ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<load_ranking>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  cmpq   $0x0,(%rdx)
  je     L63
L89:
  lea    D14(%rip),%rsi
  mov    %rbp,%rdi
  call   <fopen@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L62
L16:
  xor    %eax,%eax
  mov    $0xa,%r12d
  mov    %rsp,%rbp
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L85:
  lea    (%rdi,%rdx,8),%r15
  mov    %rbp,%rdi
  call   <atof@plt>
  mov    (%rbx),%eax
  movsd  %xmm0,(%r15)
  add    $0x1,%eax
L33:
  mov    %eax,(%rbx)
  mov    %r14,%rdx
  mov    $0xff,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L87
  movslq (%rbx),%rdx
  mov    0x0(%r13),%rdi
  cmp    %r12d,%edx
  jne    L85
  add    $0xa,%r12d
  lea    0x0(,%r12,8),%esi
  call   <realloc@plt>
  movslq (%rbx),%rdx
  mov    %rax,0x0(%r13)
  mov    %rax,%rdi
  jmp    L85
  xchg   %ax,%ax
L87:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L63:
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  jmp    L89
  nopl   0x0(%rax,%rax,1)
L62:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L16
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
type_infered: string
name: extract_symbolic_data
arguments: tag: D11
type_infered: string

tool: D14: r

tool: D11: Unable to open file: %s!!! Exiting


assistant: ```c
void load_ranking(char *fname, int *N, double **R) {
  char buff[256];
  int size = 10;
  FILE *f;
  if (*R == NULL) {
    *R = malloc(size * sizeof(double));
  }
  f = fopen(fname, "r");
  if (!f) {
    printf("Unable to open file: %s!!! Exiting\n");
    exit(1);
  }
  *N = 0;
  while (fgets(buff, 255, f)) {
    if (*N == size) {
      size += 10;
      *R = realloc(*R, size * sizeof(double));
    }
    (*R)[*N] = atof(buff);
    *N += 1;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<load_ranking>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  cmpq   $0x0,(%rdx)
  je     L59
L44:
  lea    D6(%rip),%rsi
  mov    %rbp,%rdi
  call   <fopen@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L28
L46:
  xor    %eax,%eax
  mov    $0xa,%r12d
  mov    %rsp,%rbp
  jmp    L57
  nopl   0x0(%rax,%rax,1)
L5:
  lea    (%rdi,%rdx,8),%r15
  mov    %rbp,%rdi
  call   <atof@plt>
  mov    (%rbx),%eax
  movsd  %xmm0,(%r15)
  add    $0x1,%eax
L57:
  mov    %eax,(%rbx)
  mov    %r14,%rdx
  mov    $0xff,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L33
  movslq (%rbx),%rdx
  mov    0x0(%r13),%rdi
  cmp    %r12d,%edx
  jne    L5
  add    $0xa,%r12d
  lea    0x0(,%r12,8),%esi
  call   <realloc@plt>
  movslq (%rbx),%rdx
  mov    %rax,0x0(%r13)
  mov    %rax,%rdi
  jmp    L5
  xchg   %ax,%ax
L33:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L59:
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L28:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L46
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D6
data_type: string
name: retrieve_data
arguments: data_label: D15
data_type: string

tool: D6: r

tool: D15: Unable to open file: %s!!! Exiting


assistant: ```c
void load_ranking(char *fname, int *N, double **R) {
  char buff[256];
  int size = 10;
  FILE *f;
  if (*R == NULL) {
    *R = malloc(size * sizeof(double));
  }
  f = fopen(fname, "r");
  if (!f) {
    printf("Unable to open file: %s!!! Exiting\n");
    exit(1);
  }
  *N = 0;
  while (fgets(buff, 255, f)) {
    if (*N == size) {
      size += 10;
      *R = realloc(*R, size * sizeof(double));
    }
    (*R)[*N] = atof(buff);
    *N += 1;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<getLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L29
L27:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
L29:
  subl   $0x1,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L36
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L36
  cmpl   $0xa,-0x8(%rbp)
  jne    L27
L36:
  cmpl   $0xa,-0x8(%rbp)
  jne    L47
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
L47:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    D10(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <fflush@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getLine(char s[], int maxLength) {
  int c, i = 0;
  while ((--maxLength > 0) && ((c = getchar()) != EOF) && (c != '\n'))
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  fflush(stdin);
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getLine>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  cmp    $0x1,%esi
  jle    L9
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%r12d
  mov    $0x0,%ebx
L38:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D5(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L43
  cmp    $0xa,%eax
  je     L43
  add    $0x1,%ebx
  mov    %al,0x0(%rbp)
  add    $0x1,%rbp
  cmp    %r12d,%ebx
  jne    L38
  mov    %r12d,%ebx
  jmp    L12
L9:
  mov    $0x0,%ebx
L43:
  cmp    $0xa,%eax
  je     L37
L12:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    D16(%rip),%rax
  mov    (%rax),%edi
  call   <fflush@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L37:
  movslq %ebx,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  lea    0x1(%rbx),%ebx
  jmp    L12
```

assistant: ```c
int getLine(char s[], int maxLength) {
  int c, i = 0;
  while ((--maxLength > 0) && ((c = getchar()) != EOF) && (c != '\n'))
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  fflush(stdin);
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<getLine>:
  endbr64
  push   %r14
  mov    %rdi,%rdx
  mov    %rdi,%r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L65
  mov    D6(%rip),%r13
  mov    %rdi,%rbx
  xor    %r12d,%r12d
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L53:
  add    $0x1,%r12d
  cmp    $0xa,%eax
  je     L32
  mov    %al,(%rdx)
  lea    0x1(%rdx),%rbx
  cmp    %ebp,%r12d
  je     L72
L2:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rdx
  cmp    %eax,0x0(%r13)
  jne    L53
  cmp    $0xa,%eax
  je     L43
L65:
  mov    D10(%rip),%rax
  movb   $0x0,(%rdx)
  mov    (%rax),%edi
  call   <fflush@plt>
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L43:
  add    $0x1,%r12d
  nopw   0x0(%rax,%rax,1)
L32:
  movslq %r12d,%rdx
  movb   $0xa,(%rbx)
  add    %r14,%rdx
  jmp    L65
  nopl   0x0(%rax,%rax,1)
L72:
  movslq %ebp,%rdx
  mov    %ebp,%r12d
  add    %r14,%rdx
  jmp    L65
```

assistant: ```c
int getLine(char s[], int maxLength) {
  int c, i = 0;
  while ((--maxLength > 0) && ((c = getchar()) != EOF) && (c != '\n'))
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  fflush(stdin);
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getLine>:
  endbr64
  push   %r14
  mov    %rdi,%rdx
  mov    %rdi,%r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L0
  mov    D2(%rip),%r13
  mov    %rdi,%rbx
  xor    %r12d,%r12d
  jmp    L69
  nopl   0x0(%rax,%rax,1)
L50:
  add    $0x1,%r12d
  cmp    $0xa,%eax
  je     L77
  mov    %al,(%rdx)
  lea    0x1(%rdx),%rbx
  cmp    %ebp,%r12d
  je     L75
L69:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rdx
  cmp    %eax,0x0(%r13)
  jne    L50
  cmp    $0xa,%eax
  je     L24
L0:
  mov    D16(%rip),%rax
  movb   $0x0,(%rdx)
  mov    (%rax),%edi
  call   <fflush@plt>
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L24:
  add    $0x1,%r12d
  nopw   0x0(%rax,%rax,1)
L77:
  movslq %r12d,%rdx
  movb   $0xa,(%rbx)
  add    %r14,%rdx
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L75:
  movslq %ebp,%rdx
  mov    %ebp,%r12d
  add    %r14,%rdx
  jmp    L0
```

assistant: ```c
int getLine(char s[], int maxLength) {
  int c, i = 0;
  while ((--maxLength > 0) && ((c = getchar()) != EOF) && (c != '\n'))
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  fflush(stdin);
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<puissance4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <menuPuissance4@plt>
  mov    $0x0,%eax
  call   <initPuissance4@plt>
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
L6:
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  cmpl   $0x0,-0x14(%rbp)
  jne    L64
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <IAPuissance4@plt>
  jmp    L34
L64:
  cmpl   $0x1,-0x14(%rbp)
  jne    L34
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <jouerCoupHumainPuissance4@plt>
L34:
  mov    $0x1,%edi
  call   <gagnerPuissance4@plt>
  test   %eax,%eax
  jne    L5
  mov    $0x0,%eax
  call   <puissance4pasRempli@plt>
  test   %rax,%rax
  je     L5
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  cmpl   $0x0,-0x14(%rbp)
  jne    L45
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <jouerCoupHumainPuissance4@plt>
  jmp    L84
L45:
  cmpl   $0x1,-0x14(%rbp)
  jne    L84
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x1,%edi
  call   <jouerCoupHumainPuissance4@plt>
L84:
  addl   $0x1,-0xc(%rbp)
L5:
  mov    $0x0,%edi
  call   <gagnerPuissance4@plt>
  test   %eax,%eax
  jne    L71
  mov    $0x0,%eax
  call   <puissance4pasRempli@plt>
  test   %rax,%rax
  jne    L6
L71:
  cmpl   $0x0,-0x14(%rbp)
  jne    L116
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <afficherGagnantPuissance4Solo@plt>
  jmp    L52
L116:
  cmpl   $0x1,-0x14(%rbp)
  jne    L52
  mov    $0x0,%eax
  call   <afficherGagnantPuissance4Multi@plt>
L52:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D22
datatype: string

tool: D22: clear

assistant: ```c
void puissance4(int modeDeJeux) {
  int nbCoups;
  int niveauDifficulte;
  nbCoups = 0;
  menuPuissance4(modeDeJeux, &niveauDifficulte);
  initPuissance4();
  system("clear");
  do {
    system("clear");
    if (modeDeJeux == 0)
      IAPuissance4(niveauDifficulte);
    else if (modeDeJeux == 1)
      jouerCoupHumainPuissance4(2, modeDeJeux);
    if (!gagnerPuissance4(1) && puissance4pasRempli()) {
      system("clear");
      if (modeDeJeux == 0)
        jouerCoupHumainPuissance4(2, modeDeJeux);
      else if (modeDeJeux == 1)
        jouerCoupHumainPuissance4(1, modeDeJeux);
      nbCoups++;
    }
  } while (!gagnerPuissance4(0) && puissance4pasRempli());
  if (modeDeJeux == 0)
    afficherGagnantPuissance4Solo(nbCoups, modeDeJeux, niveauDifficulte);
  else if (modeDeJeux == 1)
    afficherGagnantPuissance4Multi();
}
```
user: Translate the below assembly code into C programming language:

```asm
<puissance4>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <menuPuissance4@plt>
  mov    $0x0,%eax
  call   <initPuissance4@plt>
  lea    D12(%rip),%rdi
  call   <system@plt>
  mov    $0x0,%r12d
  lea    D12(%rip),%rbp
  jmp    L156
L1:
  mov    0x4(%rsp),%edi
  call   <IAPuissance4@plt>
L137:
  mov    $0x1,%edi
  call   <gagnerPuissance4@plt>
  test   %eax,%eax
  je     L20
L171:
  mov    $0x0,%edi
  call   <gagnerPuissance4@plt>
  test   %eax,%eax
  jne    L10
  call   <puissance4pasRempli@plt>
  test   %rax,%rax
  je     L10
L156:
  mov    %rbp,%rdi
  call   <system@plt>
  test   %ebx,%ebx
  je     L1
  cmp    $0x1,%ebx
  jne    L137
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <jouerCoupHumainPuissance4@plt>
  jmp    L137
L20:
  call   <puissance4pasRempli@plt>
  test   %rax,%rax
  je     L171
  mov    %rbp,%rdi
  call   <system@plt>
  test   %ebx,%ebx
  je     L178
  cmp    $0x1,%ebx
  je     L118
L57:
  add    $0x1,%r12d
  jmp    L171
L178:
  mov    $0x0,%esi
  mov    $0x2,%edi
  call   <jouerCoupHumainPuissance4@plt>
  jmp    L57
L118:
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <jouerCoupHumainPuissance4@plt>
  jmp    L57
L10:
  test   %ebx,%ebx
  je     L3
  cmp    $0x1,%ebx
  je     L91
L169:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L172
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  mov    0x4(%rsp),%edx
  mov    $0x0,%esi
  mov    %r12d,%edi
  call   <afficherGagnantPuissance4Solo@plt>
  jmp    L169
L91:
  mov    $0x0,%eax
  call   <afficherGagnantPuissance4Multi@plt>
  jmp    L169
L172:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D12
datatype: string

tool: D12: clear

assistant: ```c
void puissance4(int modeDeJeux) {
  int nbCoups;
  int niveauDifficulte;
  nbCoups = 0;
  menuPuissance4(modeDeJeux, &niveauDifficulte);
  initPuissance4();
  system("clear");
  do {
    system("clear");
    if (modeDeJeux == 0)
      IAPuissance4(niveauDifficulte);
    else if (modeDeJeux == 1)
      jouerCoupHumainPuissance4(2, modeDeJeux);
    if (!gagnerPuissance4(1) && puissance4pasRempli()) {
      system("clear");
      if (modeDeJeux == 0)
        jouerCoupHumainPuissance4(2, modeDeJeux);
      else if (modeDeJeux == 1)
        jouerCoupHumainPuissance4(1, modeDeJeux);
      nbCoups++;
    }
  } while (!gagnerPuissance4(0) && puissance4pasRempli());
  if (modeDeJeux == 0)
    afficherGagnantPuissance4Solo(nbCoups, modeDeJeux, niveauDifficulte);
  else if (modeDeJeux == 1)
    afficherGagnantPuissance4Multi();
}
```
user: What is the c source code of the assembly code below:

```asm
<puissance4>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <menuPuissance4@plt>
  xor    %eax,%eax
  call   <initPuissance4@plt>
  mov    %rbp,%rdi
  call   <system@plt>
  jmp    L153
  nopl   0x0(%rax,%rax,1)
L15:
  cmp    $0x1,%ebx
  je     L116
L137:
  mov    $0x1,%edi
  call   <gagnerPuissance4@plt>
  test   %eax,%eax
  je     L114
L150:
  xor    %edi,%edi
  call   <gagnerPuissance4@plt>
  test   %eax,%eax
  jne    L25
L85:
  call   <puissance4pasRempli@plt>
  test   %rax,%rax
  je     L25
L153:
  mov    %rbp,%rdi
  call   <system@plt>
  test   %ebx,%ebx
  jne    L15
  mov    0x4(%rsp),%edi
  call   <IAPuissance4@plt>
  mov    $0x1,%edi
  call   <gagnerPuissance4@plt>
  test   %eax,%eax
  jne    L150
L114:
  call   <puissance4pasRempli@plt>
  test   %rax,%rax
  je     L150
  mov    %rbp,%rdi
  call   <system@plt>
  test   %ebx,%ebx
  je     L92
  cmp    $0x1,%ebx
  je     L145
L140:
  xor    %edi,%edi
  add    $0x1,%r12d
  call   <gagnerPuissance4@plt>
  test   %eax,%eax
  je     L85
  nopl   0x0(%rax)
L25:
  test   %ebx,%ebx
  je     L159
  cmp    $0x1,%ebx
  je     L61
L122:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L116:
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <jouerCoupHumainPuissance4@plt>
  jmp    L137
  nopl   0x0(%rax)
L92:
  xor    %esi,%esi
  mov    $0x2,%edi
  call   <jouerCoupHumainPuissance4@plt>
  jmp    L140
  cs nopw 0x0(%rax,%rax,1)
L159:
  mov    0x4(%rsp),%edx
  xor    %esi,%esi
  mov    %r12d,%edi
  call   <afficherGagnantPuissance4Solo@plt>
  jmp    L122
L145:
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <jouerCoupHumainPuissance4@plt>
  jmp    L140
  nopl   0x0(%rax)
L61:
  xor    %eax,%eax
  call   <afficherGagnantPuissance4Multi@plt>
  jmp    L122
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D2
datatype: string

tool: D2: clear

assistant: ```c
void puissance4(int modeDeJeux) {
  int nbCoups;
  int niveauDifficulte;
  nbCoups = 0;
  menuPuissance4(modeDeJeux, &niveauDifficulte);
  initPuissance4();
  system("clear");
  do {
    system("clear");
    if (modeDeJeux == 0)
      IAPuissance4(niveauDifficulte);
    else if (modeDeJeux == 1)
      jouerCoupHumainPuissance4(2, modeDeJeux);
    if (!gagnerPuissance4(1) && puissance4pasRempli()) {
      system("clear");
      if (modeDeJeux == 0)
        jouerCoupHumainPuissance4(2, modeDeJeux);
      else if (modeDeJeux == 1)
        jouerCoupHumainPuissance4(1, modeDeJeux);
      nbCoups++;
    }
  } while (!gagnerPuissance4(0) && puissance4pasRempli());
  if (modeDeJeux == 0)
    afficherGagnantPuissance4Solo(nbCoups, modeDeJeux, niveauDifficulte);
  else if (modeDeJeux == 1)
    afficherGagnantPuissance4Multi();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parser_skip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <parser_skip_whitespace@plt>
  test   %rax,%rax
  je     L3
  mov    $0xffffffff,%eax
  jmp    L21
L3:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <mls@plt>
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jne    L31
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  jmp    L21
L31:
  mov    $0xffffffff,%eax
L21:
  leave
  ret
```

assistant: ```c
int parser_skip(int buf, int *p, int skip_char) {
  int c;
  if (parser_skip_whitespace(buf, p))
    return -1;
  c = (*(char *)mls((buf), (*p)));
  if (c == skip_char) {
    (*p)++;
    return 0;
  }
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<parser_skip>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%rbx
  mov    %edx,%r12d
  call   <parser_skip_whitespace@plt>
  test   %rax,%rax
  jne    L22
  mov    (%rbx),%esi
  mov    %ebp,%edi
  call   <mls@plt>
  movsbl (%rax),%eax
  cmp    %r12d,%eax
  jne    L31
  addl   $0x1,(%rbx)
  mov    $0x0,%eax
L35:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  mov    $0xffffffff,%eax
  jmp    L35
L31:
  mov    $0xffffffff,%eax
  jmp    L35
```

assistant: ```c
int parser_skip(int buf, int *p, int skip_char) {
  int c;
  if (parser_skip_whitespace(buf, p))
    return -1;
  c = (*(char *)mls((buf), (*p)));
  if (c == skip_char) {
    (*p)++;
    return 0;
  }
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parser_skip>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  call   <parser_skip_whitespace@plt>
  test   %rax,%rax
  jne    L13
  mov    (%rbx),%esi
  mov    %ebp,%edi
  call   <mls@plt>
  movsbl (%rax),%eax
  cmp    %r12d,%eax
  jne    L13
  addl   $0x1,(%rbx)
  xor    %eax,%eax
L23:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L13:
  mov    $0xffffffff,%eax
  jmp    L23
```

assistant: ```c
int parser_skip(int buf, int *p, int skip_char) {
  int c;
  if (parser_skip_whitespace(buf, p))
    return -1;
  c = (*(char *)mls((buf), (*p)));
  if (c == skip_char) {
    (*p)++;
    return 0;
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<parser_skip>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  call   <parser_skip_whitespace@plt>
  test   %rax,%rax
  jne    L20
  mov    (%rbx),%esi
  mov    %ebp,%edi
  call   <mls@plt>
  movsbl (%rax),%eax
  cmp    %r12d,%eax
  jne    L20
  addl   $0x1,(%rbx)
  xor    %eax,%eax
L21:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L20:
  mov    $0xffffffff,%eax
  jmp    L21
```

assistant: ```c
int parser_skip(int buf, int *p, int skip_char) {
  int c;
  if (parser_skip_whitespace(buf, p))
    return -1;
  c = (*(char *)mls((buf), (*p)));
  if (c == skip_char) {
    (*p)++;
    return 0;
  }
  return -1;
}
```