ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<BattleOne>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%ecx
  movslq %ecx,%rax
  imul   $0x2e8ba2e9,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  lea    0xf(%rdx),%eax
  imul   -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  mov    %edx,%ecx
  sar    $0x5,%ecx
  sar    $0x1f,%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jle    L1
  mov    $0xffffffff,%eax
  jmp    L17
L1:
  mov    -0x10(%rbp),%rax
  movl   $0x0,(%rax)
  mov    $0x1,%eax
L17:
  leave
  ret
```

assistant: ```c
int BattleOne(int atkForce, int defForce, int *defSurvivors) {
  *defSurvivors = defForce - (atkForce * (15 + rand() % 11)) / 100;
  if (*defSurvivors > 0)
    return -1;
  else {
    *defSurvivors = 0;
    return 1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<BattleOne>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %esi,%ebp
  mov    %rdx,%rbx
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x2e8ba2e9,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  lea    (%rdx,%rcx,2),%edx
  sub    %edx,%eax
  add    $0xf,%eax
  imul   %r12d,%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  sar    $0x1f,%eax
  sub    %edx,%eax
  add    %ebp,%eax
  mov    $0xffffffff,%edx
  test   %eax,%eax
  jle    L17
L13:
  mov    %eax,(%rbx)
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  mov    $0x0,%eax
  mov    $0x1,%edx
  jmp    L13
```

assistant: ```c
int BattleOne(int atkForce, int defForce, int *defSurvivors) {
  *defSurvivors = defForce - (atkForce * (15 + rand() % 11)) / 100;
  if (*defSurvivors > 0)
    return -1;
  else {
    *defSurvivors = 0;
    return 1;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<BattleOne>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdx,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  call   <rand@plt>
  mov    $0xffffffff,%r8d
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x2e8ba2e9,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  lea    (%rdx,%rcx,2),%edx
  sub    %edx,%eax
  add    $0xf,%eax
  imul   %ebp,%eax
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  sub    %edx,%eax
  add    %ebx,%eax
  test   %eax,%eax
  jg     L8
  xor    %eax,%eax
  mov    $0x1,%r8d
L8:
  mov    %eax,(%r12)
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int BattleOne(int atkForce, int defForce, int *defSurvivors) {
  *defSurvivors = defForce - (atkForce * (15 + rand() % 11)) / 100;
  if (*defSurvivors > 0)
    return -1;
  else {
    *defSurvivors = 0;
    return 1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<BattleOne>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdx,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  call   <rand@plt>
  mov    $0xffffffff,%r8d
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x2e8ba2e9,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  lea    (%rdx,%rcx,2),%edx
  sub    %edx,%eax
  add    $0xf,%eax
  imul   %ebp,%eax
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  sub    %edx,%eax
  add    %ebx,%eax
  test   %eax,%eax
  jg     L5
  xor    %eax,%eax
  mov    $0x1,%r8d
L5:
  mov    %eax,(%r12)
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int BattleOne(int atkForce, int defForce, int *defSurvivors) {
  *defSurvivors = defForce - (atkForce * (15 + rand() % 11)) / 100;
  if (*defSurvivors > 0)
    return -1;
  else {
    *defSurvivors = 0;
    return 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    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0xc,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    %rdi,%rdx
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  movl   $0xffffffff,-0x70(%rbp)
  movl   $0x3,-0x6c(%rbp)
  movl   $0x4,-0x68(%rbp)
  movl   $0x2,-0x60(%rbp)
  movl   $0x3,-0x5c(%rbp)
  movl   $0x5,-0x58(%rbp)
  movl   $0x7,-0x54(%rbp)
  movl   $0xfffffffb,-0x50(%rbp)
  movl   $0x2,-0x4c(%rbp)
  movl   $0xffffffef,-0x48(%rbp)
  movl   $0x3,-0x44(%rbp)
  movl   $0x5,-0x40(%rbp)
  movl   $0x1,-0x3c(%rbp)
  movl   $0x9,-0x34(%rbp)
  movl   $0xfffffff8,-0x30(%rbp)
  movl   $0xfffffff9,-0x2c(%rbp)
  movl   $0x1,-0x28(%rbp)
  movl   $0x6,-0x24(%rbp)
  movl   $0x3,-0x20(%rbp)
  movl   $0x2,-0x1c(%rbp)
  movl   $0x4,-0x18(%rbp)
  movl   $0x2,-0x14(%rbp)
  lea    -0x74(%rbp),%r8
  lea    -0x78(%rbp),%rdi
  lea    -0x7c(%rbp),%rcx
  lea    -0x80(%rbp),%rdx
  lea    -0x84(%rbp),%rsi
  lea    -0x70(%rbp),%rax
  mov    %r8,%r9
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <findmaxrect@plt>
  mov    -0x78(%rbp),%edi
  mov    -0x7c(%rbp),%esi
  mov    -0x80(%rbp),%ecx
  mov    -0x84(%rbp),%edx
  mov    -0x74(%rbp),%eax
  mov    %edi,%r9d
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
data_type: string

tool: D5: Max sum is %d and startcol = %d, endcol = %d, startrow = %d, endrow = %d

assistant: ```c
void main() {
  int a[5][5] = {{-1, 3, 4, 0, 2}, {3, 5, 7, -5, 2}, {-7 - 10, 3, 5, 1}, {9, -8, -7, 1, 6}, {3, 2, 4, 2, 0}};
  int startcol, endcol, startrow, endrow, maxsum;
  findmaxrect(a, &startcol, &endcol, &startrow, &endrow, &maxsum);
  printf("Max sum is %d and startcol = %d, endcol = %d, startrow = %d, endrow = %d", maxsum, startcol, endcol, startrow, endrow);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  lea    0x28(%rsp),%rdi
  mov    $0xb,%ecx
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  movl   $0xffffffff,0x20(%rsp)
  movl   $0x3,0x24(%rsp)
  movl   $0x4,0x28(%rsp)
  movl   $0x2,0x30(%rsp)
  movl   $0x3,0x34(%rsp)
  movl   $0x5,0x38(%rsp)
  movl   $0x7,0x3c(%rsp)
  movl   $0xfffffffb,0x40(%rsp)
  movl   $0x2,0x44(%rsp)
  movl   $0xffffffef,0x48(%rsp)
  movl   $0x3,0x4c(%rsp)
  movl   $0x5,0x50(%rsp)
  movl   $0x1,0x54(%rsp)
  movl   $0x9,0x5c(%rsp)
  movl   $0xfffffff8,0x60(%rsp)
  movl   $0xfffffff9,0x64(%rsp)
  movl   $0x1,0x68(%rsp)
  movl   $0x6,0x6c(%rsp)
  movl   $0x3,0x70(%rsp)
  movl   $0x2,0x74(%rsp)
  movl   $0x4,0x78(%rsp)
  movl   $0x2,0x7c(%rsp)
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    0x20(%rsp),%rdi
  lea    0x1c(%rsp),%r9
  lea    0x18(%rsp),%r8
  call   <findmaxrect@plt>
  mov    0x18(%rsp),%r9d
  mov    0x14(%rsp),%r8d
  mov    0x10(%rsp),%ecx
  mov    0xc(%rsp),%edx
  mov    0x1c(%rsp),%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x98,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

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

tool: D8: Max sum is %d and startcol = %d, endcol = %d, startrow = %d, endrow = %d

assistant: ```c
void main() {
  int a[5][5] = {{-1, 3, 4, 0, 2}, {3, 5, 7, -5, 2}, {-7 - 10, 3, 5, 1}, {9, -8, -7, 1, 6}, {3, 2, 4, 2, 0}};
  int startcol, endcol, startrow, endrow, maxsum;
  findmaxrect(a, &startcol, &endcol, &startrow, &endrow, &maxsum);
  printf("Max sum is %d and startcol = %d, endcol = %d, startrow = %d, endrow = %d", maxsum, startcol, endcol, startrow, endrow);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  movabs $0x3ffffffff,%rax
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  movq   $0x4,0x28(%rsp)
  lea    0xc(%rsp),%rsi
  lea    0x20(%rsp),%rdi
  movl   $0x0,0x80(%rsp)
  lea    0x1c(%rsp),%r9
  lea    0x18(%rsp),%r8
  mov    %rax,0x20(%rsp)
  movabs $0x300000002,%rax
  mov    %rax,0x30(%rsp)
  movabs $0x700000005,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x2fffffffb,%rax
  mov    %rax,0x40(%rsp)
  movabs $0x3ffffffef,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x100000005,%rax
  mov    %rax,0x50(%rsp)
  movabs $0x900000000,%rax
  mov    %rax,0x58(%rsp)
  movabs $0xfffffff9fffffff8,%rax
  mov    %rax,0x60(%rsp)
  movabs $0x600000001,%rax
  mov    %rax,0x68(%rsp)
  movabs $0x200000003,%rax
  mov    %rax,0x70(%rsp)
  add    $0x1,%rax
  mov    %rax,0x78(%rsp)
  call   <findmaxrect@plt>
  mov    0x18(%rsp),%r9d
  mov    0x14(%rsp),%r8d
  lea    D1(%rip),%rdi
  mov    0x10(%rsp),%ecx
  mov    0xc(%rsp),%edx
  mov    0x1c(%rsp),%esi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x98,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

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

tool: D1: Max sum is %d and startcol = %d, endcol = %d, startrow = %d, endrow = %d

assistant: ```c
void main() {
  int a[5][5] = {{-1, 3, 4, 0, 2}, {3, 5, 7, -5, 2}, {-7 - 10, 3, 5, 1}, {9, -8, -7, 1, 6}, {3, 2, 4, 2, 0}};
  int startcol, endcol, startrow, endrow, maxsum;
  findmaxrect(a, &startcol, &endcol, &startrow, &endrow, &maxsum);
  printf("Max sum is %d and startcol = %d, endcol = %d, startrow = %d, endrow = %d", maxsum, startcol, endcol, startrow, endrow);
}
```
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    $0x98,%rsp
  mov    $0xb,%ecx
  movdqa D68(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  lea    0x28(%rsp),%rdi
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    0x1c(%rsp),%r9
  lea    0x18(%rsp),%r8
  rep stos %rax,%es:(%rdi)
  movaps %xmm0,0x30(%rsp)
  lea    0x14(%rsp),%rcx
  movdqa D28(%rip),%xmm0
  mov    D22(%rip),%rax
  movl   $0x4,0x28(%rsp)
  movaps %xmm0,0x40(%rsp)
  movdqa D18(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  mov    D5(%rip),%rax
  movups %xmm0,0x5c(%rsp)
  movdqa D19(%rip),%xmm0
  mov    %rax,0x50(%rsp)
  movl   $0x0,(%rdi)
  lea    0x20(%rsp),%rdi
  movl   $0x2,0x7c(%rsp)
  movups %xmm0,0x6c(%rsp)
  call   <findmaxrect@plt>
  mov    0x18(%rsp),%r9d
  mov    0x14(%rsp),%r8d
  lea    D59(%rip),%rdi
  mov    0x10(%rsp),%ecx
  mov    0xc(%rsp),%edx
  mov    0x1c(%rsp),%esi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x98,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D68
datatype: byte[16]
name: retrieve_data
arguments: label: D28
datatype: byte[16]
name: retrieve_data
arguments: label: D22
datatype: byte[8]
name: retrieve_data
arguments: label: D18
datatype: byte[16]
name: retrieve_data
arguments: label: D5
datatype: byte[8]
name: retrieve_data
arguments: label: D19
datatype: byte[16]
name: retrieve_data
arguments: label: D59
datatype: string

tool: {"D68": [0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: {"D28": [0xfb,0xff,0xff,0xff,0x02,0x00,0x00,0x00,0xef,0xff,0xff,0xff,0x03,0x00,0x00,0x00]}

tool: {"D22": [0xff,0xff,0xff,0xff,0x03,0x00,0x00,0x00]}

tool: {"D18": [0x09,0x00,0x00,0x00,0xf8,0xff,0xff,0xff,0xf9,0xff,0xff,0xff,0x01,0x00,0x00,0x00]}

tool: {"D5": [0x05,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D19": [0x06,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D59: Max sum is %d and startcol = %d, endcol = %d, startrow = %d, endrow = %d

assistant: ```c
void main() {
  int a[5][5] = {{-1, 3, 4, 0, 2}, {3, 5, 7, -5, 2}, {-7 - 10, 3, 5, 1}, {9, -8, -7, 1, 6}, {3, 2, 4, 2, 0}};
  int startcol, endcol, startrow, endrow, maxsum;
  findmaxrect(a, &startcol, &endcol, &startrow, &endrow, &maxsum);
  printf("Max sum is %d and startcol = %d, endcol = %d, startrow = %d, endrow = %d", maxsum, startcol, endcol, startrow, endrow);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<yodel_checksum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void yodel_checksum(uint8_t ch, uint8_t *csap, uint8_t *csbp) {
  *csap += ch;
  *csbp += *csap;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<yodel_checksum>:
  endbr64
  add    (%rsi),%rdi
  mov    %rdi,(%rsi)
  add    %rdi,(%rdx)
  ret
```

assistant: ```c
void yodel_checksum(uint8_t ch, uint8_t *csap, uint8_t *csbp) {
  *csap += ch;
  *csbp += *csap;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<yodel_checksum>:
  endbr64
  add    (%rsi),%rdi
  mov    %rdi,(%rsi)
  add    %rdi,(%rdx)
  ret
```

assistant: ```c
void yodel_checksum(uint8_t ch, uint8_t *csap, uint8_t *csbp) {
  *csap += ch;
  *csbp += *csap;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<yodel_checksum>:
  endbr64
  add    (%rsi),%rdi
  mov    %rdi,(%rsi)
  add    %rdi,(%rdx)
  ret
```

assistant: ```c
void yodel_checksum(uint8_t ch, uint8_t *csap, uint8_t *csbp) {
  *csap += ch;
  *csbp += *csap;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<zero_first_column>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rcx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <extended_gcd@plt>
  mov    -0x10(%rbp),%rdi
  mov    -0x18(%rbp),%rsi
  mov    -0x20(%rbp),%r9
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%r8
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%rax
  push   %rdi
  push   %rsi
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <set_zero@plt>
  add    $0x10,%rsp
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void zero_first_column(long L[3][3], const int j, long A[3][3]) {
  long vals[3];
  extended_gcd(vals, A[0][0], A[j][0]);
  set_zero(L, 0, j, A[0][0], A[j][0], vals[0], vals[1], vals[2]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<zero_first_column>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %rdx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq %esi,%rax
  lea    (%rax,%rax,2),%rax
  lea    (%rdx,%rax,8),%r13
  mov    %rsp,%rdi
  mov    0x0(%r13),%rdx
  mov    (%rbx),%rsi
  call   <extended_gcd@plt>
  push   0x10(%rsp)
  push   0x10(%rsp)
  mov    0x10(%rsp),%r9
  mov    0x0(%r13),%r8
  mov    (%rbx),%rcx
  mov    %ebp,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <set_zero@plt>
  add    $0x10,%rsp
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void zero_first_column(long L[3][3], const int j, long A[3][3]) {
  long vals[3];
  extended_gcd(vals, A[0][0], A[j][0]);
  set_zero(L, 0, j, A[0][0], A[j][0], vals[0], vals[1], vals[2]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<zero_first_column>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq %esi,%rax
  mov    (%rbx),%rsi
  mov    %rsp,%rdi
  lea    (%rax,%rax,2),%rax
  lea    (%rdx,%rax,8),%r13
  mov    0x0(%r13),%rdx
  call   <extended_gcd@plt>
  push   0x10(%rsp)
  mov    (%rbx),%rcx
  mov    %r12d,%edx
  push   0x10(%rsp)
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    0x10(%rsp),%r9
  mov    0x0(%r13),%r8
  call   <set_zero@plt>
  pop    %rax
  pop    %rdx
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void zero_first_column(long L[3][3], const int j, long A[3][3]) {
  long vals[3];
  extended_gcd(vals, A[0][0], A[j][0]);
  set_zero(L, 0, j, A[0][0], A[j][0], vals[0], vals[1], vals[2]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<zero_first_column>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq %esi,%rax
  mov    (%rbx),%rsi
  mov    %rsp,%rdi
  lea    (%rax,%rax,2),%rax
  lea    (%rdx,%rax,8),%r13
  mov    0x0(%r13),%rdx
  call   <extended_gcd@plt>
  push   0x10(%rsp)
  mov    (%rbx),%rcx
  mov    %r12d,%edx
  push   0x10(%rsp)
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    0x10(%rsp),%r9
  mov    0x0(%r13),%r8
  call   <set_zero@plt>
  pop    %rax
  pop    %rdx
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void zero_first_column(long L[3][3], const int j, long A[3][3]) {
  long vals[3];
  extended_gcd(vals, A[0][0], A[j][0]);
  set_zero(L, 0, j, A[0][0], A[j][0], vals[0], vals[1], vals[2]);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movq   $0x1,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L11
L7:
  shlq   -0x8(%rbp)
  movabs $0x2540be3ff,%rax
  cmp    %rax,-0x8(%rbp)
  jbe    L20
  mov    -0x8(%rbp),%rcx
  movabs $0xdbe6fecebdedd5bf,%rdx
  mov    %rcx,%rax
  mul    %rdx
  shr    $0x21,%rdx
  movabs $0x2540be400,%rax
  imul   %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L20:
  addl   $0x1,-0xc(%rbp)
L11:
  cmpl   $0x777bb8,-0xc(%rbp)
  jle    L7
  mov    -0x8(%rbp),%rax
  imul   $0x6f11,%rax,%rax
  lea    0x1(%rax),%rcx
  movabs $0xdbe6fecebdedd5bf,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x21,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movabs $0x2540be400,%rdx
  imul   %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D0: The last ten digits of 28433*2^7830457+1 are %llu


assistant: ```c
int main() {
  unsigned long long result = 1;
  int i;
  for (i = 0; i < 7830457; i++) {
    result *= 2;
    if (result >= 10000000000) {
      result %= 10000000000;
    }
  }
  result = (result * 28433 + 1) % 10000000000;
  printf("The last ten digits of 28433*2^7830457+1 are %llu\n", result);
  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
  mov    $0x777bb9,%esi
  mov    $0x2,%ecx
  movabs $0x2540be3ff,%rdi
  movabs $0xdbe6fecebdedd5bf,%r9
  movabs $0x2540be400,%r8
  jmp    L2
L37:
  mov    %rcx,%rax
  mul    %r9
  shr    $0x21,%rdx
  imul   %r8,%rdx
  sub    %rdx,%rcx
L2:
  sub    $0x1,%esi
  je     L20
  add    %rcx,%rcx
  cmp    %rdi,%rcx
  ja     L37
  jmp    L2
L20:
  imul   $0x6f11,%rcx,%rcx
  lea    0x1(%rcx),%rsi
  movabs $0xdbe6fecebdedd5bf,%rdx
  mov    %rsi,%rax
  mul    %rdx
  shr    $0x21,%rdx
  movabs $0x2540be400,%rax
  imul   %rax,%rdx
  sub    %rdx,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D3
type_guessed: string

tool: D3: The last ten digits of 28433*2^7830457+1 are %llu


assistant: ```c
int main() {
  unsigned long long result = 1;
  int i;
  for (i = 0; i < 7830457; i++) {
    result *= 2;
    if (result >= 10000000000) {
      result %= 10000000000;
    }
  }
  result = (result * 28433 + 1) % 10000000000;
  printf("The last ten digits of 28433*2^7830457+1 are %llu\n", result);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x777bb9,%esi
  mov    $0x2,%ecx
  movabs $0x2540be3ff,%rdi
  movabs $0xdbe6fecebdedd5bf,%r9
  movabs $0x2540be400,%r8
L26:
  sub    $0x1,%esi
  je     L29
L0:
  add    %rcx,%rcx
  cmp    %rdi,%rcx
  jbe    L26
  mov    %rcx,%rax
  mul    %r9
  shr    $0x21,%rdx
  imul   %r8,%rdx
  sub    %rdx,%rcx
  sub    $0x1,%esi
  jne    L0
L29:
  imul   $0x6f11,%rcx,%rcx
  lea    D7(%rip),%rdi
  movabs $0xdbe6fecebdedd5bf,%rdx
  lea    0x1(%rcx),%rsi
  mov    %rsi,%rax
  mul    %rdx
  movabs $0x2540be400,%rax
  shr    $0x21,%rdx
  imul   %rax,%rdx
  sub    %rdx,%rsi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D7
type_guessed: string

tool: D7: The last ten digits of 28433*2^7830457+1 are %llu


assistant: ```c
int main() {
  unsigned long long result = 1;
  int i;
  for (i = 0; i < 7830457; i++) {
    result *= 2;
    if (result >= 10000000000) {
      result %= 10000000000;
    }
  }
  result = (result * 28433 + 1) % 10000000000;
  printf("The last ten digits of 28433*2^7830457+1 are %llu\n", result);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x777bb8,%esi
  mov    $0x2,%ecx
  movabs $0x2540be3ff,%rdi
  movabs $0xdbe6fecebdedd5bf,%r9
  movabs $0x2540be400,%r8
L9:
  add    %rcx,%rcx
  cmp    %rdi,%rcx
  jbe    L4
  mov    %rcx,%rax
  mul    %r9
  shr    $0x21,%rdx
  imul   %r8,%rdx
  sub    %rdx,%rcx
L4:
  sub    $0x1,%esi
  jne    L9
  imul   $0x6f11,%rcx,%rcx
  lea    D7(%rip),%rdi
  movabs $0xdbe6fecebdedd5bf,%rdx
  lea    0x1(%rcx),%rsi
  mov    %rsi,%rax
  mul    %rdx
  movabs $0x2540be400,%rax
  shr    $0x21,%rdx
  imul   %rax,%rdx
  sub    %rdx,%rsi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type: string

tool: D7: The last ten digits of 28433*2^7830457+1 are %llu


assistant: ```c
int main() {
  unsigned long long result = 1;
  int i;
  for (i = 0; i < 7830457; i++) {
    result *= 2;
    if (result >= 10000000000) {
      result %= 10000000000;
    }
  }
  result = (result * 28433 + 1) % 10000000000;
  printf("The last ten digits of 28433*2^7830457+1 are %llu\n", result);
  return 0;
}
```