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

```asm
<doOutb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x3,-0x14(%rbp)
  jne    L26
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <isNumber@plt>
  test   %rax,%rax
  je     L26
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <isNumber@plt>
  test   %rax,%rax
  je     L26
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <stringToInt@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <stringToInt@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <printValue@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x8(%rbp),%eax
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <ioperm@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <outb@plt>
  mov    -0x8(%rbp),%eax
  mov    $0x0,%edx
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <ioperm@plt>
  mov    $0x0,%eax
  jmp    L3
L26:
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L3:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
type_guessed: string
name: read_data
arguments: data_tag: D10
type_guessed: string

tool: D0: outb 0x%X: 

tool: D10: modo de uso: outb <numero da porta> <valor>

assistant: ```c
int doOutb(int argc, char *argv[]) {
  if (argc == 3 && isNumber(argv[1]) && isNumber(argv[2])) {
    int puerto, valor;
    puerto = stringToInt(argv[1]);
    valor = stringToInt(argv[2]);
    printf("outb 0x%X: ", puerto);
    printValue(valor);
    printf("\n");
    ioperm(puerto, 1, 1);
    outb(valor, puerto);
    ioperm(puerto, 1, 0);
    return 0;
  }
  printf("modo de uso: outb <numero da porta> <valor>\n");
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<doOutb>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  je     L32
L38:
  lea    D16(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L32:
  mov    %rsi,%rbx
  mov    0x8(%rsi),%rdi
  call   <isNumber@plt>
  test   %rax,%rax
  je     L38
  mov    0x10(%rbx),%rdi
  call   <isNumber@plt>
  test   %rax,%rax
  je     L38
  mov    0x8(%rbx),%rdi
  call   <stringToInt@plt>
  mov    %eax,%ebp
  mov    0x10(%rbx),%rdi
  call   <stringToInt@plt>
  mov    %eax,%ebx
  mov    %ebp,%esi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%edi
  call   <printValue@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <ioperm@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <outb@plt>
  mov    $0x0,%edx
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <ioperm@plt>
  mov    $0x0,%eax
  jmp    L3
```

tool_calls: name: disassemble_data
arguments: data_label: D16
type_guessed: string
name: disassemble_data
arguments: data_label: D17
type_guessed: string

tool: D16: modo de uso: outb <numero da porta> <valor>

tool: D17: outb 0x%X: 

assistant: ```c
int doOutb(int argc, char *argv[]) {
  if (argc == 3 && isNumber(argv[1]) && isNumber(argv[2])) {
    int puerto, valor;
    puerto = stringToInt(argv[1]);
    valor = stringToInt(argv[2]);
    printf("outb 0x%X: ", puerto);
    printValue(valor);
    printf("\n");
    ioperm(puerto, 1, 1);
    outb(valor, puerto);
    ioperm(puerto, 1, 0);
    return 0;
  }
  printf("modo de uso: outb <numero da porta> <valor>\n");
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<doOutb>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x3,%edi
  je     L13
L39:
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L20:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  call   <isNumber@plt>
  test   %rax,%rax
  je     L39
  mov    0x10(%rbx),%rdi
  call   <isNumber@plt>
  test   %rax,%rax
  je     L39
  mov    0x8(%rbx),%rdi
  call   <stringToInt@plt>
  mov    0x10(%rbx),%rdi
  mov    %eax,%ebp
  call   <stringToInt@plt>
  mov    %ebp,%esi
  lea    D19(%rip),%rdi
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%edi
  call   <printValue@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <ioperm@plt>
  mov    %ebp,%esi
  mov    %r12d,%edi
  call   <outb@plt>
  xor    %edx,%edx
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <ioperm@plt>
  xor    %eax,%eax
  jmp    L20
```

tool_calls: name: unpack_data
arguments: tag: D14
type_guessed: string
name: unpack_data
arguments: tag: D19
type_guessed: string

tool: D14: modo de uso: outb <numero da porta> <valor>

tool: D19: outb 0x%X: 

assistant: ```c
int doOutb(int argc, char *argv[]) {
  if (argc == 3 && isNumber(argv[1]) && isNumber(argv[2])) {
    int puerto, valor;
    puerto = stringToInt(argv[1]);
    valor = stringToInt(argv[2]);
    printf("outb 0x%X: ", puerto);
    printValue(valor);
    printf("\n");
    ioperm(puerto, 1, 1);
    outb(valor, puerto);
    ioperm(puerto, 1, 0);
    return 0;
  }
  printf("modo de uso: outb <numero da porta> <valor>\n");
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<doOutb>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x3,%edi
  je     L13
L16:
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L32:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  call   <isNumber@plt>
  test   %rax,%rax
  je     L16
  mov    0x10(%rbx),%rdi
  call   <isNumber@plt>
  test   %rax,%rax
  je     L16
  mov    0x8(%rbx),%rdi
  call   <stringToInt@plt>
  mov    0x10(%rbx),%rdi
  mov    %eax,%ebp
  call   <stringToInt@plt>
  mov    %ebp,%esi
  lea    D0(%rip),%rdi
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%edi
  call   <printValue@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <ioperm@plt>
  mov    %ebp,%esi
  mov    %r12d,%edi
  call   <outb@plt>
  xor    %edx,%edx
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <ioperm@plt>
  xor    %eax,%eax
  jmp    L32
```

tool_calls: name: read_data
arguments: label: D11
datatype: string
name: read_data
arguments: label: D0
datatype: string

tool: D11: modo de uso: outb <numero da porta> <valor>

tool: D0: outb 0x%X: 

assistant: ```c
int doOutb(int argc, char *argv[]) {
  if (argc == 3 && isNumber(argv[1]) && isNumber(argv[2])) {
    int puerto, valor;
    puerto = stringToInt(argv[1]);
    valor = stringToInt(argv[2]);
    printf("outb 0x%X: ", puerto);
    printValue(valor);
    printf("\n");
    ioperm(puerto, 1, 1);
    outb(valor, puerto);
    ioperm(puerto, 1, 0);
    return 0;
  }
  printf("modo de uso: outb <numero da porta> <valor>\n");
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jle    L127
  mov    -0x18(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jl     L109
L127:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jge    L121
  mov    -0x18(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jle    L121
L109:
  mov    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L117
L121:
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  cmp    %eax,%edx
  jle    L63
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jl     L41
L63:
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  cmp    %eax,%edx
  jge    L37
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jle    L37
L41:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L117
L37:
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%eax
  cmp    %eax,%edx
  jle    L53
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jl     L58
L53:
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%eax
  cmp    %eax,%edx
  jge    L117
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jle    L117
L58:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L117:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L79
  call   <__stack_chk_fail@plt>
L79:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D8
data_type: string
name: parse_data
arguments: label: D25
data_type: string
name: parse_data
arguments: label: D21
data_type: string

tool: D8: Please enter 3 numbers separated by spaces > 

tool: D25: %d%d%d

tool: D21: %d is the median


assistant: ```c
int main(void) {
  int num1, num2, num3, median;
  printf("Please enter 3 numbers separated by spaces > ");
  scanf("%d%d%d", &num1, &num2, &num3);
  if ((num1 > num2 && num1 < num3) || (num1 < num2 && num1 > num3))
    median = num1;
  else if ((num2 > num1 && num2 < num3) || (num2 < num1 && num2 > num3))
    median = num2;
  else if ((num3 > num1 && num3 < num2) || (num3 < num1 && num3 > num2))
    median = num3;
  printf("%d is the median\n", median);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x10(%rsp),%eax
  cmp    %eax,%esi
  jle    L31
  cmp    0x14(%rsp),%esi
  jl     L151
  mov    0x14(%rsp),%edx
  cmp    %edx,%eax
  jg     L0
L39:
  cmp    %edx,%eax
  jge    L151
  cmp    %edx,%esi
  jle    L151
L17:
  mov    %edx,%esi
  jmp    L151
L31:
  jge    L70
  mov    0x14(%rsp),%edx
  cmp    %edx,%esi
  jg     L151
  cmp    %edx,%eax
  jl     L143
  mov    0x14(%rsp),%edx
  cmp    %edx,%eax
  jle    L151
  cmp    %edx,%esi
  jl     L17
  jmp    L151
L70:
  mov    0x14(%rsp),%edx
  cmp    %edx,%eax
  jle    L39
  cmp    %edx,%esi
  jl     L17
  jmp    L39
L143:
  mov    %eax,%esi
L151:
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L114
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L0:
  mov    %eax,%esi
  jmp    L151
L114:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
data_type: string
name: disassemble_data
arguments: data_tag: D0
data_type: string
name: disassemble_data
arguments: data_tag: D6
data_type: string

tool: D4: Please enter 3 numbers separated by spaces > 

tool: D0: %d%d%d

tool: D6: %d is the median


assistant: ```c
int main(void) {
  int num1, num2, num3, median;
  printf("Please enter 3 numbers separated by spaces > ");
  scanf("%d%d%d", &num1, &num2, &num3);
  if ((num1 > num2 && num1 < num3) || (num1 < num2 && num1 > num3))
    median = num1;
  else if ((num2 > num1 && num2 < num3) || (num2 < num1 && num2 > num3))
    median = num2;
  else if ((num3 > num1 && num3 < num2) || (num3 < num1 && num3 > num2))
    median = num3;
  printf("%d is the median\n", median);
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rcx
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x10(%rsp),%eax
  mov    0x14(%rsp),%edx
  cmp    %eax,%esi
  jle    L56
  cmp    %edx,%esi
  jge    L105
L14:
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L60
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L56:
  jge    L32
  cmp    %edx,%esi
  jg     L14
  cmp    %edx,%eax
  jge    L128
L16:
  mov    %eax,%esi
  jmp    L14
L105:
  cmp    %edx,%eax
  jg     L16
L118:
  cmp    %edx,%esi
  jle    L14
  cmp    %edx,%eax
  cmovl  %edx,%esi
  jmp    L14
L128:
  cmp    %edx,%esi
  cmovl  %edx,%esi
  jmp    L14
L32:
  cmp    %edx,%esi
  jge    L118
  mov    %edx,%esi
  jmp    L14
L60:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D14
type_infered: string
name: parse_data
arguments: label: D26
type_infered: string
name: parse_data
arguments: label: D4
type_infered: string

tool: D14: Please enter 3 numbers separated by spaces > 

tool: D26: %d%d%d

tool: D4: %d is the median


assistant: ```c
int main(void) {
  int num1, num2, num3, median;
  printf("Please enter 3 numbers separated by spaces > ");
  scanf("%d%d%d", &num1, &num2, &num3);
  if ((num1 > num2 && num1 < num3) || (num1 < num2 && num1 > num3))
    median = num1;
  else if ((num2 > num1 && num2 < num3) || (num2 < num1 && num2 > num3))
    median = num2;
  else if ((num3 > num1 && num3 < num2) || (num3 < num1 && num3 > num2))
    median = num3;
  printf("%d is the median\n", median);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D8(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rcx
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x10(%rsp),%eax
  mov    0x14(%rsp),%edx
  cmp    %eax,%esi
  jle    L55
  cmp    %edx,%esi
  jge    L52
L115:
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L55:
  jge    L114
  cmp    %edx,%esi
  jg     L115
  cmp    %edx,%eax
  jge    L116
L99:
  mov    %eax,%esi
  jmp    L115
L52:
  cmp    %edx,%eax
  jg     L99
L12:
  cmp    %edx,%esi
  jle    L115
  cmp    %edx,%eax
  cmovl  %edx,%esi
  jmp    L115
L116:
  cmp    %edx,%esi
  cmovl  %edx,%esi
  jmp    L115
L114:
  cmp    %edx,%esi
  jge    L12
  mov    %edx,%esi
  jmp    L115
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D8
type_infered: string
name: read_data
arguments: data_label: D26
type_infered: string
name: read_data
arguments: data_label: D18
type_infered: string

tool: D8: Please enter 3 numbers separated by spaces > 

tool: D26: %d%d%d

tool: D18: %d is the median


assistant: ```c
int main(void) {
  int num1, num2, num3, median;
  printf("Please enter 3 numbers separated by spaces > ");
  scanf("%d%d%d", &num1, &num2, &num3);
  if ((num1 > num2 && num1 < num3) || (num1 < num2 && num1 > num3))
    median = num1;
  else if ((num2 > num1 && num2 < num3) || (num2 < num1 && num2 > num3))
    median = num2;
  else if ((num3 > num1 && num3 < num2) || (num3 < num1 && num3 > num2))
    median = num3;
  printf("%d is the median\n", median);
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<HMM_ATanF>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movd   %eax,%xmm0
  call   <atanf@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  leave
  ret
```

assistant: ```c
float HMM_ATanF(float Radians) {
  float Result = atanf(Radians);
  return (Result);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<HMM_ATanF>:
  endbr64
  sub    $0x8,%rsp
  call   <atanf@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
float HMM_ATanF(float Radians) {
  float Result = atanf(Radians);
  return (Result);
}
```
user: Translate the below assembly code into C programming language:

```asm
<HMM_ATanF>:
  endbr64
  jmp    1050 <atanf@plt>
```

assistant: ```c
float HMM_ATanF(float Radians) {
  float Result = atanf(Radians);
  return (Result);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<HMM_ATanF>:
  endbr64
  jmp    1050 <atanf@plt>
```

assistant: ```c
float HMM_ATanF(float Radians) {
  float Result = atanf(Radians);
  return (Result);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<uart_puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L0
L11:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <uart_putc@plt>
L0:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L11
  nop
  nop
  leave
  ret
```

assistant: ```c
void uart_puts(void *uart_base, const char *string) {
  while (*string != 0) {
    uart_putc(uart_base, *string++);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<uart_puts>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  movzbl (%rsi),%esi
  test   %sil,%sil
  je     L6
L1:
  add    $0x1,%rbx
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <uart_putc@plt>
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L1
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void uart_puts(void *uart_base, const char *string) {
  while (*string != 0) {
    uart_putc(uart_base, *string++);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<uart_puts>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movsbl (%rsi),%esi
  test   %sil,%sil
  je     L4
  mov    %rdi,%rbp
  nopl   0x0(%rax,%rax,1)
L3:
  add    $0x1,%rbx
  mov    %rbp,%rdi
  call   <uart_putc@plt>
  movsbl (%rbx),%esi
  test   %sil,%sil
  jne    L3
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void uart_puts(void *uart_base, const char *string) {
  while (*string != 0) {
    uart_putc(uart_base, *string++);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<uart_puts>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movsbl (%rsi),%esi
  test   %sil,%sil
  je     L14
  mov    %rdi,%rbp
  nopl   0x0(%rax,%rax,1)
L4:
  add    $0x1,%rbx
  mov    %rbp,%rdi
  call   <uart_putc@plt>
  movsbl (%rbx),%esi
  test   %sil,%sil
  jne    L4
L14:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void uart_puts(void *uart_base, const char *string) {
  while (*string != 0) {
    uart_putc(uart_base, *string++);
  }
}
```
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
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x7,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L73:
  cmp    %rdx,%rsp
  je     L74
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L73
L74:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L65
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L65:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L36
L8:
  mov    -0x20(%rbp),%rax
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  mov    -0x34(%rbp),%ecx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x34(%rbp)
L36:
  mov    -0x34(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L8
  movl   $0x0,-0x30(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L39
L43:
  mov    -0x20(%rbp),%rax
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  add    %eax,-0x30(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    $0x0,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L39:
  mov    -0x34(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L43
  mov    -0x30(%rbp),%eax
  mov    %eax,%edx
  mov    $0x0,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsp
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D21
type_guessed: string
name: read_data
arguments: label: D14
type_guessed: string
name: read_data
arguments: label: D24
type_guessed: string

tool: D21: thread %d suma a[%d] / 

tool: D14: 
* thread %d suma= %d

tool: D24: 
 suma = %d


assistant: ```c
main() {
  int i, n = 7;
  int a[n], suma;
  for (i = 0; i < n; i++)
    a[i] = i;
#pragma omp parallel private(suma)
  {
    suma = 0;
#pragma omp for
    for (i = 0; i < n; i++) {
      suma = suma + a[i];
      printf("thread %d suma a[%d] / ", 0, i);
    }
    printf("\n* thread %d suma= %d", 0, suma);
  }
  printf("\n suma = %d\n", suma);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x1,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0x3,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x5,0x14(%rsp)
  movl   $0x6,0x18(%rsp)
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  lea    D20(%rip),%r12
L2:
  add    (%rsp,%rbx,4),%ebp
  mov    %ebx,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x7,%rbx
  jne    L2
  mov    %ebp,%edx
  mov    $0x0,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D20
datatype: string
name: unpack_data
arguments: data_label: D4
datatype: string
name: unpack_data
arguments: data_label: D16
datatype: string

tool: D20: thread %d suma a[%d] / 

tool: D4: 
* thread %d suma= %d

tool: D16: 
 suma = %d


assistant: ```c
main() {
  int i, n = 7;
  int a[n], suma;
  for (i = 0; i < n; i++)
    a[i] = i;
#pragma omp parallel private(suma)
  {
    suma = 0;
#pragma omp for
    for (i = 0; i < n; i++) {
      suma = suma + a[i];
      printf("thread %d suma a[%d] / ", 0, i);
    }
    printf("\n* thread %d suma= %d", 0, suma);
  }
  printf("\n suma = %d\n", suma);
}
```