
  mov    -0x8(%rbp),%rax
  and    $0x7,%eax
  test   %rax,%rax
  je     L11
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L59
  jmp    L11
L41:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,(%rax)
  subq   $0x8,-0x28(%rbp)
  addq   $0x8,-0x10(%rbp)
  addq   $0x8,-0x8(%rbp)
L11:
  cmpq   $0x7,-0x28(%rbp)
  ja     L41
  jmp    L18
L29:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L18:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L29
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void memcpy(void *dest, const void *src, size_t size) {
  char *destc = (char *)dest;
  const char *srcc = (const char *)src;
  if ((uintptr_t)srcc % sizeof(size_t) == (uintptr_t)destc % sizeof(size_t)) {
    while ((uintptr_t)srcc % sizeof(size_t) && size--) {
      *destc++ = *srcc++;
    }
    for (; size >= sizeof(size_t); size -= sizeof(size_t), destc += sizeof(size_t), srcc += sizeof(size_t)) {
      *(size_t *)destc = *(size_t *)srcc;
    }
  }
  while (size--) {
    *destc++ = *srcc++;
  }
}
```