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
<removePrefixedStrings>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    $0x8,%edi
  call   <allocateMemory@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L28
L19:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stringStartsWith@plt>
  test   %eax,%eax
  jne    L2
  addl   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <reallocateMemory@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  sub    $0x1,%edx
  mov    %edx,%edx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%rax
  mov    %rax,(%rdx)
L2:
  addl   $0x1,-0xc(%rbp)
L28:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jb     L19
  mov    -0x20(%rbp),%rax
  mov    -0x10(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void removePrefixedStrings(char ***strings, unsigned *numStrings, const char *prefix) {
  char **stringsWithoutPrefix = allocateMemory(sizeof(char *));
  unsigned numStringsWithoutPrefix = 0;
  for (unsigned i = 0; i < *numStrings; i++) {
    if (!stringStartsWith((*strings)[i], prefix)) {
      ++numStringsWithoutPrefix;
      stringsWithoutPrefix = reallocateMemory(stringsWithoutPrefix, numStringsWithoutPrefix * sizeof(char *));
      stringsWithoutPrefix[numStringsWithoutPrefix - 1] = (*strings)[i];
    }
  }
  (*numStrings) = numStringsWithoutPrefix;
  (*strings) = stringsWithoutPrefix;
}
```
user: Translate the below assembly code into C programming language:

```asm
<removePrefixedStrings>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r14
  mov    %rdx,(%rsp)
  mov    $0x8,%edi
  call   <allocateMemory@plt>
  mov    %rax,0x8(%rsp)
  mov    (%r14),%r12d
  test   %r12d,%r12d
  je     L0
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  jmp    L28
L10:
  add    $0x1,%ebx
  cmp    %ebx,(%r14)
  jbe    L0
L28:
  mov    %ebx,%ebp
  mov    0x0(%r13),%rax
  mov    (%rax,%rbp,8),%rdi
  mov    (%rsp),%rsi
  call   <stringStartsWith@plt>
  test   %eax,%eax
  jne    L10
  lea    0x1(%r12),%r15d
  lea    0x0(,%r15,8),%esi
  mov    0x8(%rsp),%rdi
  call   <reallocateMemory@plt>
  mov    %rax,%rcx
  mov    %rax,0x8(%rsp)
  mov    0x0(%r13),%rax
  mov    (%rax,%rbp,8),%rax
  mov    %r12d,%r12d
  mov    %rax,(%rcx,%r12,8)
  mov    %r15d,%r12d
  jmp    L10
L0:
  mov    %r12d,(%r14)
  mov    0x8(%rsp),%rax
  mov    %rax,0x0(%r13)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void removePrefixedStrings(char ***strings, unsigned *numStrings, const char *prefix) {
  char **stringsWithoutPrefix = allocateMemory(sizeof(char *));
  unsigned numStringsWithoutPrefix = 0;
  for (unsigned i = 0; i < *numStrings; i++) {
    if (!stringStartsWith((*strings)[i], prefix)) {
      ++numStringsWithoutPrefix;
      stringsWithoutPrefix = reallocateMemory(stringsWithoutPrefix, numStringsWithoutPrefix * sizeof(char *));
      stringsWithoutPrefix[numStringsWithoutPrefix - 1] = (*strings)[i];
    }
  }
  (*numStrings) = numStringsWithoutPrefix;
  (*strings) = stringsWithoutPrefix;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<removePrefixedStrings>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rdi,%r14
  mov    $0x8,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %rsi,(%rsp)
  call   <allocateMemory@plt>
  mov    (%rbx),%r13d
  mov    %rax,%r12
  test   %r13d,%r13d
  je     L34
  xor    %ebx,%ebx
  xor    %r13d,%r13d
  jmp    L16
  nopl   0x0(%rax,%rax,1)
L47:
  mov    (%rsp),%rax
  add    $0x1,%ebx
  cmp    %ebx,(%rax)
  jbe    L34
L16:
  mov    (%r14),%rax
  mov    %ebx,%ebp
  mov    %r15,%rsi
  mov    (%rax,%rbp,8),%rdi
  call   <stringStartsWith@plt>
  test   %eax,%eax
  jne    L47
  lea    0x1(%r13),%edx
  mov    %r12,%rdi
  add    $0x1,%ebx
  lea    0x0(,%rdx,8),%esi
  mov    %edx,0xc(%rsp)
  call   <reallocateMemory@plt>
  mov    0xc(%rsp),%edx
  mov    %rax,%r12
  mov    (%r14),%rax
  mov    (%rax,%rbp,8),%rax
  mov    %rax,(%r12,%r13,8)
  mov    (%rsp),%rax
  mov    %edx,%r13d
  cmp    %ebx,(%rax)
  ja     L16
L34:
  mov    (%rsp),%rax
  mov    %r13d,(%rax)
  mov    %r12,(%r14)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void removePrefixedStrings(char ***strings, unsigned *numStrings, const char *prefix) {
  char **stringsWithoutPrefix = allocateMemory(sizeof(char *));
  unsigned numStringsWithoutPrefix = 0;
  for (unsigned i = 0; i < *numStrings; i++) {
    if (!stringStartsWith((*strings)[i], prefix)) {
      ++numStringsWithoutPrefix;
      stringsWithoutPrefix = reallocateMemory(stringsWithoutPrefix, numStringsWithoutPrefix * sizeof(char *));
      stringsWithoutPrefix[numStringsWithoutPrefix - 1] = (*strings)[i];
    }
  }
  (*numStrings) = numStringsWithoutPrefix;
  (*strings) = stringsWithoutPrefix;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<removePrefixedStrings>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    $0x8,%edi
  call   <allocateMemory@plt>
  mov    0x0(%r13),%r12d
  mov    %rax,%rbp
  test   %r12d,%r12d
  je     L30
  mov    (%rbx),%rax
  xor    %r14d,%r14d
  xor    %r12d,%r12d
  xchg   %ax,%ax
L0:
  mov    %r14d,%ebx
  mov    %r15,%rsi
  mov    (%rax,%rbx,8),%rdi
  call   <stringStartsWith@plt>
  test   %eax,%eax
  je     L16
  add    $0x1,%r14d
  cmp    %r14d,0x0(%r13)
  jbe    L30
  mov    (%rsp),%rax
  mov    (%rax),%rax
  jmp    L0
  cs nopw 0x0(%rax,%rax,1)
L16:
  lea    0x1(%r12),%ecx
  mov    %rbp,%rdi
  add    $0x1,%r14d
  lea    0x0(,%rcx,8),%esi
  mov    %ecx,0xc(%rsp)
  call   <reallocateMemory@plt>
  cmp    %r14d,0x0(%r13)
  mov    0xc(%rsp),%ecx
  mov    %rax,%rbp
  mov    (%rsp),%rax
  mov    (%rax),%rax
  mov    (%rax,%rbx,8),%rsi
  mov    %rsi,0x0(%rbp,%r12,8)
  jbe    L21
  mov    %ecx,%r12d
  jmp    L0
  xchg   %ax,%ax
L21:
  mov    %ecx,%r12d
L30:
  mov    (%rsp),%rax
  mov    %r12d,0x0(%r13)
  mov    %rbp,(%rax)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void removePrefixedStrings(char ***strings, unsigned *numStrings, const char *prefix) {
  char **stringsWithoutPrefix = allocateMemory(sizeof(char *));
  unsigned numStringsWithoutPrefix = 0;
  for (unsigned i = 0; i < *numStrings; i++) {
    if (!stringStartsWith((*strings)[i], prefix)) {
      ++numStringsWithoutPrefix;
      stringsWithoutPrefix = reallocateMemory(stringsWithoutPrefix, numStringsWithoutPrefix * sizeof(char *));
      stringsWithoutPrefix[numStringsWithoutPrefix - 1] = (*strings)[i];
    }
  }
  (*numStrings) = numStringsWithoutPrefix;
  (*strings) = stringsWithoutPrefix;
}
```
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
  sub    $0x10,%rsp
  movl   $0x1,-0x4(%rbp)
  jmp    L34
L80:
  cmpl   $0xb,-0x4(%rbp)
  je     L10
  cmpl   $0xb,-0x4(%rbp)
  jg     L51
  cmpl   $0x4,-0x4(%rbp)
  je     L61
  cmpl   $0x7,-0x4(%rbp)
  je     L18
  jmp    L51
L61:
  mov    $0x4,%edi
  call   <growTree@plt>
  jmp    L51
L18:
  mov    $0x7,%edi
  call   <growTree@plt>
  jmp    L51
L10:
  mov    $0xb,%edi
  call   <growTree@plt>
  nop
L51:
  addl   $0x1,-0x4(%rbp)
L34:
  cmpl   $0xb,-0x4(%rbp)
  jle    L80
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int size;
  for (size = 1; size <= 11; size++) {
    switch (size) {
    case 4:
      growTree(4);
      break;
    case 7:
      growTree(7);
      break;
    case 11:
      growTree(11);
      break;
      break;
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2,%ebx
  jmp    L42
L63:
  mov    $0x4,%edi
  call   <growTree@plt>
L40:
  add    $0x1,%ebx
L42:
  lea    -0x1(%rbx),%eax
  cmp    $0x8,%ebx
  je     L20
  cmp    $0xb,%eax
  je     L12
  cmp    $0x4,%eax
  je     L63
  cmp    $0xb,%ebx
  jle    L40
  jmp    L0
L20:
  mov    $0x7,%edi
  call   <growTree@plt>
  jmp    L40
L12:
  mov    $0xb,%edi
  call   <growTree@plt>
L0:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int size;
  for (size = 1; size <= 11; size++) {
    switch (size) {
    case 4:
      growTree(4);
      break;
    case 7:
      growTree(7);
      break;
    case 11:
      growTree(11);
      break;
      break;
    }
  }
  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   %rbx
  mov    $0x2,%ebx
  jmp    L2
  nopl   0x0(%rax)
L57:
  cmp    $0x4,%eax
  je     L68
  cmp    $0xc,%ebx
  je     L15
L14:
  add    $0x1,%ebx
L2:
  lea    -0x1(%rbx),%eax
  cmp    $0x8,%ebx
  je     L50
  cmp    $0xb,%eax
  jne    L57
  mov    $0xb,%edi
  call   <growTree@plt>
L15:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L68:
  mov    $0x4,%edi
  call   <growTree@plt>
  jmp    L14
  nopl   0x0(%rax)
L50:
  mov    $0x7,%edi
  call   <growTree@plt>
  jmp    L14
```

assistant: ```c
int main() {
  int size;
  for (size = 1; size <= 11; size++) {
    switch (size) {
    case 4:
      growTree(4);
      break;
    case 7:
      growTree(7);
      break;
    case 11:
      growTree(11);
      break;
      break;
    }
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <growTree@plt>
  mov    $0x7,%edi
  call   <growTree@plt>
  mov    $0xb,%edi
  call   <growTree@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int size;
  for (size = 1; size <= 11; size++) {
    switch (size) {
    case 4:
      growTree(4);
      break;
    case 7:
      growTree(7);
      break;
    case 11:
      growTree(11);
      break;
      break;
    }
  }
  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   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
datatype: string

tool: D6: %i


assistant: ```c
int main() {
  int x = 5;
  int *y = &x;
  printf("%i\n", *y);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D7: %i


assistant: ```c
int main() {
  int x = 5;
  int *y = &x;
  printf("%i\n", *y);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
data_type: string

tool: D1: %i


assistant: ```c
int main() {
  int x = 5;
  int *y = &x;
  printf("%i\n", *y);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D3
type: string

tool: D3: %i


assistant: ```c
int main() {
  int x = 5;
  int *y = &x;
  printf("%i\n", *y);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strindex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L6
L78:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L75:
  cmpl   $0x0,-0x4(%rbp)
  jle    L37
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L37
  mov    -0xc(%rbp),%eax
  jmp    L77
L37:
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L3
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L75
L3:
  addl   $0x1,-0xc(%rbp)
L6:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L78
  mov    $0xffffffff,%eax
L77:
  pop    %rbp
  ret
```

assistant: ```c
int strindex(char s[], char t[]) {
  int i, j, k;
  for (i = 0; s[i] != '\0'; i++) {
    for (j = i, k = 0; t[k] != '\0' && s[j] == t[k]; j++, k++)
      if (k > 0 && t[k] == '\0')
        return i;
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strindex>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L19
  movzbl (%rsi),%r8d
  mov    $0x0,%ecx
  jmp    L67
L13:
  movzbl 0x1(%rdi,%rcx,1),%eax
  add    $0x1,%rcx
  sub    $0x1,%rsi
  test   %al,%al
  je     L19
L67:
  mov    %rcx,%rax
  mov    %r8d,%edx
  test   %r8b,%r8b
  je     L13
L63:
  cmp    %dl,(%rdi,%rax,1)
  jne    L13
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L63
  jmp    L13
L19:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int strindex(char s[], char t[]) {
  int i, j, k;
  for (i = 0; s[i] != '\0'; i++) {
    for (j = i, k = 0; t[k] != '\0' && s[j] == t[k]; j++, k++)
      if (k > 0 && t[k] == '\0')
        return i;
  }
  return -1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strindex>:
  endbr64
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int strindex(char s[], char t[]) {
  int i, j, k;
  for (i = 0; s[i] != '\0'; i++) {
    for (j = i, k = 0; t[k] != '\0' && s[j] == t[k]; j++, k++)
      if (k > 0 && t[k] == '\0')
        return i;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below:

```asm
<strindex>:
  endbr64
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int strindex(char s[], char t[]) {
  int i, j, k;
  for (i = 0; s[i] != '\0'; i++) {
    for (j = i, k = 0; t[k] != '\0' && s[j] == t[k]; j++, k++)
      if (k > 0 && t[k] == '\0')
        return i;
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<roman_to_int>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbx
  call   <string_length@plt>
  test   %eax,%eax
  jle    L157
  lea    -0x1(%rax),%edx
  movslq %edx,%r12
  add    %rbx,%r12
  cltq
  lea    -0x2(%rbx,%rax,1),%rax
  mov    %edx,%edx
  sub    %rdx,%rax
  mov    %rax,0x18(%rsp)
  movl   $0x0,0x14(%rsp)
  movl   $0x0,0x10(%rsp)
  mov    $0x0,%ebp
  mov    $0xffffffff,%ebx
  mov    $0x0,%r15d
  jmp    L101
L75:
  movl   $0x1,0x10(%rsp)
L26:
  cmp    %r13d,%ebx
  jl     L172
  add    %ebx,%r15d
  movl   $0x0,0x14(%rsp)
L110:
  sub    $0x1,%r12
  cmp    0x18(%rsp),%r12
  je     L65
L101:
  mov    %bpl,0xf(%rsp)
  movzbl (%r12),%ebp
  movsbl %bpl,%r14d
  mov    %ebx,%r13d
  mov    %r14d,%edi
  call   <alg_value@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L178
  cmp    0xf(%rsp),%bpl
  je     L39
  cmp    %r13d,%eax
  jge    L75
  cmpl   $0x1,0x10(%rsp)
  jle    L75
  mov    $0xffffffff,%ebx
  jmp    L178
L39:
  mov    %r14d,%edi
  call   <alg_can_repeat@plt>
  test   %eax,%eax
  je     L51
  addl   $0x1,0x10(%rsp)
  mov    0x10(%rsp),%ecx
  cmp    $0x1,%ecx
  setg   %al
  test   %al,0x14(%rsp)
  jne    L46
  cmp    $0x3,%ecx
  jle    L26
L46:
  mov    $0xffffffff,%ebx
L178:
  mov    %ebx,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L172:
  movsbl 0xf(%rsp),%esi
  mov    %r14d,%edi
  call   <alg_can_come_before@plt>
  test   %eax,%eax
  je     L134
  sub    %ebx,%r15d
  movl   $0x1,0x14(%rsp)
  jmp    L110
L65:
  mov    %r15d,%ebx
  jmp    L178
L157:
  mov    $0xffffffff,%ebx
  jmp    L178
L51:
  mov    $0xffffffff,%ebx
  jmp    L178
L134:
  mov    $0xffffffff,%ebx
  jmp    L178
```

assistant: ```c
int roman_to_int(const char *input) {
  int length;
  int i;
  int finalValue = 0;
  int valueNow = -1;
  int valueBefore = -1;
  char algNow = '\0';
  char algBefore = '\0';
  int repetitions = 0;
  bool isSubtracting = false;
  length = string_length(input);
  if (length <= 0)
    return -1;
  for (i = length - 1; i >= 0; --i) {
    algNow = input[i];
    valueNow = alg_value(algNow);
    if (valueNow == -1)
      return -1;
    if (algBefore != algNow) {
      if (valueNow < valueBefore && repetitions > 1)
        return -1;
      repetitions = 1;
    } else {
      if (!alg_can_repeat(algNow))
        return -1;
      ++repetitions;
    }
    if (isSubtracting && repetitions >= 2)
      return -1;
    if (repetitions > 3)
      return -1;
    if (valueNow < valueBefore) {
      if (!alg_can_come_before(algNow, algBefore))
        return -1;
      finalValue -= valueNow;
      isSubtracting = true;
    } else {
      finalValue += valueNow;
      isSubtracting = false;
    }
    algBefore = algNow;
    valueBefore = valueNow;
  }
  return finalValue;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<roman_to_int>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  call   <string_length@plt>
  test   %eax,%eax
  jle    L92
  lea    -0x1(%rax),%edx
  cltq
  movl   $0x0,0x10(%rsp)
  movslq %edx,%r8
  lea    -0x2(%rbx,%rax,1),%r14
  mov    %edx,%edx
  movl   $0x0,0xc(%rsp)
  sub    %rdx,%r14
  lea    (%rbx,%r8,1),%r15
  movl   $0x0,0x14(%rsp)
  mov    $0xffffffff,%ebx
  mov    %r14,0x18(%rsp)
  xor    %r14d,%r14d
  jmp    L97
  nop
L13:
  add    %ebx,0x14(%rsp)
  movl   $0x0,0x10(%rsp)
L129:
  sub    $0x1,%r15
  cmp    0x18(%rsp),%r15
  je     L112
L97:
  movsbl (%r15),%r13d
  mov    %r14d,%ebp
  mov    %ebx,%r12d
  mov    %r13d,%edi
  mov    %r13d,%r14d
  call   <alg_value@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L92
  cmp    %bpl,%r13b
  je     L114
  cmp    %r12d,%eax
  jge    L16
  cmpl   $0x1,0xc(%rsp)
  jg     L92
L16:
  movl   $0x1,0xc(%rsp)
L120:
  cmp    %r12d,%ebx
  jge    L13
  movsbl %bpl,%esi
  mov    %r13d,%edi
  call   <alg_can_come_before@plt>
  test   %eax,%eax
  je     L92
  sub    %ebx,0x14(%rsp)
  movl   $0x1,0x10(%rsp)
  jmp    L129
  nopl   0x0(%rax,%rax,1)
L114:
  mov    %r13d,%edi
  call   <alg_can_repeat@plt>
  test   %eax,%eax
  je     L92
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%ecx
  cmp    $0x1,%ecx
  setg   %al
  test   %al,0x10(%rsp)
  jne    L92
  cmp    $0x4,%ecx
  jne    L120
  xchg   %ax,%ax
L92:
  movl   $0xffffffff,0x14(%rsp)
L112:
  mov    0x14(%rsp),%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int roman_to_int(const char *input) {
  int length;
  int i;
  int finalValue = 0;
  int valueNow = -1;
  int valueBefore = -1;
  char algNow = '\0';
  char algBefore = '\0';
  int repetitions = 0;
  bool isSubtracting = false;
  length = string_length(input);
  if (length <= 0)
    return -1;
  for (i = length - 1; i >= 0; --i) {
    algNow = input[i];
    valueNow = alg_value(algNow);
    if (valueNow == -1)
      return -1;
    if (algBefore != algNow) {
      if (valueNow < valueBefore && repetitions > 1)
        return -1;
      repetitions = 1;
    } else {
      if (!alg_can_repeat(algNow))
        return -1;
      ++repetitions;
    }
    if (isSubtracting && repetitions >= 2)
      return -1;
    if (repetitions > 3)
      return -1;
    if (valueNow < valueBefore) {
      if (!alg_can_come_before(algNow, algBefore))
        return -1;
      finalValue -= valueNow;
      isSubtracting = true;
    } else {
      finalValue += valueNow;
      isSubtracting = false;
    }
    algBefore = algNow;
    valueBefore = valueNow;
  }
  return finalValue;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<roman_to_int>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  call   <string_length@plt>
  test   %eax,%eax
  jle    L123
  lea    -0x1(%rax),%edx
  cltq
  movl   $0x0,0x10(%rsp)
  movslq %edx,%r8
  lea    -0x2(%rbx,%rax,1),%r14
  mov    %edx,%edx
  movl   $0x0,0xc(%rsp)
  sub    %rdx,%r14
  lea    (%rbx,%r8,1),%r15
  movl   $0x0,0x14(%rsp)
  mov    $0xffffffff,%ebx
  mov    %r14,0x18(%rsp)
  xor    %r14d,%r14d
  jmp    L122
  nop
L4:
  add    %ebx,0x14(%rsp)
  movl   $0x0,0x10(%rsp)
L90:
  sub    $0x1,%r15
  cmp    0x18(%rsp),%r15
  je     L56
L122:
  movsbl (%r15),%r13d
  mov    %r14d,%ebp
  mov    %ebx,%r12d
  mov    %r13d,%edi
  mov    %r13d,%r14d
  call   <alg_value@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L123
  cmp    %bpl,%r13b
  je     L100
  cmp    %r12d,%eax
  jge    L17
  cmpl   $0x1,0xc(%rsp)
  jg     L123
L17:
  movl   $0x1,0xc(%rsp)
L51:
  cmp    %r12d,%ebx
  jge    L4
  movsbl %bpl,%esi
  mov    %r13d,%edi
  call   <alg_can_come_before@plt>
  test   %eax,%eax
  je     L123
  sub    %ebx,0x14(%rsp)
  movl   $0x1,0x10(%rsp)
  jmp    L90
  nopl   0x0(%rax,%rax,1)
L100:
  mov    %r13d,%edi
  call   <alg_can_repeat@plt>
  test   %eax,%eax
  je     L123
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%ecx
  cmp    $0x1,%ecx
  setg   %al
  test   %al,0x10(%rsp)
  jne    L123
  cmp    $0x4,%ecx
  jne    L51
  xchg   %ax,%ax
L123:
  movl   $0xffffffff,0x14(%rsp)
L56:
  mov    0x14(%rsp),%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int roman_to_int(const char *input) {
  int length;
  int i;
  int finalValue = 0;
  int valueNow = -1;
  int valueBefore = -1;
  char algNow = '\0';
  char algBefore = '\0';
  int repetitions = 0;
  bool isSubtracting = false;
  length = string_length(input);
  if (length <= 0)
    return -1;
  for (i = length - 1; i >= 0; --i) {
    algNow = input[i];
    valueNow = alg_value(algNow);
    if (valueNow == -1)
      return -1;
    if (algBefore != algNow) {
      if (valueNow < valueBefore && repetitions > 1)
        return -1;
      repetitions = 1;
    } else {
      if (!alg_can_repeat(algNow))
        return -1;
      ++repetitions;
    }
    if (isSubtracting && repetitions >= 2)
      return -1;
    if (repetitions > 3)
      return -1;
    if (valueNow < valueBefore) {
      if (!alg_can_come_before(algNow, algBefore))
        return -1;
      finalValue -= valueNow;
      isSubtracting = true;
    } else {
      finalValue += valueNow;
      isSubtracting = false;
    }
    algBefore = algNow;
    valueBefore = valueNow;
  }
  return finalValue;
}
```