ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<display_matrix>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %r9,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%rdi
  mov    %rdx,0x8(%rsp)
  test   %rdi,%rdi
  je     L37
  call   <puts@plt>
L37:
  cmp    0x8(%rsp),%r13
  jg     L22
  lea    (%rbx,%r13,8),%rbx
L20:
  mov    %r14,%r15
  cmp    %rbp,%r14
  jg     L5
  nopl   0x0(%rax,%rax,1)
L2:
  mov    (%rbx),%rax
  mov    %r12,%rdi
  movss  (%rax,%r15,4),%xmm0
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jge    L2
L5:
  mov    $0xa,%edi
  add    $0x1,%r13
  add    $0x8,%rbx
  call   <putchar@plt>
  cmp    %r13,0x8(%rsp)
  jge    L20
L22:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void display_matrix(float **m, long nrl, long nrh, long ncl, long nch, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = nrl; i <= nrh; i++) {
    for (j = ncl; j <= nch; j++) {
      printf(format, m[i][j]);
    }
    putchar('\n');
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<display_matrix>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %r9,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%rdi
  mov    %rdx,0x8(%rsp)
  test   %rdi,%rdi
  je     L45
  call   <puts@plt>
L45:
  cmp    0x8(%rsp),%r13
  jg     L21
  lea    (%rbx,%r13,8),%rbx
L5:
  mov    %r14,%r15
  cmp    %rbp,%r14
  jg     L22
  nopl   0x0(%rax,%rax,1)
L42:
  mov    (%rbx),%rax
  mov    %r12,%rdi
  movss  (%rax,%r15,4),%xmm0
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jge    L42
L22:
  mov    $0xa,%edi
  add    $0x1,%r13
  add    $0x8,%rbx
  call   <putchar@plt>
  cmp    %r13,0x8(%rsp)
  jge    L5
L21:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void display_matrix(float **m, long nrl, long nrh, long ncl, long nch, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = nrl; i <= nrh; i++) {
    for (j = ncl; j <= nch; j++) {
      printf(format, m[i][j]);
    }
    putchar('\n');
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<dump_row_ues>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x2c,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L17
L8:
  mov    $0x1703,%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <jedec_bit_get@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0xc(%rbp)
L17:
  cmpl   $0x3f,-0xc(%rbp)
  jle    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
data_type: string

tool: D6: %2d: 

assistant: ```c
void dump_row_ues(struct jedec *jed) {
  int i;
  printf("%2d: ", 44);
  for (i = 0; i < 64; i++) {
    int v, fuse;
    fuse = 5891 - i;
    v = jedec_bit_get(jed, fuse);
    printf("%c", '0' + v);
  }
  printf("\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<dump_row_ues>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x2c,%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1703,%ebx
L8:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <jedec_bit_get@plt>
  lea    0x30(%rax),%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  cmp    $0x16c3,%ebx
  jne    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type_infered: string

tool: D5: %2d: 

assistant: ```c
void dump_row_ues(struct jedec *jed) {
  int i;
  printf("%2d: ", 44);
  for (i = 0; i < 64; i++) {
    int v, fuse;
    fuse = 5891 - i;
    v = jedec_bit_get(jed, fuse);
    printf("%c", '0' + v);
  }
  printf("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dump_row_ues>:
  endbr64
  push   %rbp
  mov    $0x2c,%esi
  mov    %rdi,%rbp
  xor    %eax,%eax
  push   %rbx
  lea    D3(%rip),%rdi
  mov    $0x1703,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  nopl   (%rax)
L1:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  sub    $0x1,%ebx
  call   <jedec_bit_get@plt>
  lea    0x30(%rax),%edi
  call   <putchar@plt>
  cmp    $0x16c3,%ebx
  jne    L1
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
type_guessed: string

tool: D3: %2d: 

assistant: ```c
void dump_row_ues(struct jedec *jed) {
  int i;
  printf("%2d: ", 44);
  for (i = 0; i < 64; i++) {
    int v, fuse;
    fuse = 5891 - i;
    v = jedec_bit_get(jed, fuse);
    printf("%c", '0' + v);
  }
  printf("\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dump_row_ues>:
  endbr64
  push   %rbp
  mov    $0x2c,%esi
  mov    %rdi,%rbp
  xor    %eax,%eax
  push   %rbx
  lea    D9(%rip),%rdi
  mov    $0x1703,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  nopl   (%rax)
L1:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  sub    $0x1,%ebx
  call   <jedec_bit_get@plt>
  lea    0x30(%rax),%edi
  call   <putchar@plt>
  cmp    $0x16c3,%ebx
  jne    L1
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <putchar@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D9
type_guessed: string

tool: D9: %2d: 

assistant: ```c
void dump_row_ues(struct jedec *jed) {
  int i;
  printf("%2d: ", 44);
  for (i = 0; i < 64; i++) {
    int v, fuse;
    fuse = 5891 - i;
    v = jedec_bit_get(jed, fuse);
    printf("%c", '0' + v);
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<skipWordBackwards>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3f2c2e20,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L17
L38:
  subl   $0x1,-0x14(%rbp)
L17:
  cmpl   $0x0,-0x14(%rbp)
  jle    L36
  mov    -0x14(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  lea    -0xd(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L38
L36:
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
int skipWordBackwards(char *string, int position) {
  char separators[] = " .,?";
  int i;
  for (i = position; i > 0 && strchr(separators, string[i - 1]) == NULL; i--)
    ;
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<skipWordBackwards>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x3f2c2e20,0x3(%rsp)
  movb   $0x0,0x7(%rsp)
  test   %esi,%esi
  jle    L27
  movslq %esi,%rax
  lea    -0x1(%rdi,%rax,1),%rbp
  lea    0x3(%rsp),%r12
L33:
  movsbl 0x0(%rbp),%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L27
  sub    $0x1,%rbp
  sub    $0x1,%ebx
  jne    L33
L27:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int skipWordBackwards(char *string, int position) {
  char separators[] = " .,?";
  int i;
  for (i = position; i > 0 && strchr(separators, string[i - 1]) == NULL; i--)
    ;
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<skipWordBackwards>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x3f2c2e20,0x3(%rsp)
  movb   $0x0,0x7(%rsp)
  test   %esi,%esi
  jle    L24
  movslq %esi,%rax
  lea    0x3(%rsp),%rbp
  lea    -0x1(%rdi,%rax,1),%rbx
  jmp    L46
  nop
L6:
  sub    $0x1,%rbx
  sub    $0x1,%r12d
  je     L24
L46:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L6
L24:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int skipWordBackwards(char *string, int position) {
  char separators[] = " .,?";
  int i;
  for (i = position; i > 0 && strchr(separators, string[i - 1]) == NULL; i--)
    ;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<skipWordBackwards>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x3f2c2e20,0x3(%rsp)
  movb   $0x0,0x7(%rsp)
  test   %esi,%esi
  jle    L38
  movslq %esi,%rax
  lea    0x3(%rsp),%rbp
  lea    -0x1(%rdi,%rax,1),%rbx
  jmp    L15
  nop
L49:
  sub    $0x1,%rbx
  sub    $0x1,%r12d
  je     L38
L15:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L49
L38:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int skipWordBackwards(char *string, int position) {
  char separators[] = " .,?";
  int i;
  for (i = position; i > 0 && strchr(separators, string[i - 1]) == NULL; i--)
    ;
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<showRules>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x46,%edi
  call   <startBuffer@plt>
  mov    $0x0,%edi
  call   <setLineAlign@plt>
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <output@plt>
  mov    $0xffffffff,%edi
  call   <setLineAlign@plt>
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <output@plt>
  mov    $0x0,%edi
  call   <setLineAlign@plt>
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <output@plt>
  mov    $0x0,%eax
  call   <flushBuffer@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D20
type: string
name: parse_data
arguments: tag: D29
type: string
name: parse_data
arguments: tag: D13
type: string

tool: D20: [ RULES ]



tool: D29: This a implementation of the famous two player game "Connect Four".
So you'll need a friend to play this game :) To win you simply have to
connect four of your chips in a row. This row can be horizontal,
vertical or diagonal,it doesn't matter as long as there are no "enemy"
chips between your chips.
The game will automatically end if one you won and take you to the
"Hall of Shame"



tool: D13: [ Press any key to return to main menu ]


assistant: ```c
void showRules() {
  startBuffer(70);
  setLineAlign(0);
  output("[ RULES ]\n\n");
  setLineAlign(-1);
  output("This a implementation of the famous two player game \"Connect Four\".\nSo you'll need a friend to play this game :) To win you simply have to\nconnect four of your chips in a row. This row can be horizontal,\nvertical or diagonal,it doesn't matter as long as there are no \"enemy\"\nchips between your chips.\nThe game will automatically end if one you won and take you to the\n\"Hall of Shame\"\n\n");
  setLineAlign(0);
  output("[ Press any key to return to main menu ]\n");
  flushBuffer();
  getch();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<showRules>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x46,%edi
  call   <startBuffer@plt>
  mov    $0x0,%edi
  call   <setLineAlign@plt>
  lea    D7(%rip),%rdi
  call   <output@plt>
  mov    $0xffffffff,%edi
  call   <setLineAlign@plt>
  lea    D10(%rip),%rdi
  call   <output@plt>
  mov    $0x0,%edi
  call   <setLineAlign@plt>
  lea    D23(%rip),%rdi
  call   <output@plt>
  mov    $0x0,%eax
  call   <flushBuffer@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type_infered: string
name: unpack_data
arguments: data_tag: D10
type_infered: string
name: unpack_data
arguments: data_tag: D23
type_infered: string

tool: D7: [ RULES ]



tool: D10: This a implementation of the famous two player game "Connect Four".
So you'll need a friend to play this game :) To win you simply have to
connect four of your chips in a row. This row can be horizontal,
vertical or diagonal,it doesn't matter as long as there are no "enemy"
chips between your chips.
The game will automatically end if one you won and take you to the
"Hall of Shame"



tool: D23: [ Press any key to return to main menu ]


assistant: ```c
void showRules() {
  startBuffer(70);
  setLineAlign(0);
  output("[ RULES ]\n\n");
  setLineAlign(-1);
  output("This a implementation of the famous two player game \"Connect Four\".\nSo you'll need a friend to play this game :) To win you simply have to\nconnect four of your chips in a row. This row can be horizontal,\nvertical or diagonal,it doesn't matter as long as there are no \"enemy\"\nchips between your chips.\nThe game will automatically end if one you won and take you to the\n\"Hall of Shame\"\n\n");
  setLineAlign(0);
  output("[ Press any key to return to main menu ]\n");
  flushBuffer();
  getch();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<showRules>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x46,%edi
  call   <startBuffer@plt>
  xor    %edi,%edi
  call   <setLineAlign@plt>
  lea    D26(%rip),%rdi
  call   <output@plt>
  mov    $0xffffffff,%edi
  call   <setLineAlign@plt>
  lea    D15(%rip),%rdi
  call   <output@plt>
  xor    %edi,%edi
  call   <setLineAlign@plt>
  lea    D24(%rip),%rdi
  call   <output@plt>
  xor    %eax,%eax
  call   <flushBuffer@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10c0 <getch@plt>
```

tool_calls: name: parse_data
arguments: tag: D26
datatype: string
name: parse_data
arguments: tag: D15
datatype: string
name: parse_data
arguments: tag: D24
datatype: string

tool: D26: [ RULES ]



tool: D15: This a implementation of the famous two player game "Connect Four".
So you'll need a friend to play this game :) To win you simply have to
connect four of your chips in a row. This row can be horizontal,
vertical or diagonal,it doesn't matter as long as there are no "enemy"
chips between your chips.
The game will automatically end if one you won and take you to the
"Hall of Shame"



tool: D24: [ Press any key to return to main menu ]


assistant: ```c
void showRules() {
  startBuffer(70);
  setLineAlign(0);
  output("[ RULES ]\n\n");
  setLineAlign(-1);
  output("This a implementation of the famous two player game \"Connect Four\".\nSo you'll need a friend to play this game :) To win you simply have to\nconnect four of your chips in a row. This row can be horizontal,\nvertical or diagonal,it doesn't matter as long as there are no \"enemy\"\nchips between your chips.\nThe game will automatically end if one you won and take you to the\n\"Hall of Shame\"\n\n");
  setLineAlign(0);
  output("[ Press any key to return to main menu ]\n");
  flushBuffer();
  getch();
}
```
user: What is the c source code of the assembly code below?

```asm
<showRules>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x46,%edi
  call   <startBuffer@plt>
  xor    %edi,%edi
  call   <setLineAlign@plt>
  lea    D19(%rip),%rdi
  call   <output@plt>
  mov    $0xffffffff,%edi
  call   <setLineAlign@plt>
  lea    D28(%rip),%rdi
  call   <output@plt>
  xor    %edi,%edi
  call   <setLineAlign@plt>
  lea    D2(%rip),%rdi
  call   <output@plt>
  xor    %eax,%eax
  call   <flushBuffer@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10c0 <getch@plt>
```

tool_calls: name: read_data
arguments: label: D19
datatype: string
name: read_data
arguments: label: D28
datatype: string
name: read_data
arguments: label: D2
datatype: string

tool: D19: [ RULES ]



tool: D28: This a implementation of the famous two player game "Connect Four".
So you'll need a friend to play this game :) To win you simply have to
connect four of your chips in a row. This row can be horizontal,
vertical or diagonal,it doesn't matter as long as there are no "enemy"
chips between your chips.
The game will automatically end if one you won and take you to the
"Hall of Shame"



tool: D2: [ Press any key to return to main menu ]


assistant: ```c
void showRules() {
  startBuffer(70);
  setLineAlign(0);
  output("[ RULES ]\n\n");
  setLineAlign(-1);
  output("This a implementation of the famous two player game \"Connect Four\".\nSo you'll need a friend to play this game :) To win you simply have to\nconnect four of your chips in a row. This row can be horizontal,\nvertical or diagonal,it doesn't matter as long as there are no \"enemy\"\nchips between your chips.\nThe game will automatically end if one you won and take you to the\n\"Hall of Shame\"\n\n");
  setLineAlign(0);
  output("[ Press any key to return to main menu ]\n");
  flushBuffer();
  getch();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<insert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L13
L3:
  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    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L13:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L3
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void insert(int a[], int len, int n, int v) {
  int i = n, temp = v, swap;
  while (i <= len) {
    swap = a[i];
    a[i] = temp;
    temp = swap;
    i++;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<insert>:
  endbr64
  cmp    %esi,%edx
  jg     L6
  movslq %edx,%r8
  lea    (%rdi,%r8,4),%rax
  sub    %edx,%esi
  add    %r8,%rsi
  lea    0x4(%rdi,%rsi,4),%rsi
L18:
  mov    %ecx,%edx
  mov    (%rax),%ecx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L18
L6:
  ret
```

assistant: ```c
void insert(int a[], int len, int n, int v) {
  int i = n, temp = v, swap;
  while (i <= len) {
    swap = a[i];
    a[i] = temp;
    temp = swap;
    i++;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<insert>:
  endbr64
  cmp    %esi,%edx
  jg     L12
  movslq %edx,%r8
  sub    %edx,%esi
  add    %r8,%rsi
  lea    (%rdi,%r8,4),%rax
  lea    0x4(%rdi,%rsi,4),%rsi
  nopl   0x0(%rax)
L8:
  mov    %ecx,%edx
  add    $0x4,%rax
  mov    -0x4(%rax),%ecx
  mov    %edx,-0x4(%rax)
  cmp    %rsi,%rax
  jne    L8
L12:
  ret
```

assistant: ```c
void insert(int a[], int len, int n, int v) {
  int i = n, temp = v, swap;
  while (i <= len) {
    swap = a[i];
    a[i] = temp;
    temp = swap;
    i++;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<insert>:
  endbr64
  cmp    %esi,%edx
  jg     L5
  movslq %edx,%r8
  sub    %edx,%esi
  add    %r8,%rsi
  lea    (%rdi,%r8,4),%rax
  lea    0x4(%rdi,%rsi,4),%rsi
  nopl   0x0(%rax)
L4:
  mov    %ecx,%edx
  add    $0x4,%rax
  mov    -0x4(%rax),%ecx
  mov    %edx,-0x4(%rax)
  cmp    %rsi,%rax
  jne    L4
L5:
  ret
```

assistant: ```c
void insert(int a[], int len, int n, int v) {
  int i = n, temp = v, swap;
  while (i <= len) {
    swap = a[i];
    a[i] = temp;
    temp = swap;
    i++;
  }
}
```
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
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x34(%rbp)
  movl   $0x0,-0x3c(%rbp)
  jmp    L6
L10:
  pxor   %xmm1,%xmm1
  cvtsi2ssl -0x3c(%rbp),%xmm1
  mov    -0x3c(%rbp),%eax
  add    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  D0(%rip),%xmm2
  divss  %xmm2,%xmm0
  addss  %xmm1,%xmm0
  mov    -0x3c(%rbp),%eax
  cltq
  movss  %xmm0,-0x30(%rbp,%rax,4)
  mov    -0x3c(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  movd   %eax,%xmm0
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x3c(%rbp),%eax
  cltq
  movss  -0x30(%rbp,%rax,4),%xmm0
  movss  -0x38(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x38(%rbp)
  addl   $0x1,-0x3c(%rbp)
L6:
  cmpl   $0x9,-0x3c(%rbp)
  jle    L10
  movss  -0x38(%rbp),%xmm0
  movss  D0(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  movd   %eax,%xmm0
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D0
datatype: byte[4]
name: unpack_data
arguments: tag: D18
datatype: string
name: unpack_data
arguments: tag: D20
datatype: string

tool: {"D0": [0x00,0x00,0x20,0x41]}

tool: D18: %2f 

tool: D20: The average is %2f 


assistant: ```c
int main(void) {
  int i;
  float sum;
  float average = 0;
  float numbers[10];
  for (i = 0; i < 10; ++i) {
    numbers[i] = i + (float)(i + 1) / 10;
    printf("%2f ", numbers[i]);
    sum += numbers[i];
  }
  average = sum / 10;
  printf("The average is %2f \n", average);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    $0x0,%ebx
  lea    D30(%rip),%rbp
L4:
  pxor   %xmm1,%xmm1
  cvtsi2ss %ebx,%xmm1
  add    $0x1,%ebx
  pxor   %xmm0,%xmm0
  cvtsi2ss %ebx,%xmm0
  divss  D29(%rip),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,0xc(%rsp)
  mov    %rbp,%rdi
  call   <printf@plt>
  movss  0x8(%rsp),%xmm2
  addss  0xc(%rsp),%xmm2
  movss  %xmm2,0x8(%rsp)
  cmp    $0xa,%ebx
  jne    L4
  movaps %xmm2,%xmm0
  divss  D29(%rip),%xmm0
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D30
type_guessed: string
name: unpack_data
arguments: data_label: D29
type_guessed: byte[4]
name: unpack_data
arguments: data_label: D25
type_guessed: string

tool: D30: %2f 

tool: {"D29": [0x00,0x00,0x20,0x41]}

tool: D25: The average is %2f 


assistant: ```c
int main(void) {
  int i;
  float sum;
  float average = 0;
  float numbers[10];
  for (i = 0; i < 10; ++i) {
    numbers[i] = i + (float)(i + 1) / 10;
    printf("%2f ", numbers[i]);
    sum += numbers[i];
  }
  average = sum / 10;
  printf("The average is %2f \n", average);
  return 0;
}
```