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

```asm
<main>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rdi
  lea    D13(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D2(%rip),%rdi
  call   <puts@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%esi
  mov    0xc(%rsp),%edi
  mov    0x14(%rsp),%edx
  call   <_vote@plt>
  lea    D18(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D8
data_type: string
name: retrieve_data
arguments: data_label: D13
data_type: string
name: retrieve_data
arguments: data_label: D38
data_type: string
name: retrieve_data
arguments: data_label: D2
data_type: string
name: retrieve_data
arguments: data_label: D18
data_type: string

tool: D8: First vote [0/1]

tool: D13: %d

tool: D38: Second vote

tool: D2: Third vote

tool: D18: Result is %d


assistant: ```c
int main(int argc, char **argv) {
  int a, b, c, result;
  printf("First vote [0/1]\n");
  scanf("%d", &a);
  printf("Second vote\n");
  scanf("%d", &b);
  printf("Third vote\n");
  scanf("%d", &c);
  result = _vote(a, b, c);
  printf("Result is %d\n", result);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D40(%rip),%rdi
  lea    D37(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D7(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D3(%rip),%rdi
  call   <puts@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%esi
  mov    0xc(%rsp),%edi
  mov    0x14(%rsp),%edx
  call   <_vote@plt>
  lea    D36(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D40
data_type: string
name: read_data
arguments: tag: D37
data_type: string
name: read_data
arguments: tag: D7
data_type: string
name: read_data
arguments: tag: D3
data_type: string
name: read_data
arguments: tag: D36
data_type: string

tool: D40: First vote [0/1]

tool: D37: %d

tool: D7: Second vote

tool: D3: Third vote

tool: D36: Result is %d


assistant: ```c
int main(int argc, char **argv) {
  int a, b, c, result;
  printf("First vote [0/1]\n");
  scanf("%d", &a);
  printf("Second vote\n");
  scanf("%d", &b);
  printf("Third vote\n");
  scanf("%d", &c);
  result = _vote(a, b, c);
  printf("Result is %d\n", result);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<my_count_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L18:
  addl   $0x1,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
L0:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L76
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L60
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jle    L18
L60:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L15
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jle    L18
L15:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L76
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L18
L76:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int my_count_char(char *s, int *i) {
  int cpt;
  cpt = 0;
  while (s[*i] && ((s[*i] >= 'a' && s[*i] <= 'z') || (s[*i] >= 'A' && s[*i] <= 'Z') || (s[*i] >= '0' && s[*i] <= '9'))) {
    cpt++;
    *i = *i + 1;
  }
  return (cpt);
}
```
user: What is the c source code of the assembly code below:

```asm
<my_count_char>:
  endbr64
  mov    (%rsi),%edx
  movslq %edx,%rax
  movzbl (%rdi,%rax,1),%eax
  test   %al,%al
  je     L38
  add    $0x1,%edx
  movslq %edx,%rdx
  mov    $0x0,%r8d
  jmp    L29
L32:
  add    $0x1,%r8d
  mov    %edx,(%rsi)
  add    $0x1,%rdx
  movzbl -0x1(%rdi,%rdx,1),%eax
  test   %al,%al
  je     L35
L29:
  mov    %eax,%ecx
  and    $0xffffffdf,%ecx
  sub    $0x41,%ecx
  cmp    $0x19,%cl
  jbe    L32
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L32
L35:
  mov    %r8d,%eax
  ret
L38:
  mov    $0x0,%r8d
  jmp    L35
```

assistant: ```c
int my_count_char(char *s, int *i) {
  int cpt;
  cpt = 0;
  while (s[*i] && ((s[*i] >= 'a' && s[*i] <= 'z') || (s[*i] >= 'A' && s[*i] <= 'Z') || (s[*i] >= '0' && s[*i] <= '9'))) {
    cpt++;
    *i = *i + 1;
  }
  return (cpt);
}
```
user: Translate the below assembly code into C programming language:

```asm
<my_count_char>:
  endbr64
  movslq (%rsi),%rax
  mov    %rax,%rdx
  movzbl (%rdi,%rax,1),%eax
  test   %al,%al
  je     L29
  add    $0x1,%edx
  xor    %r8d,%r8d
  movslq %edx,%rdx
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %eax,%ecx
  and    $0xffffffdf,%ecx
  sub    $0x41,%ecx
  cmp    $0x19,%cl
  jbe    L49
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L43
L49:
  mov    %edx,(%rsi)
  add    $0x1,%rdx
  add    $0x1,%r8d
  movzbl -0x1(%rdi,%rdx,1),%eax
  test   %al,%al
  jne    L8
L43:
  mov    %r8d,%eax
  ret
L29:
  xor    %r8d,%r8d
  jmp    L43
```

assistant: ```c
int my_count_char(char *s, int *i) {
  int cpt;
  cpt = 0;
  while (s[*i] && ((s[*i] >= 'a' && s[*i] <= 'z') || (s[*i] >= 'A' && s[*i] <= 'Z') || (s[*i] >= '0' && s[*i] <= '9'))) {
    cpt++;
    *i = *i + 1;
  }
  return (cpt);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<my_count_char>:
  endbr64
  movslq (%rsi),%rax
  mov    %rax,%rdx
  movzbl (%rdi,%rax,1),%eax
  test   %al,%al
  je     L46
  add    $0x1,%edx
  xor    %r8d,%r8d
  movslq %edx,%rdx
  nopl   0x0(%rax,%rax,1)
L35:
  mov    %eax,%ecx
  and    $0xffffffdf,%ecx
  sub    $0x41,%ecx
  cmp    $0x19,%cl
  jbe    L48
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L33
L48:
  mov    %edx,(%rsi)
  add    $0x1,%rdx
  add    $0x1,%r8d
  movzbl -0x1(%rdi,%rdx,1),%eax
  test   %al,%al
  jne    L35
L33:
  mov    %r8d,%eax
  ret
L46:
  xor    %r8d,%r8d
  jmp    L33
```

assistant: ```c
int my_count_char(char *s, int *i) {
  int cpt;
  cpt = 0;
  while (s[*i] && ((s[*i] >= 'a' && s[*i] <= 'z') || (s[*i] >= 'A' && s[*i] <= 'Z') || (s[*i] >= '0' && s[*i] <= '9'))) {
    cpt++;
    *i = *i + 1;
  }
  return (cpt);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<interpolate_vertical>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  mov    %rcx,%r10
  lea    (%rdx,%rdx,1),%edx
  lea    (%rdx,%rbx,1),%eax
  add    %esi,%eax
  mov    %eax,%r11d
  sub    %edx,%r11d
  mov    %r11d,%ecx
  sub    %edx,%ecx
  cltq
  movslq (%rdi,%rax,8),%rbp
  movslq %r11d,%r11
  movslq (%rdi,%r11,8),%rax
  mov    (%r10,%rbp,4),%r11d
  sub    (%r10,%rax,4),%r11d
  mov    %r11d,%eax
  neg    %eax
  cmovs  %r11d,%eax
  movslq %ecx,%r11
  movslq (%rdi,%r11,8),%r11
  sub    %edx,%ecx
  movslq %ecx,%rcx
  movslq (%rdi,%rcx,8),%rcx
  mov    (%r10,%r11,4),%r11d
  sub    (%r10,%rcx,4),%r11d
  mov    %r11d,%ecx
  neg    %ecx
  cmovs  %r11d,%ecx
  mov    %eax,%r11d
  add    %ecx,%r11d
  jne    L8
  add    %esi,%edx
  movslq %edx,%rdx
  mov    (%rdi,%rdx,8),%rax
  movslq %esi,%rsi
  mov    %rax,(%rdi,%rsi,8)
L20:
  pop    %rbx
  pop    %rbp
  ret
L8:
  lea    (%rdx,%rsi,1),%ebp
  shl    $0x2,%ebx
  mov    %ebp,%edx
  sub    %ebx,%edx
  movslq %edx,%rdx
  mov    (%rdi,%rdx,8),%rbx
  shl    $0x8,%eax
  cltd
  idiv   %r11d
  imul   (%r10,%rbx,4),%eax
  sar    $0x8,%eax
  mov    %eax,%ebx
  movslq %ebp,%rdx
  mov    (%rdi,%rdx,8),%rbp
  mov    %ecx,%eax
  shl    $0x8,%eax
  cltd
  idiv   %r11d
  imul   (%r10,%rbp,4),%eax
  sar    $0x8,%eax
  add    %ebx,%eax
  mov    $0x0,%edx
  test   %eax,%eax
  jle    L9
  movslq %eax,%rdx
  shl    $0x2,%rdx
  cmp    $0x6ffff,%eax
  mov    $0x1bfffc,%eax
  cmovge %rax,%rdx
L9:
  movslq %esi,%rsi
  add    (%r8,%rdx,1),%r9d
  movslq %r9d,%r9
  mov    %r9,(%rdi,%rsi,8)
  jmp    L20
```

assistant: ```c
void interpolate_vertical(uint16_t *image_data, int i, int w, int *raw2ev, int *ev2raw, int black) {
  int gv1 = image_data[i + w * 3];
  int gv2 = image_data[i + w];
  int gv3 = image_data[i - w];
  int gv4 = image_data[i - w * 3];
  int dv1 = ((raw2ev[gv1] - raw2ev[gv2]) > 0 ? (raw2ev[gv1] - raw2ev[gv2]) : -(raw2ev[gv1] - raw2ev[gv2]));
  int dv2 = ((raw2ev[gv3] - raw2ev[gv4]) > 0 ? (raw2ev[gv3] - raw2ev[gv4]) : -(raw2ev[gv3] - raw2ev[gv4]));
  int sum = dv1 + dv2;
  if (sum == 0) {
    image_data[i] = image_data[i + w * 2];
  } else {
    int cv1 = ((sum - dv1) << 8) / sum;
    int cv2 = ((sum - dv2) << 8) / sum;
    int ev_corr = ((raw2ev[image_data[i + w * 2]] * cv1) >> 8) + ((raw2ev[image_data[i - w * 2]] * cv2) >> 8);
    image_data[i] = ev2raw[(((((((ev_corr)) < ((14 * 32768 - 1))) ? ((ev_corr)) : ((14 * 32768 - 1)))) > ((0))) ? (((((ev_corr)) < ((14 * 32768 - 1))) ? ((ev_corr)) : ((14 * 32768 - 1)))) : ((0)))] + black;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<interpolate_vertical>:
  endbr64
  mov    %rcx,%r10
  lea    (%rdx,%rdx,1),%ecx
  push   %r12
  mov    %rdi,%r12
  lea    (%rcx,%rdx,1),%r11d
  push   %rbp
  lea    (%r11,%rsi,1),%eax
  push   %rbx
  cltq
  movslq (%rdi,%rax,8),%rdi
  lea    (%rdx,%rsi,1),%eax
  cltq
  movslq (%r12,%rax,8),%rax
  mov    (%r10,%rdi,4),%edi
  sub    (%r10,%rax,4),%edi
  mov    %edi,%eax
  neg    %eax
  cmovs  %edi,%eax
  mov    %esi,%edi
  sub    %edx,%edi
  movslq %edi,%rdx
  mov    %eax,%ebp
  movslq (%r12,%rdx,8),%rbx
  mov    %esi,%edx
  sub    %r11d,%edx
  movslq %edx,%rdx
  movslq (%r12,%rdx,8),%rdi
  mov    (%r10,%rbx,4),%edx
  sub    (%r10,%rdi,4),%edx
  mov    %edx,%edi
  neg    %edi
  cmovs  %edx,%edi
  lea    (%rsi,%rcx,1),%edx
  movslq %edx,%rdx
  mov    (%r12,%rdx,8),%rbx
  movslq %esi,%rdx
  lea    (%r12,%rdx,8),%r11
  add    %edi,%ebp
  je     L29
  shl    $0x8,%eax
  sub    %ecx,%esi
  cltd
  movslq %esi,%rsi
  idiv   %ebp
  mov    (%r12,%rsi,8),%rcx
  imul   (%r10,%rcx,4),%eax
  sar    $0x8,%eax
  mov    %eax,%ecx
  mov    %edi,%eax
  shl    $0x8,%eax
  cltd
  idiv   %ebp
  imul   (%r10,%rbx,4),%eax
  sar    $0x8,%eax
  add    %ecx,%eax
  test   %eax,%eax
  jle    L31
  cmp    $0x6fffe,%eax
  jg     L10
  cltq
  lea    (%r8,%rax,4),%r8
L31:
  add    (%r8),%r9d
  movslq %r9d,%rbx
L29:
  mov    %rbx,(%r11)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L10:
  add    $0x1bfffc,%r8
  jmp    L31
```

assistant: ```c
void interpolate_vertical(uint16_t *image_data, int i, int w, int *raw2ev, int *ev2raw, int black) {
  int gv1 = image_data[i + w * 3];
  int gv2 = image_data[i + w];
  int gv3 = image_data[i - w];
  int gv4 = image_data[i - w * 3];
  int dv1 = ((raw2ev[gv1] - raw2ev[gv2]) > 0 ? (raw2ev[gv1] - raw2ev[gv2]) : -(raw2ev[gv1] - raw2ev[gv2]));
  int dv2 = ((raw2ev[gv3] - raw2ev[gv4]) > 0 ? (raw2ev[gv3] - raw2ev[gv4]) : -(raw2ev[gv3] - raw2ev[gv4]));
  int sum = dv1 + dv2;
  if (sum == 0) {
    image_data[i] = image_data[i + w * 2];
  } else {
    int cv1 = ((sum - dv1) << 8) / sum;
    int cv2 = ((sum - dv2) << 8) / sum;
    int ev_corr = ((raw2ev[image_data[i + w * 2]] * cv1) >> 8) + ((raw2ev[image_data[i - w * 2]] * cv2) >> 8);
    image_data[i] = ev2raw[(((((((ev_corr)) < ((14 * 32768 - 1))) ? ((ev_corr)) : ((14 * 32768 - 1)))) > ((0))) ? (((((ev_corr)) < ((14 * 32768 - 1))) ? ((ev_corr)) : ((14 * 32768 - 1)))) : ((0)))] + black;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<interpolate_vertical>:
  endbr64
  mov    %rcx,%r10
  lea    (%rdx,%rdx,1),%ecx
  push   %r12
  mov    %rdi,%r12
  lea    (%rcx,%rdx,1),%r11d
  push   %rbp
  lea    (%r11,%rsi,1),%eax
  push   %rbx
  cltq
  movslq (%rdi,%rax,8),%rdi
  lea    (%rdx,%rsi,1),%eax
  cltq
  movslq (%r12,%rax,8),%rax
  mov    (%r10,%rdi,4),%edi
  sub    (%r10,%rax,4),%edi
  mov    %edi,%eax
  neg    %eax
  cmovs  %edi,%eax
  mov    %esi,%edi
  sub    %edx,%edi
  movslq %edi,%rdx
  mov    %eax,%ebp
  movslq (%r12,%rdx,8),%rbx
  mov    %esi,%edx
  sub    %r11d,%edx
  movslq %edx,%rdx
  movslq (%r12,%rdx,8),%rdi
  mov    (%r10,%rbx,4),%edx
  sub    (%r10,%rdi,4),%edx
  mov    %edx,%edi
  neg    %edi
  cmovs  %edx,%edi
  lea    (%rsi,%rcx,1),%edx
  movslq %edx,%rdx
  mov    (%r12,%rdx,8),%rbx
  movslq %esi,%rdx
  lea    (%r12,%rdx,8),%r11
  add    %edi,%ebp
  je     L35
  shl    $0x8,%eax
  sub    %ecx,%esi
  cltd
  movslq %esi,%rsi
  idiv   %ebp
  mov    (%r12,%rsi,8),%rcx
  imul   (%r10,%rcx,4),%eax
  sar    $0x8,%eax
  mov    %eax,%ecx
  mov    %edi,%eax
  shl    $0x8,%eax
  cltd
  idiv   %ebp
  imul   (%r10,%rbx,4),%eax
  sar    $0x8,%eax
  add    %ecx,%eax
  test   %eax,%eax
  jle    L22
  cmp    $0x6fffe,%eax
  jg     L12
  cltq
  lea    (%r8,%rax,4),%r8
L22:
  add    (%r8),%r9d
  movslq %r9d,%rbx
L35:
  mov    %rbx,(%r11)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L12:
  add    $0x1bfffc,%r8
  jmp    L22
```

assistant: ```c
void interpolate_vertical(uint16_t *image_data, int i, int w, int *raw2ev, int *ev2raw, int black) {
  int gv1 = image_data[i + w * 3];
  int gv2 = image_data[i + w];
  int gv3 = image_data[i - w];
  int gv4 = image_data[i - w * 3];
  int dv1 = ((raw2ev[gv1] - raw2ev[gv2]) > 0 ? (raw2ev[gv1] - raw2ev[gv2]) : -(raw2ev[gv1] - raw2ev[gv2]));
  int dv2 = ((raw2ev[gv3] - raw2ev[gv4]) > 0 ? (raw2ev[gv3] - raw2ev[gv4]) : -(raw2ev[gv3] - raw2ev[gv4]));
  int sum = dv1 + dv2;
  if (sum == 0) {
    image_data[i] = image_data[i + w * 2];
  } else {
    int cv1 = ((sum - dv1) << 8) / sum;
    int cv2 = ((sum - dv2) << 8) / sum;
    int ev_corr = ((raw2ev[image_data[i + w * 2]] * cv1) >> 8) + ((raw2ev[image_data[i - w * 2]] * cv2) >> 8);
    image_data[i] = ev2raw[(((((((ev_corr)) < ((14 * 32768 - 1))) ? ((ev_corr)) : ((14 * 32768 - 1)))) > ((0))) ? (((((ev_corr)) < ((14 * 32768 - 1))) ? ((ev_corr)) : ((14 * 32768 - 1)))) : ((0)))] + black;
  }
}
```
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
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  1 < 2;
  3 > 4;
  5 <= 6;
  7 >= 8;
  9 == 10;
  11 != 12;
  13 && 0;
  15 || 0;
  !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
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  1 < 2;
  3 > 4;
  5 <= 6;
  7 >= 8;
  9 == 10;
  11 != 12;
  13 && 0;
  15 || 0;
  !0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

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

assistant: ```c
int main() {
  1 < 2;
  3 > 4;
  5 <= 6;
  7 >= 8;
  9 == 10;
  11 != 12;
  13 && 0;
  15 || 0;
  !0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

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

assistant: ```c
int main() {
  1 < 2;
  3 > 4;
  5 <= 6;
  7 >= 8;
  9 == 10;
  11 != 12;
  13 && 0;
  15 || 0;
  !0;
}
```
user: What is the c source code of the assembly code below?

```asm
<converter_string_base>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  jne    L32
  mov    $0x0,%eax
  jmp    L0
L32:
  mov    -0x18(%rbp),%rax
  lea    -0x18(%rbp),%rcx
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
  mov    -0x4(%rbp),%eax
  jmp    L0
L6:
  mov    $0x0,%eax
L0:
  leave
  ret
```

assistant: ```c
uint8_t converter_string_base(char *string) {
  if (!string)
    return 0;
  uint8_t base = strtol(string, &string, 10);
  return *string == '\0' ? base : 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<converter_string_base>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L7
  lea    0x8(%rsp),%rsi
  mov    $0xa,%edx
  call   <strtol@plt>
  mov    0x8(%rsp),%rdx
  cmpb   $0x0,(%rdx)
  mov    $0x0,%edx
  cmovne %edx,%eax
L7:
  add    $0x18,%rsp
  ret
```

assistant: ```c
uint8_t converter_string_base(char *string) {
  if (!string)
    return 0;
  uint8_t base = strtol(string, &string, 10);
  return *string == '\0' ? base : 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<converter_string_base>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  test   %rdi,%rdi
  je     L0
  mov    $0xa,%edx
  lea    0x8(%rsp),%rsi
  call   <strtol@plt>
  mov    0x8(%rsp),%rdx
  cmpb   $0x0,(%rdx)
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
uint8_t converter_string_base(char *string) {
  if (!string)
    return 0;
  uint8_t base = strtol(string, &string, 10);
  return *string == '\0' ? base : 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<myrandom>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  imul   $0x191,%rax,%rax
  add    $0x2a1,%rax
  cqto
  idivq  -0x20(%rbp)
  mov    %rdx,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long myrandom(long num, long nrange) {
  long i;
  i = (num * 401 + 673) % nrange;
  return (i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<myrandom>:
  endbr64
  imul   $0x191,%rdi,%rax
  add    $0x2a1,%rax
  cqto
  idiv   %rsi
  mov    %rdx,%rax
  ret
```

assistant: ```c
long myrandom(long num, long nrange) {
  long i;
  i = (num * 401 + 673) % nrange;
  return (i);
}
```
user: What is the c source code of the assembly code below:

```asm
<myrandom>:
  endbr64
  imul   $0x191,%rdi,%rax
  add    $0x2a1,%rax
  cqto
  idiv   %rsi
  mov    %rdx,%rax
  ret
```

assistant: ```c
long myrandom(long num, long nrange) {
  long i;
  i = (num * 401 + 673) % nrange;
  return (i);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<myrandom>:
  endbr64
  imul   $0x191,%rdi,%rax
  add    $0x2a1,%rax
  cqto
  idiv   %rsi
  mov    %rdx,%rax
  ret
```

assistant: ```c
long myrandom(long num, long nrange) {
  long i;
  i = (num * 401 + 673) % nrange;
  return (i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<change_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  jle    L3
  cmpl   $0x6,-0x4(%rbp)
  jg     L3
  addl   $0x1,-0x8(%rbp)
  jmp    L10
L3:
  cmpl   $0xa,-0x4(%rbp)
  jne    L10
  subl   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
datatype: string

tool: D1: Current count: %i


assistant: ```c
int change_count(int val, int count) {
  if ((val > 2) && (val < 7)) {
    count++;
  } else if (val == 10) {
    count--;
  }
  printf("Current count: %i\n", count);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<change_count>:
  endbr64
  sub    $0x8,%rsp
  lea    -0x3(%rdi),%eax
  cmp    $0x3,%eax
  ja     L4
  add    $0x1,%esi
L3:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
L4:
  cmp    $0xa,%edi
  sete   %al
  movzbl %al,%eax
  sub    %eax,%esi
  jmp    L3
```

tool_calls: name: parse_data
arguments: tag: D6
type_infered: string

tool: D6: Current count: %i


assistant: ```c
int change_count(int val, int count) {
  if ((val > 2) && (val < 7)) {
    count++;
  } else if (val == 10) {
    count--;
  }
  printf("Current count: %i\n", count);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<change_count>:
  endbr64
  lea    -0x3(%rdi),%eax
  cmp    $0x3,%eax
  ja     L3
  add    $0x1,%esi
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  cmp    $0xa,%edi
  lea    D8(%rip),%rdi
  sete   %al
  sub    %eax,%esi
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type_guessed: string

tool: D8: Current count: %i


assistant: ```c
int change_count(int val, int count) {
  if ((val > 2) && (val < 7)) {
    count++;
  } else if (val == 10) {
    count--;
  }
  printf("Current count: %i\n", count);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<change_count>:
  endbr64
  lea    -0x3(%rdi),%eax
  cmp    $0x3,%eax
  ja     L0
  add    $0x1,%esi
  lea    D7(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %eax,%eax
  cmp    $0xa,%edi
  lea    D7(%rip),%rdi
  sete   %al
  sub    %eax,%esi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: tag: D7
type_infered: string

tool: D7: Current count: %i


assistant: ```c
int change_count(int val, int count) {
  if ((val > 2) && (val < 7)) {
    count++;
  } else if (val == 10) {
    count--;
  }
  printf("Current count: %i\n", count);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<vector_col4_printf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8d,-0x34(%rbp)
  mov    %r9,-0x40(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x40(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x40(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x40(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x40(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L10
  nop
  nop
  leave
  ret
```

assistant: ```c
void vector_col4_printf(double *v1, double *v2, double *v3, double *v4, int length, char *format) {
  int i;
  for (i = 0; i < length; i++) {
    printf(format, v1[i]);
    printf(" ");
    printf(format, v2[i]);
    printf(" ");
    printf(format, v3[i]);
    printf(" ");
    printf(format, v4[i]);
    printf("\n");
  }
}
```