ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<bin_search_diagonal>:
  endbr64
  push   %r13
  mov    %rsi,%r10
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x28(%rsp),%rdi
  mov    0x30(%rsp),%ebp
  lea    (%rcx,%rdi,1),%rsi
  sub    %r9,%rsi
  cmp    %r8,%rsi
  jg     L33
  mov    %rcx,%rbx
  cmp    %rdi,%r9
  jg     L80
  cmp    %rsi,%rcx
  jg     L80
  mov    %rdx,%r11
  jmp    L60
  nopw   0x0(%rax,%rax,1)
L21:
  lea    0x1(%rax),%rcx
  lea    0x1(%rdx),%r9
  cmp    %rcx,%rsi
  jl     L0
L71:
  cmp    %r9,%rdi
  jl     L0
L60:
  mov    %rsi,%rax
  mov    %rdi,%rdx
  sub    %rcx,%rax
  sub    %r9,%rdx
  sar    %rax
  sar    %rdx
  add    %rcx,%rax
  add    %r9,%rdx
  mov    %rax,%r13
  lea    0x0(,%rdx,4),%r12
  imul   %r10,%r13
  lea    (%r12,%r13,4),%r12
  cmp    %ebp,(%r11,%r12,1)
  je     L23
  jl     L21
  lea    -0x1(%rax),%rsi
  lea    -0x1(%rdx),%rdi
  cmp    %rcx,%rsi
  jge    L71
L0:
  xor    %eax,%eax
  cmp    %rsi,%rbx
  jg     L33
  mov    %r8,%rax
  sub    %rbx,%rsi
  sub    %rbx,%rax
  cmp    %rcx,%r8
  cmovge %rsi,%rax
L33:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  sub    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L80:
  mov    %rbx,%rcx
  jmp    L0
```

assistant: ```c
ssize_t bin_search_diagonal(size_t M, size_t N, int matrix[M][N], ssize_t row_l, ssize_t row_r, ssize_t col_l, ssize_t col_r, int val) {
  ssize_t x_l = row_l;
  ssize_t y_l = col_l;
  ssize_t x_r = row_l + col_r - col_l;
  ssize_t y_r = col_r;
  if (x_r > row_r || y_r > col_r) {
    return 0;
  }
  while (x_l <= x_r && y_l <= y_r) {
    ssize_t mid_x = x_l + (x_r - x_l) / 2;
    ssize_t mid_y = y_l + (y_r - y_l) / 2;
    if (matrix[mid_x][mid_y] == val) {
      return mid_x - row_l;
    } else if (matrix[mid_x][mid_y] < val) {
      x_l = mid_x + 1;
      y_l = mid_y + 1;
    } else {
      x_r = mid_x - 1;
      y_r = mid_y - 1;
    }
  }
  if (x_r < row_l) {
    return 0;
  } else if (x_l > row_r) {
    return row_r - row_l;
  } else {
    return x_r - row_l;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<randperm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L26
L35:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L35
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L10
L9:
  mov    $0x0,%eax
  call   <lrand48@plt>
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,-0x8(%rbp)
  subl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L10:
  cmpl   $0x1,-0xc(%rbp)
  jg     L9
  nop
  nop
  leave
  ret
```

assistant: ```c
void randperm(int perm[], int d) {
  int i;
  for (i = 0; i < d; i++)
    perm[i] = i;
  i = d;
  while (i > 1) {
    int j = lrand48() % i;
    i--;
    int t = perm[i];
    perm[i] = perm[j];
    perm[j] = t;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<randperm>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  test   %esi,%esi
  jle    L39
  mov    %esi,%edx
  mov    $0x0,%eax
L2:
  mov    %eax,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L2
L39:
  cmp    $0x1,%esi
  jle    L34
  movslq %esi,%rbx
L27:
  mov    $0x0,%eax
  call   <lrand48@plt>
  mov    -0x4(%rbp,%rbx,4),%ecx
  cltd
  idiv   %ebx
  movslq %edx,%rdx
  lea    0x0(%rbp,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,-0x4(%rbp,%rbx,4)
  mov    %ecx,(%rax)
  sub    $0x1,%rbx
  cmp    $0x1,%ebx
  jg     L27
L34:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void randperm(int perm[], int d) {
  int i;
  for (i = 0; i < d; i++)
    perm[i] = i;
  i = d;
  while (i > 1) {
    int j = lrand48() % i;
    i--;
    int t = perm[i];
    perm[i] = perm[j];
    perm[j] = t;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<randperm>:
  endbr64
  test   %esi,%esi
  jle    L23
  push   %rbp
  movslq %esi,%rax
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L27:
  mov    %ebx,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    %rax,%rbx
  jne    L27
  cmp    $0x1,%esi
  jle    L25
  nopw   0x0(%rax,%rax,1)
L21:
  xor    %eax,%eax
  call   <lrand48@plt>
  mov    -0x4(%rbp,%rbx,4),%ecx
  cltd
  idiv   %ebx
  movslq %edx,%rdx
  lea    0x0(%rbp,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,-0x4(%rbp,%rbx,4)
  sub    $0x1,%rbx
  mov    %ecx,(%rax)
  cmp    $0x1,%ebx
  jg     L21
L25:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L23:
  ret
```

assistant: ```c
void randperm(int perm[], int d) {
  int i;
  for (i = 0; i < d; i++)
    perm[i] = i;
  i = d;
  while (i > 1) {
    int j = lrand48() % i;
    i--;
    int t = perm[i];
    perm[i] = perm[j];
    perm[j] = t;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<randperm>:
  endbr64
  test   %esi,%esi
  jle    L62
  push   %rbp
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%eax
  jbe    L61
  mov    %esi,%edx
  movdqa D7(%rip),%xmm0
  movdqa D15(%rip),%xmm2
  mov    %rdi,%rax
  shr    $0x2,%edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   (%rax)
L22:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  movups %xmm1,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L22
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%sil
  je     L43
L83:
  movslq %eax,%rdx
  mov    %eax,0x0(%rbp,%rdx,4)
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%rax),%edx
  cmp    %edx,%esi
  jle    L94
  add    $0x2,%eax
  mov    %edx,0x4(%rbp,%rcx,1)
  cmp    %eax,%esi
  jle    L43
  mov    %eax,0x8(%rbp,%rcx,1)
L43:
  movslq %esi,%rbx
L34:
  xor    %eax,%eax
  call   <lrand48@plt>
  mov    -0x4(%rbp,%rbx,4),%ecx
  cltd
  idiv   %ebx
  movslq %edx,%rdx
  lea    0x0(%rbp,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,-0x4(%rbp,%rbx,4)
  sub    $0x1,%rbx
  mov    %ecx,(%rax)
  cmp    $0x1,%ebx
  jg     L34
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L62:
  ret
L61:
  xor    %eax,%eax
  jmp    L83
L94:
  cmp    $0x1,%esi
  jg     L43
  jmp    L5
```

tool_calls: name: parse_data
arguments: label: D7
type_guessed: byte[16]
name: parse_data
arguments: label: D15
type_guessed: byte[16]

tool: {"D7": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D15": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
void randperm(int perm[], int d) {
  int i;
  for (i = 0; i < d; i++)
    perm[i] = i;
  i = d;
  while (i > 1) {
    int j = lrand48() % i;
    i--;
    int t = perm[i];
    perm[i] = perm[j];
    perm[j] = t;
  }
}
```
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
  movl   $0x0,-0x10(%rbp)
  jmp    L26
L30:
  mov    -0x10(%rbp),%edx
  movslq %edx,%rax
  imul   $0x78787879,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x4,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L18
  addl   $0x1,-0xc(%rbp)
  jmp    L11
L18:
  addl   $0x1,-0x8(%rbp)
L11:
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x10(%rbp)
L26:
  cmpl   $0x3e7,-0x10(%rbp)
  jle    L30
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
main() {
  int i;
  int a, b, c;
  for (i = 0; i < 1000; i++) {
    if (i % 17)
      a++;
    else
      b++;
    c++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x3e8,%eax
L6:
  sub    $0x1,%eax
  jne    L6
  mov    $0x0,%eax
  ret
```

assistant: ```c
main() {
  int i;
  int a, b, c;
  for (i = 0; i < 1000; i++) {
    if (i % 17)
      a++;
    else
      b++;
    c++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
main() {
  int i;
  int a, b, c;
  for (i = 0; i < 1000; i++) {
    if (i % 17)
      a++;
    else
      b++;
    c++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
main() {
  int i;
  int a, b, c;
  for (i = 0; i < 1000; i++) {
    if (i % 17)
      a++;
    else
      b++;
    c++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<recv_comm_until>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x440,%rsp
  mov    %edi,-0x424(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %rdx,-0x438(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x400,-0x418(%rbp)
  movl   $0x0,-0x41c(%rbp)
  mov    -0x418(%rbp),%eax
  mov    %eax,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,%rdx
  mov    -0x430(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L48
L89:
  lea    -0x410(%rbp),%rsi
  mov    -0x424(%rbp),%eax
  mov    $0x0,%ecx
  mov    $0x400,%edx
  mov    %eax,%edi
  call   <recv@plt>
  mov    %eax,-0x414(%rbp)
  cmpl   $0x0,-0x414(%rbp)
  jns    L83
  mov    D16(%rip),%rax
  mov    (%rax),%rdx
  mov    D13(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L7
  jmp    L48
L7:
  mov    $0xfffffffe,%eax
  jmp    L71
L83:
  cmpl   $0x0,-0x414(%rbp)
  jne    L59
  mov    $0xffffffff,%eax
  jmp    L71
L59:
  mov    -0x41c(%rbp),%edx
  mov    -0x414(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x418(%rbp)
  jge    L10
  mov    -0x41c(%rbp),%edx
  mov    -0x414(%rbp),%eax
  add    %edx,%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x430(%rbp),%rax
  mov    (%rax),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,%rdx
  mov    -0x430(%rbp),%rax
  mov    %rdx,(%rax)
L10:
  mov    -0x430(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x41c(%rbp),%eax
  cltq
  lea    (%rdx,%rax,1),%rcx
  mov    -0x414(%rbp),%edx
  lea    -0x410(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x414(%rbp),%eax
  add    %eax,-0x41c(%rbp)
L48:
  mov    -0x430(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x438(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L89
  mov    $0x0,%eax
L71:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
int recv_comm_until(int socket, unsigned char **buffer, char *needle) {
  int total_bytes_read;
  int bytes_read;
  int buffer_length = 1024;
  char temp_buffer[1024];
  total_bytes_read = 0;
  *buffer = (unsigned char *)calloc(sizeof(unsigned char), buffer_length);
  while (strstr((char *)*buffer, needle) == NULL) {
    bytes_read = recv(socket, temp_buffer, 1024, 0);
    if (bytes_read < 0) {
      if (errno == EINTR) {
        continue;
      } else {
        return -2;
      }
    } else if (bytes_read == 0) {
      return -1;
    }
    if ((total_bytes_read + bytes_read) > buffer_length) {
      *buffer = (unsigned char *)realloc(*buffer, (total_bytes_read + bytes_read) * 2);
    }
    memcpy(&(*buffer)[total_bytes_read], temp_buffer, bytes_read);
    total_bytes_read += bytes_read;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<recv_comm_until>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x428,%rsp
  mov    %edi,0xc(%rsp)
  mov    %rsi,%r12
  mov    %rdx,%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  mov    $0x400,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,(%r12)
  mov    $0x0,%ebp
  lea    0x10(%rsp),%r14
  jmp    L80
L24:
  mov    D9(%rip),%rdx
  mov    D6(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L80
  mov    $0xfffffffe,%eax
  jmp    L22
L16:
  movslq %ebp,%rdi
  add    (%r12),%rdi
  mov    %ebx,%edx
  mov    %r14,%rsi
  call   <memcpy@plt>
  mov    %r13d,%ebp
L80:
  mov    %r15,%rsi
  mov    (%r12),%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L1
  mov    $0x0,%ecx
  mov    $0x400,%edx
  mov    %r14,%rsi
  mov    0xc(%rsp),%edi
  call   <recv@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L24
  je     L82
  lea    0x0(%rbp,%rax,1),%r13d
  cmp    $0x400,%r13d
  jle    L16
  lea    0x0(%r13,%r13,1),%esi
  mov    (%r12),%rdi
  call   <realloc@plt>
  mov    %rax,(%r12)
  jmp    L16
L1:
  mov    $0x0,%eax
L22:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x428,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L82:
  mov    $0xffffffff,%eax
  jmp    L22
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int recv_comm_until(int socket, unsigned char **buffer, char *needle) {
  int total_bytes_read;
  int bytes_read;
  int buffer_length = 1024;
  char temp_buffer[1024];
  total_bytes_read = 0;
  *buffer = (unsigned char *)calloc(sizeof(unsigned char), buffer_length);
  while (strstr((char *)*buffer, needle) == NULL) {
    bytes_read = recv(socket, temp_buffer, 1024, 0);
    if (bytes_read < 0) {
      if (errno == EINTR) {
        continue;
      } else {
        return -2;
      }
    } else if (bytes_read == 0) {
      return -1;
    }
    if ((total_bytes_read + bytes_read) > buffer_length) {
      *buffer = (unsigned char *)realloc(*buffer, (total_bytes_read + bytes_read) * 2);
    }
    memcpy(&(*buffer)[total_bytes_read], temp_buffer, bytes_read);
    total_bytes_read += bytes_read;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<recv_comm_until>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  mov    $0x1,%edi
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x400,%esi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x428,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r13
  call   <calloc@plt>
  mov    %rax,0x0(%rbp)
  mov    %rax,%rdi
  jmp    L35
L67:
  je     L84
  lea    (%rbx,%rax,1),%ecx
  mov    0x0(%rbp),%rdi
  cmp    $0x400,%ecx
  jg     L68
L44:
  add    %rbx,%rdi
  mov    %r12d,%edx
  mov    %r13,%rsi
  mov    %ecx,0xc(%rsp)
  call   <memcpy@plt>
  mov    0xc(%rsp),%ecx
  movslq %ecx,%rbx
L83:
  mov    0x0(%rbp),%rdi
L35:
  mov    %r14,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L40
  xor    %ecx,%ecx
  mov    $0x400,%edx
  mov    %r13,%rsi
  mov    %r15d,%edi
  call   <recv@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jns    L67
  mov    D9(%rip),%rax
  mov    D15(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L83
  mov    $0xfffffffe,%eax
  jmp    L78
  nopl   (%rax)
L68:
  lea    (%rcx,%rcx,1),%esi
  mov    %ecx,0xc(%rsp)
  call   <realloc@plt>
  mov    0xc(%rsp),%ecx
  mov    %rax,0x0(%rbp)
  mov    %rax,%rdi
  jmp    L44
  nopl   0x0(%rax)
L40:
  xor    %eax,%eax
L78:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L66
  add    $0x428,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L84:
  mov    $0xffffffff,%eax
  jmp    L78
L66:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int recv_comm_until(int socket, unsigned char **buffer, char *needle) {
  int total_bytes_read;
  int bytes_read;
  int buffer_length = 1024;
  char temp_buffer[1024];
  total_bytes_read = 0;
  *buffer = (unsigned char *)calloc(sizeof(unsigned char), buffer_length);
  while (strstr((char *)*buffer, needle) == NULL) {
    bytes_read = recv(socket, temp_buffer, 1024, 0);
    if (bytes_read < 0) {
      if (errno == EINTR) {
        continue;
      } else {
        return -2;
      }
    } else if (bytes_read == 0) {
      return -1;
    }
    if ((total_bytes_read + bytes_read) > buffer_length) {
      *buffer = (unsigned char *)realloc(*buffer, (total_bytes_read + bytes_read) * 2);
    }
    memcpy(&(*buffer)[total_bytes_read], temp_buffer, bytes_read);
    total_bytes_read += bytes_read;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<recv_comm_until>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  mov    $0x1,%edi
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x400,%esi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x428,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r13
  call   <calloc@plt>
  mov    %rax,0x0(%rbp)
  mov    %rax,%rdi
  jmp    L18
L51:
  je     L76
  lea    (%rbx,%rax,1),%ecx
  mov    0x0(%rbp),%rdi
  cmp    $0x400,%ecx
  jg     L13
L78:
  add    %rbx,%rdi
  mov    %r12d,%edx
  mov    %r13,%rsi
  mov    %ecx,0xc(%rsp)
  call   <memcpy@plt>
  mov    0xc(%rsp),%ecx
  movslq %ecx,%rbx
L98:
  mov    0x0(%rbp),%rdi
L18:
  mov    %r14,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L26
  xor    %ecx,%ecx
  mov    $0x400,%edx
  mov    %r13,%rsi
  mov    %r15d,%edi
  call   <recv@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jns    L51
  mov    D6(%rip),%rax
  mov    D8(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L98
  mov    $0xfffffffe,%eax
  jmp    L59
  nopl   (%rax)
L13:
  lea    (%rcx,%rcx,1),%esi
  mov    %ecx,0xc(%rsp)
  call   <realloc@plt>
  mov    0xc(%rsp),%ecx
  mov    %rax,0x0(%rbp)
  mov    %rax,%rdi
  jmp    L78
  nopl   0x0(%rax)
L26:
  xor    %eax,%eax
L59:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L75
  add    $0x428,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L76:
  mov    $0xffffffff,%eax
  jmp    L59
L75:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int recv_comm_until(int socket, unsigned char **buffer, char *needle) {
  int total_bytes_read;
  int bytes_read;
  int buffer_length = 1024;
  char temp_buffer[1024];
  total_bytes_read = 0;
  *buffer = (unsigned char *)calloc(sizeof(unsigned char), buffer_length);
  while (strstr((char *)*buffer, needle) == NULL) {
    bytes_read = recv(socket, temp_buffer, 1024, 0);
    if (bytes_read < 0) {
      if (errno == EINTR) {
        continue;
      } else {
        return -2;
      }
    } else if (bytes_read == 0) {
      return -1;
    }
    if ((total_bytes_read + bytes_read) > buffer_length) {
      *buffer = (unsigned char *)realloc(*buffer, (total_bytes_read + bytes_read) * 2);
    }
    memcpy(&(*buffer)[total_bytes_read], temp_buffer, bytes_read);
    total_bytes_read += bytes_read;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<readint_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
  jmp    L10
L29:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x24(%rbp),%eax
  cltq
  mov    %dl,-0x20(%rbp,%rax,1)
  addl   $0x1,-0x24(%rbp)
L10:
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L29
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  cltq
  movb   $0x0,-0x20(%rbp,%rax,1)
  mov    -0x48(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  lea    D9(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sscanf@plt>
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type: string

tool: D9: %d

assistant: ```c
readint_(a, len, ix) char *a;
int *ix, *len;
{
  char b[20];
  int i;
  for (i = 0; i < *len; i++)
    b[i] = a[i];
  b[*len] = 0;
  sscanf(b, "%d", ix);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<readint_>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    (%rsi),%r8d
  test   %r8d,%r8d
  jle    L11
  mov    %r8d,%esi
L8:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,(%rsp,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L8
L11:
  movslq %r8d,%r8
  movb   $0x0,(%rsp,%r8,1)
  mov    %rsp,%rdi
  lea    D0(%rip),%rsi
  mov    $0x0,%eax
  call   <sscanf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D0
type_infered: string

tool: D0: %d

assistant: ```c
readint_(a, len, ix) char *a;
int *ix, *len;
{
  char b[20];
  int i;
  for (i = 0; i < *len; i++)
    b[i] = a[i];
  b[*len] = 0;
  sscanf(b, "%d", ix);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<readint_>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    (%rsi),%eax
  mov    %rsp,%r8
  movslq %eax,%rbx
  test   %eax,%eax
  jle    L7
  mov    %rdi,%rsi
  mov    %rbx,%rdx
  mov    %r8,%rdi
  call   <memcpy@plt>
  mov    %rax,%r8
L7:
  xor    %eax,%eax
  mov    %r12,%rdx
  lea    D8(%rip),%rsi
  mov    %r8,%rdi
  movb   $0x0,(%rsp,%rbx,1)
  call   <sscanf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type: string

tool: D8: %d

assistant: ```c
readint_(a, len, ix) char *a;
int *ix, *len;
{
  char b[20];
  int i;
  for (i = 0; i < *len; i++)
    b[i] = a[i];
  b[*len] = 0;
  sscanf(b, "%d", ix);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readint_>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    (%rsi),%eax
  mov    %rsp,%r8
  movslq %eax,%rbx
  test   %eax,%eax
  jle    L17
  mov    %rdi,%rsi
  mov    %rbx,%rdx
  mov    %r8,%rdi
  call   <memcpy@plt>
  mov    %rax,%r8
L17:
  xor    %eax,%eax
  mov    %r12,%rdx
  lea    D0(%rip),%rsi
  mov    %r8,%rdi
  movb   $0x0,(%rsp,%rbx,1)
  call   <sscanf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D0
type_guessed: string

tool: D0: %d

assistant: ```c
readint_(a, len, ix) char *a;
int *ix, *len;
{
  char b[20];
  int i;
  for (i = 0; i < *len; i++)
    b[i] = a[i];
  b[*len] = 0;
  sscanf(b, "%d", ix);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <printName@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  printName();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <printName@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  printName();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <printName@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  printName();
  return 0;
}
```
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
  xor    %eax,%eax
  call   <printName@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  printName();
  return 0;
}
```