ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xf0(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x105(%rbp),%rax
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0xb0(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0xb0(%rbp),%rax
  mov    %rax,%rdi
  call   <strl@plt>
  mov    %eax,-0xf4(%rbp)
  movl   $0x0,-0x104(%rbp)
  movl   $0x0,-0x100(%rbp)
  jmp    L54
L81:
  mov    -0x104(%rbp),%eax
  cltq
  movzbl -0xf0(%rbp,%rax,1),%edx
  mov    -0x104(%rbp),%eax
  cltq
  mov    %dl,-0x70(%rbp,%rax,1)
  addl   $0x1,-0x100(%rbp)
  addl   $0x1,-0x104(%rbp)
L54:
  cmpl   $0x31,-0x104(%rbp)
  jg     L57
  mov    -0x104(%rbp),%eax
  cltq
  movzbl -0xf0(%rbp,%rax,1),%edx
  movzbl -0x105(%rbp),%eax
  cmp    %al,%dl
  jne    L81
L57:
  movl   $0x0,-0xfc(%rbp)
  jmp    L96
L65:
  mov    -0xfc(%rbp),%eax
  cltq
  movzbl -0xb0(%rbp,%rax,1),%edx
  mov    -0x104(%rbp),%eax
  cltq
  mov    %dl,-0x70(%rbp,%rax,1)
  addl   $0x1,-0xfc(%rbp)
  addl   $0x1,-0x104(%rbp)
L96:
  mov    -0xfc(%rbp),%eax
  cmp    -0xf4(%rbp),%eax
  jge    L56
  mov    -0xfc(%rbp),%eax
  cltq
  movzbl -0xb0(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L65
L56:
  movl   $0x0,-0xf8(%rbp)
  jmp    L25
L73:
  mov    -0x100(%rbp),%edx
  mov    -0xf8(%rbp),%eax
  add    %edx,%eax
  cltq
  movzbl -0xf0(%rbp,%rax,1),%edx
  mov    -0x104(%rbp),%eax
  cltq
  mov    %dl,-0x70(%rbp,%rax,1)
  addl   $0x1,-0xf8(%rbp)
  addl   $0x1,-0x104(%rbp)
L25:
  cmpl   $0x31,-0xf8(%rbp)
  jg     L17
  mov    -0x100(%rbp),%edx
  mov    -0xf8(%rbp),%eax
  add    %edx,%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L73
L17:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L42
  call   <__stack_chk_fail@plt>
L42:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D17
datatype: string
name: parse_data
arguments: data_label: D37
datatype: string
name: parse_data
arguments: data_label: D5
datatype: string

tool: D17: %s


tool: D37: %c


tool: D5: %s

assistant: ```c
int main() {
  char str1[50], str2[50], ch, str[100];
  scanf("%s\n", str1);
  scanf("%c\n", &ch);
  scanf("%s\n", str2);
  int m = strl(str2);
  int i = 0, co = 0;
  while ((i < 50) && (str1[i] != ch)) {
    str[i] = str1[i];
    co = co + 1;
    i = i + 1;
  }
  int j = 0;
  while ((j < m) && (str2[j] != '\0')) {
    str[i] = str2[j];
    j = j + 1;
    i = i + 1;
  }
  int k = 0;
  while ((k < 50) && (str[co + k] != '\0')) {
    str[i] = str1[co + k];
    k = k + 1;
    i = i + 1;
  }
  printf("%s", str);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x108,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  lea    D23(%rip),%rbp
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0xf(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  lea    0x50(%rsp),%rbx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strl@plt>
  movzbl 0xf(%rsp),%edi
  mov    $0x0,%ecx
  mov    $0x0,%edx
L27:
  movzbl 0x10(%rsp,%rcx,1),%esi
  cmp    %dil,%sil
  je     L32
  mov    %sil,0x90(%rsp,%rcx,1)
  add    $0x1,%edx
  add    $0x1,%rcx
  cmp    $0x32,%edx
  jne    L27
  mov    %edx,%r8d
  jmp    L22
L32:
  mov    %edx,%r8d
L22:
  test   %eax,%eax
  jle    L64
  movslq %edx,%rdx
  mov    %eax,%eax
  lea    (%rax,%rdx,1),%rsi
  mov    %rdx,%rax
  lea    0x50(%rsp),%rdi
  sub    %rdx,%rdi
L13:
  mov    %eax,%edx
  movzbl (%rdi,%rax,1),%ecx
  test   %cl,%cl
  je     L64
  mov    %cl,0x90(%rsp,%rax,1)
  lea    0x1(%rax),%edx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L13
L64:
  mov    $0x0,%eax
  movslq %r8d,%rdi
  lea    0x90(%rsp),%rcx
  lea    (%rcx,%rdi,1),%rsi
  movslq %edx,%rdx
  add    %rdx,%rcx
  lea    0x10(%rsp,%rdi,1),%rdi
L59:
  cmpb   $0x0,(%rsi,%rax,1)
  je     L14
  movzbl (%rdi,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    $0x32,%rax
  jne    L59
L14:
  lea    0x90(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    0xf8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  mov    $0x0,%eax
  add    $0x108,%rsp
  pop    %rbx
  pop    %rbp
  ret
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D23
type_guessed: string
name: parse_data
arguments: data_label: D29
type_guessed: string
name: parse_data
arguments: data_label: D22
type_guessed: string

tool: D23: %s


tool: D29: %c


tool: D22: %s

assistant: ```c
int main() {
  char str1[50], str2[50], ch, str[100];
  scanf("%s\n", str1);
  scanf("%c\n", &ch);
  scanf("%s\n", str2);
  int m = strl(str2);
  int i = 0, co = 0;
  while ((i < 50) && (str1[i] != ch)) {
    str[i] = str1[i];
    co = co + 1;
    i = i + 1;
  }
  int j = 0;
  while ((j < m) && (str2[j] != '\0')) {
    str[i] = str2[j];
    j = j + 1;
    i = i + 1;
  }
  int k = 0;
  while ((k < 50) && (str[co + k] != '\0')) {
    str[i] = str1[co + k];
    k = k + 1;
    i = i + 1;
  }
  printf("%s", str);
  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
  push   %r12
  lea    D29(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x100,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x50(%rsp),%rbp
  mov    %rbx,%rsi
  call   <scanf@plt>
  lea    0xf(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strl@plt>
  movzbl 0xf(%rsp),%r8d
  xor    %edx,%edx
  lea    0x90(%rsp),%rsi
  jmp    L34
  nopl   0x0(%rax)
L129:
  mov    %cl,(%rsi,%rdx,1)
  add    $0x1,%rdx
  cmp    $0x32,%rdx
  je     L118
L34:
  movzbl (%rbx,%rdx,1),%ecx
  movslq %edx,%rdi
  cmp    %r8b,%cl
  jne    L129
L94:
  test   %eax,%eax
  jle    L65
  cltq
  mov    %rbp,%r9
  lea    (%rax,%rdi,1),%r8
  sub    %rdi,%r9
  mov    %rdi,%rax
  jmp    L122
  xchg   %ax,%ax
L13:
  mov    %cl,(%rsi,%rax,1)
  lea    0x1(%rax),%edx
  add    $0x1,%rax
  cmp    %r8,%rax
  je     L89
L122:
  movzbl (%r9,%rax,1),%ecx
  mov    %eax,%edx
  test   %cl,%cl
  jne    L13
L89:
  movslq %edx,%rdx
  xor    %eax,%eax
  lea    (%rsi,%rdi,1),%r8
  add    %rdi,%rbx
  add    %rsi,%rdx
  jmp    L81
  nopl   0x0(%rax,%rax,1)
L70:
  movzbl (%rbx,%rax,1),%ecx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    $0x32,%rax
  je     L128
L81:
  cmpb   $0x0,(%r8,%rax,1)
  jne    L70
L128:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0xf8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x100,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L118:
  mov    $0x32,%edi
  jmp    L94
L65:
  movslq %edi,%rdi
  mov    %rdi,%rdx
  jmp    L89
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D29
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D8
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D19
type_guessed: string

tool: D29: %s


tool: D8: %c


tool: D19: %s

assistant: ```c
int main() {
  char str1[50], str2[50], ch, str[100];
  scanf("%s\n", str1);
  scanf("%c\n", &ch);
  scanf("%s\n", str2);
  int m = strl(str2);
  int i = 0, co = 0;
  while ((i < 50) && (str1[i] != ch)) {
    str[i] = str1[i];
    co = co + 1;
    i = i + 1;
  }
  int j = 0;
  while ((j < m) && (str2[j] != '\0')) {
    str[i] = str2[j];
    j = j + 1;
    i = i + 1;
  }
  int k = 0;
  while ((k < 50) && (str[co + k] != '\0')) {
    str[i] = str1[co + k];
    k = k + 1;
    i = i + 1;
  }
  printf("%s", str);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D17(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x100,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x50(%rsp),%rbp
  mov    %rbx,%rsi
  call   <scanf@plt>
  lea    0xf(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strl@plt>
  movzbl 0xf(%rsp),%r8d
  xor    %edx,%edx
  lea    0x90(%rsp),%rsi
  jmp    L121
  nopl   0x0(%rax)
L80:
  mov    %cl,(%rsi,%rdx,1)
  add    $0x1,%rdx
  cmp    $0x32,%rdx
  je     L55
L121:
  movzbl (%rbx,%rdx,1),%ecx
  movslq %edx,%rdi
  cmp    %r8b,%cl
  jne    L80
L65:
  test   %eax,%eax
  jle    L22
  cltq
  mov    %rbp,%r9
  lea    (%rax,%rdi,1),%r8
  sub    %rdi,%r9
  mov    %rdi,%rax
  jmp    L100
  xchg   %ax,%ax
L17:
  mov    %cl,(%rsi,%rax,1)
  lea    0x1(%rax),%edx
  add    $0x1,%rax
  cmp    %r8,%rax
  je     L99
L100:
  movzbl (%r9,%rax,1),%ecx
  mov    %eax,%edx
  test   %cl,%cl
  jne    L17
L99:
  movslq %edx,%rdx
  xor    %eax,%eax
  lea    (%rsi,%rdi,1),%r8
  add    %rdi,%rbx
  add    %rsi,%rdx
  jmp    L114
  nopl   0x0(%rax,%rax,1)
L113:
  movzbl (%rbx,%rax,1),%ecx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    $0x32,%rax
  je     L28
L114:
  cmpb   $0x0,(%r8,%rax,1)
  jne    L113
L28:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0xf8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L79
  add    $0x100,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L55:
  mov    $0x32,%edi
  jmp    L65
L22:
  movslq %edi,%rdi
  mov    %rdi,%rdx
  jmp    L99
L79:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D17
type: string
name: read_data
arguments: tag: D7
type: string
name: read_data
arguments: tag: D10
type: string

tool: D17: %s


tool: D7: %c


tool: D10: %s

assistant: ```c
int main() {
  char str1[50], str2[50], ch, str[100];
  scanf("%s\n", str1);
  scanf("%c\n", &ch);
  scanf("%s\n", str2);
  int m = strl(str2);
  int i = 0, co = 0;
  while ((i < 50) && (str1[i] != ch)) {
    str[i] = str1[i];
    co = co + 1;
    i = i + 1;
  }
  int j = 0;
  while ((j < m) && (str2[j] != '\0')) {
    str[i] = str2[j];
    j = j + 1;
    i = i + 1;
  }
  int k = 0;
  while ((k < 50) && (str[co + k] != '\0')) {
    str[i] = str1[co + k];
    k = k + 1;
    i = i + 1;
  }
  printf("%s", str);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bc_iso4217_numeric>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0xb2,-0x8(%rbp)
  jmp    L56
L42:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rdx
  lea    D2(%rip),%rax
  movzwl (%rdx,%rax,1),%eax
  cwtl
  cmp    %eax,-0x14(%rbp)
  jle    L60
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L56
L60:
  mov    -0x4(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rdx
  lea    D2(%rip),%rax
  movzwl (%rdx,%rax,1),%eax
  cwtl
  cmp    %eax,-0x14(%rbp)
  jge    L20
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L56
L20:
  mov    $0x1,%eax
  jmp    L39
L56:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L42
  mov    $0x0,%eax
L39:
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type_infered: string

tool: D2: 

assistant: ```c
int bc_iso4217_numeric(int cc) {
  static const short codes[179] = {
      8, 12, 32, 36, 44, 48, 50, 51, 52, 60, 64, 68, 72, 84, 90, 96, 104, 108, 116, 124, 132, 136, 144, 152, 156, 170, 174, 188, 191, 192, 203, 208, 214, 222, 230, 232, 238, 242, 262, 270, 292, 320, 324, 328, 332, 340, 344, 348, 352, 356, 360, 364, 368, 376, 388, 392, 398, 400, 404, 408, 410, 414, 417, 418, 422, 426, 430, 434, 446, 454, 458, 462, 480, 484, 496, 498, 504, 512, 516, 524, 532, 533, 548, 554, 558, 566, 578, 586, 590, 598, 600, 604, 608, 634, 643, 646, 654, 682, 690, 694, 702, 704, 706, 710, 728, 748, 752, 756, 760, 764, 776, 780, 784, 788, 800, 807, 818, 826, 834, 840, 858, 860, 882, 886, 901, 927, 928, 929, 930, 931, 932, 933, 934, 936, 938, 940, 941, 943, 944, 946, 947, 948, 949, 950, 951, 952, 953, 955, 956, 957, 958, 959, 960, 961, 962, 963, 964, 965, 967, 968, 969, 970, 971, 972, 973, 975, 976, 977, 978, 979, 980, 981, 984, 985, 986, 990, 994, 997, 999,
  };
  int s = 0, e = ((int)(sizeof(codes) / sizeof((codes)[0]))) - 1;
  while (s <= e) {
    int m = (s + e) / 2;
    if (codes[m] < cc) {
      s = m + 1;
    } else if (codes[m] > cc) {
      e = m - 1;
    } else {
      return 1;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<bc_iso4217_numeric>:
  endbr64
  mov    $0xb2,%esi
  mov    $0x0,%ecx
  lea    D0(%rip),%r8
  jmp    L30
L27:
  jle    L6
  lea    -0x1(%rax),%esi
L48:
  cmp    %esi,%ecx
  jg     L34
L30:
  lea    (%rcx,%rsi,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  movslq %eax,%rdx
  movswl (%r8,%rdx,2),%edx
  cmp    %edi,%edx
  jge    L27
  lea    0x1(%rax),%ecx
  jmp    L48
L34:
  mov    $0x0,%eax
  ret
L6:
  mov    $0x1,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
data_type: string

tool: D0: 

assistant: ```c
int bc_iso4217_numeric(int cc) {
  static const short codes[179] = {
      8, 12, 32, 36, 44, 48, 50, 51, 52, 60, 64, 68, 72, 84, 90, 96, 104, 108, 116, 124, 132, 136, 144, 152, 156, 170, 174, 188, 191, 192, 203, 208, 214, 222, 230, 232, 238, 242, 262, 270, 292, 320, 324, 328, 332, 340, 344, 348, 352, 356, 360, 364, 368, 376, 388, 392, 398, 400, 404, 408, 410, 414, 417, 418, 422, 426, 430, 434, 446, 454, 458, 462, 480, 484, 496, 498, 504, 512, 516, 524, 532, 533, 548, 554, 558, 566, 578, 586, 590, 598, 600, 604, 608, 634, 643, 646, 654, 682, 690, 694, 702, 704, 706, 710, 728, 748, 752, 756, 760, 764, 776, 780, 784, 788, 800, 807, 818, 826, 834, 840, 858, 860, 882, 886, 901, 927, 928, 929, 930, 931, 932, 933, 934, 936, 938, 940, 941, 943, 944, 946, 947, 948, 949, 950, 951, 952, 953, 955, 956, 957, 958, 959, 960, 961, 962, 963, 964, 965, 967, 968, 969, 970, 971, 972, 973, 975, 976, 977, 978, 979, 980, 981, 984, 985, 986, 990, 994, 997, 999,
  };
  int s = 0, e = ((int)(sizeof(codes) / sizeof((codes)[0]))) - 1;
  while (s <= e) {
    int m = (s + e) / 2;
    if (codes[m] < cc) {
      s = m + 1;
    } else if (codes[m] > cc) {
      e = m - 1;
    } else {
      return 1;
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bc_iso4217_numeric>:
  endbr64
  mov    $0x256,%edx
  mov    $0xb2,%esi
  xor    %ecx,%ecx
  mov    $0xb2,%eax
  lea    D7(%rip),%r8
  jmp    L46
  xchg   %ax,%ax
L21:
  lea    0x1(%rax),%ecx
  cmp    %esi,%ecx
  jg     L7
L9:
  lea    (%rcx,%rsi,1),%eax
  mov    %eax,%edx
  sar    %edx
  movslq %edx,%rdx
  movswl (%r8,%rdx,2),%edx
L46:
  sar    %eax
  cmp    %edi,%edx
  jl     L21
  jle    L47
  lea    -0x1(%rax),%esi
  cmp    %esi,%ecx
  jle    L9
L7:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L47:
  mov    $0x1,%eax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type: string

tool: D7: 

assistant: ```c
int bc_iso4217_numeric(int cc) {
  static const short codes[179] = {
      8, 12, 32, 36, 44, 48, 50, 51, 52, 60, 64, 68, 72, 84, 90, 96, 104, 108, 116, 124, 132, 136, 144, 152, 156, 170, 174, 188, 191, 192, 203, 208, 214, 222, 230, 232, 238, 242, 262, 270, 292, 320, 324, 328, 332, 340, 344, 348, 352, 356, 360, 364, 368, 376, 388, 392, 398, 400, 404, 408, 410, 414, 417, 418, 422, 426, 430, 434, 446, 454, 458, 462, 480, 484, 496, 498, 504, 512, 516, 524, 532, 533, 548, 554, 558, 566, 578, 586, 590, 598, 600, 604, 608, 634, 643, 646, 654, 682, 690, 694, 702, 704, 706, 710, 728, 748, 752, 756, 760, 764, 776, 780, 784, 788, 800, 807, 818, 826, 834, 840, 858, 860, 882, 886, 901, 927, 928, 929, 930, 931, 932, 933, 934, 936, 938, 940, 941, 943, 944, 946, 947, 948, 949, 950, 951, 952, 953, 955, 956, 957, 958, 959, 960, 961, 962, 963, 964, 965, 967, 968, 969, 970, 971, 972, 973, 975, 976, 977, 978, 979, 980, 981, 984, 985, 986, 990, 994, 997, 999,
  };
  int s = 0, e = ((int)(sizeof(codes) / sizeof((codes)[0]))) - 1;
  while (s <= e) {
    int m = (s + e) / 2;
    if (codes[m] < cc) {
      s = m + 1;
    } else if (codes[m] > cc) {
      e = m - 1;
    } else {
      return 1;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<bc_iso4217_numeric>:
  endbr64
  mov    $0x256,%edx
  mov    $0xb2,%esi
  xor    %ecx,%ecx
  mov    $0xb2,%eax
  lea    D4(%rip),%r8
  jmp    L28
  xchg   %ax,%ax
L41:
  lea    0x1(%rax),%ecx
  cmp    %esi,%ecx
  jg     L2
L34:
  lea    (%rcx,%rsi,1),%eax
  mov    %eax,%edx
  sar    %edx
  movslq %edx,%rdx
  movswl (%r8,%rdx,2),%edx
L28:
  sar    %eax
  cmp    %edi,%edx
  jl     L41
  jle    L13
  lea    -0x1(%rax),%esi
  cmp    %esi,%ecx
  jle    L34
L2:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  mov    $0x1,%eax
  ret
```

tool_calls: name: read_data
arguments: tag: D4
type_guessed: string

tool: D4: 

assistant: ```c
int bc_iso4217_numeric(int cc) {
  static const short codes[179] = {
      8, 12, 32, 36, 44, 48, 50, 51, 52, 60, 64, 68, 72, 84, 90, 96, 104, 108, 116, 124, 132, 136, 144, 152, 156, 170, 174, 188, 191, 192, 203, 208, 214, 222, 230, 232, 238, 242, 262, 270, 292, 320, 324, 328, 332, 340, 344, 348, 352, 356, 360, 364, 368, 376, 388, 392, 398, 400, 404, 408, 410, 414, 417, 418, 422, 426, 430, 434, 446, 454, 458, 462, 480, 484, 496, 498, 504, 512, 516, 524, 532, 533, 548, 554, 558, 566, 578, 586, 590, 598, 600, 604, 608, 634, 643, 646, 654, 682, 690, 694, 702, 704, 706, 710, 728, 748, 752, 756, 760, 764, 776, 780, 784, 788, 800, 807, 818, 826, 834, 840, 858, 860, 882, 886, 901, 927, 928, 929, 930, 931, 932, 933, 934, 936, 938, 940, 941, 943, 944, 946, 947, 948, 949, 950, 951, 952, 953, 955, 956, 957, 958, 959, 960, 961, 962, 963, 964, 965, 967, 968, 969, 970, 971, 972, 973, 975, 976, 977, 978, 979, 980, 981, 984, 985, 986, 990, 994, 997, 999,
  };
  int s = 0, e = ((int)(sizeof(codes) / sizeof((codes)[0]))) - 1;
  while (s <= e) {
    int m = (s + e) / 2;
    if (codes[m] < cc) {
      s = m + 1;
    } else if (codes[m] > cc) {
      e = m - 1;
    } else {
      return 1;
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<add_base_qscores>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  je     L39
  addl   $0x1,-0x10(%rbp)
  jmp    L42
L30:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x10(%rbp)
  jge    L55
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
L55:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x8(%rbp)
L42:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L30
  jmp    L32
L0:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x8(%rbp)
L39:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L0
L32:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void add_base_qscores(char *s, int base, int cap) {
  if (cap) {
    cap = cap + 1;
    while (*s) {
      if (*s > cap)
        *s = cap;
      *s = *s + base - 1;
      ++s;
    }
  } else
    while (*s) {
      *s = *s + base - 1;
      ++s;
    }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<add_base_qscores>:
  endbr64
  test   %edx,%edx
  jne    L7
  movzbl (%rdi),%eax
  test   %al,%al
  je     L23
L61:
  lea    -0x1(%rax,%rsi,1),%eax
  mov    %al,(%rdi)
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L61
L63:
  ret
L23:
  ret
L7:
  add    $0x1,%edx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L63
  sub    $0x1,%esi
  jmp    L53
L30:
  add    %sil,(%rdi)
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L79
L53:
  movsbl %al,%eax
  cmp    %edx,%eax
  jle    L30
  mov    %dl,(%rdi)
  jmp    L30
L79:
  ret
```

assistant: ```c
void add_base_qscores(char *s, int base, int cap) {
  if (cap) {
    cap = cap + 1;
    while (*s) {
      if (*s > cap)
        *s = cap;
      *s = *s + base - 1;
      ++s;
    }
  } else
    while (*s) {
      *s = *s + base - 1;
      ++s;
    }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<add_base_qscores>:
  endbr64
  movsbl (%rdi),%eax
  mov    %rdi,%rcx
  test   %edx,%edx
  jne    L38
  test   %al,%al
  je     L10
  nopw   0x0(%rax,%rax,1)
L6:
  lea    -0x1(%rax,%rsi,1),%eax
  add    $0x1,%rcx
  mov    %al,-0x1(%rcx)
  movzbl (%rcx),%eax
  test   %al,%al
  jne    L6
L39:
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  add    $0x1,%edx
  test   %al,%al
  je     L39
  sub    $0x1,%esi
  mov    %edx,%r8d
  nopl   (%rax)
L1:
  mov    %r8d,%edi
  cmp    %edx,%eax
  jg     L3
  movzbl (%rcx),%edi
L3:
  add    %esi,%edi
  add    $0x1,%rcx
  mov    %dil,-0x1(%rcx)
  movsbl (%rcx),%eax
  test   %al,%al
  jne    L1
  ret
  nopl   0x0(%rax)
L10:
  ret
```

assistant: ```c
void add_base_qscores(char *s, int base, int cap) {
  if (cap) {
    cap = cap + 1;
    while (*s) {
      if (*s > cap)
        *s = cap;
      *s = *s + base - 1;
      ++s;
    }
  } else
    while (*s) {
      *s = *s + base - 1;
      ++s;
    }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<add_base_qscores>:
  endbr64
  movzbl (%rdi),%eax
  test   %edx,%edx
  jne    L25
  test   %al,%al
  je     L2
  nop
L31:
  lea    -0x1(%rsi,%rax,1),%eax
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L31
L58:
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  add    $0x1,%edx
  test   %al,%al
  je     L58
  sub    $0x1,%esi
  lea    (%rsi,%rdx,1),%r8d
  cs nopw 0x0(%rax,%rax,1)
L33:
  movsbl %al,%ecx
  cmp    %ecx,%edx
  jge    L6
  mov    %r8b,(%rdi)
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L33
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  add    %esi,%eax
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L33
  ret
  nopl   0x0(%rax)
L2:
  ret
```

assistant: ```c
void add_base_qscores(char *s, int base, int cap) {
  if (cap) {
    cap = cap + 1;
    while (*s) {
      if (*s > cap)
        *s = cap;
      *s = *s + base - 1;
      ++s;
    }
  } else
    while (*s) {
      *s = *s + base - 1;
      ++s;
    }
}
```