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

```asm
<getchars>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L13:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x5(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x5(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L13
  nop
  nop
  leave
  ret
```

assistant: ```c
void getchars(char *start, int count) {
  for (int i = 0; i < count; i++) {
    char c = getchar();
    *(start + i) = c;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getchars>:
  endbr64
  test   %esi,%esi
  jle    L3
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rbp
L7:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L7
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  ret
```

assistant: ```c
void getchars(char *start, int count) {
  for (int i = 0; i < count; i++) {
    char c = getchar();
    *(start + i) = c;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<getchars>:
  endbr64
  test   %esi,%esi
  jle    L5
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x1(%rdi,%rax,1),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L13:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <getchar@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L13
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  ret
```

assistant: ```c
void getchars(char *start, int count) {
  for (int i = 0; i < count; i++) {
    char c = getchar();
    *(start + i) = c;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getchars>:
  endbr64
  test   %esi,%esi
  jle    L13
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x1(%rdi,%rax,1),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L12:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <getchar@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L12
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  ret
```

assistant: ```c
void getchars(char *start, int count) {
  for (int i = 0; i < count; i++) {
    char c = getchar();
    *(start + i) = c;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readRLEbyte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
L63:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <readbytefrommem@plt>
  mov    -0x20(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jns    L56
  mov    $0xffffffff,%eax
  jmp    L52
L56:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x2f,%eax
  jle    L76
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x39,%eax
  jg     L76
  cmpl   $0x0,-0x4(%rbp)
  jle    L54
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L40
L54:
  movl   $0x0,-0x4(%rbp)
L40:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x30,%eax
  add    %eax,-0x4(%rbp)
  jmp    L63
L76:
  cmpl   $0x0,-0x4(%rbp)
  jns    L16
  movl   $0x1,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
L52:
  leave
  ret
```

assistant: ```c
int readRLEbyte(unsigned char **memptr, int *bytebuff) {
  int rleprefix = -1;
  for (;;) {
    *bytebuff = readbytefrommem(memptr);
    if (*bytebuff < 0)
      return (-1);
    if ((*bytebuff >= '0') && (*bytebuff <= '9')) {
      if (rleprefix > 0) {
        rleprefix *= 10;
      } else {
        rleprefix = 0;
      }
      rleprefix += (*bytebuff - '0');
    } else {
      break;
    }
  }
  if (rleprefix < 0)
    rleprefix = 1;
  return (rleprefix);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readRLEbyte>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0xffffffff,%ebx
  mov    $0x0,%r13d
  jmp    L11
L49:
  lea    -0x30(%rax,%rdx,1),%ebx
L11:
  mov    %r12,%rdi
  call   <readbytefrommem@plt>
  mov    %eax,0x0(%rbp)
  test   %eax,%eax
  js     L36
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  ja     L20
  mov    %r13d,%edx
  test   %ebx,%ebx
  jle    L49
  lea    (%rbx,%rbx,4),%edx
  add    %edx,%edx
  jmp    L49
L20:
  test   %ebx,%ebx
  mov    $0x1,%eax
  cmovs  %eax,%ebx
  jmp    L19
L36:
  mov    $0xffffffff,%ebx
L19:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int readRLEbyte(unsigned char **memptr, int *bytebuff) {
  int rleprefix = -1;
  for (;;) {
    *bytebuff = readbytefrommem(memptr);
    if (*bytebuff < 0)
      return (-1);
    if ((*bytebuff >= '0') && (*bytebuff <= '9')) {
      if (rleprefix > 0) {
        rleprefix *= 10;
      } else {
        rleprefix = 0;
      }
      rleprefix += (*bytebuff - '0');
    } else {
      break;
    }
  }
  if (rleprefix < 0)
    rleprefix = 1;
  return (rleprefix);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<readRLEbyte>:
  endbr64
  push   %r12
  mov    $0xffffffff,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L29
  cs nopw 0x0(%rax,%rax,1)
L10:
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L23
  xor    %edx,%edx
  test   %r12d,%r12d
  jle    L33
  lea    (%r12,%r12,4),%edx
  add    %edx,%edx
L33:
  lea    (%rax,%rdx,1),%r12d
L29:
  mov    %rbp,%rdi
  call   <readbytefrommem@plt>
  mov    %eax,(%rbx)
  test   %eax,%eax
  jns    L10
  mov    $0xffffffff,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   (%rax)
L23:
  mov    $0x1,%eax
  cmp    $0xffffffff,%r12d
  pop    %rbx
  pop    %rbp
  cmove  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int readRLEbyte(unsigned char **memptr, int *bytebuff) {
  int rleprefix = -1;
  for (;;) {
    *bytebuff = readbytefrommem(memptr);
    if (*bytebuff < 0)
      return (-1);
    if ((*bytebuff >= '0') && (*bytebuff <= '9')) {
      if (rleprefix > 0) {
        rleprefix *= 10;
      } else {
        rleprefix = 0;
      }
      rleprefix += (*bytebuff - '0');
    } else {
      break;
    }
  }
  if (rleprefix < 0)
    rleprefix = 1;
  return (rleprefix);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<readRLEbyte>:
  endbr64
  push   %r12
  mov    $0xffffffff,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L5:
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L35
  xor    %edx,%edx
  test   %r12d,%r12d
  jle    L22
  lea    (%r12,%r12,4),%edx
  add    %edx,%edx
L22:
  lea    (%rax,%rdx,1),%r12d
L16:
  mov    %rbp,%rdi
  call   <readbytefrommem@plt>
  mov    %eax,(%rbx)
  test   %eax,%eax
  jns    L5
  mov    $0xffffffff,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   (%rax)
L35:
  mov    $0x1,%eax
  cmp    $0xffffffff,%r12d
  pop    %rbx
  pop    %rbp
  cmove  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int readRLEbyte(unsigned char **memptr, int *bytebuff) {
  int rleprefix = -1;
  for (;;) {
    *bytebuff = readbytefrommem(memptr);
    if (*bytebuff < 0)
      return (-1);
    if ((*bytebuff >= '0') && (*bytebuff <= '9')) {
      if (rleprefix > 0) {
        rleprefix *= 10;
      } else {
        rleprefix = 0;
      }
      rleprefix += (*bytebuff - '0');
    } else {
      break;
    }
  }
  if (rleprefix < 0)
    rleprefix = 1;
  return (rleprefix);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<struct_EC_POINT_point2oct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  mov    %r9,-0x50(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_start@plt>
  mov    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x50(%rbp),%rdi
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rdi,%r9
  mov    $0x0,%r8d
  mov    %rax,%rdi
  call   <EC_POINT_get_Jprojective_coordinates_GFp@plt>
  mov    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_end@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_num_bytes@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x48(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    $0x21,%eax
  sub    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <BN_bn2bin@plt>
  mov    %eax,-0x14(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x38(%rbp)
  jne    L38
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_num_bytes@plt>
  mov    %eax,-0x14(%rbp)
  mov    $0x41,%eax
  sub    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <BN_bn2bin@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x40(%rbp),%rax
  movb   $0x4,(%rax)
  movl   $0x41,-0x14(%rbp)
  jmp    L8
L38:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_is_odd@plt>
  test   %rax,%rax
  je     L19
  mov    -0x40(%rbp),%rax
  movb   $0x3,(%rax)
  jmp    L18
L19:
  mov    -0x40(%rbp),%rax
  movb   $0x2,(%rax)
L18:
  movl   $0x21,-0x14(%rbp)
L8:
  mov    -0x14(%rbp),%eax
  cltq
  leave
  ret
```

assistant: ```c
size_t struct_EC_POINT_point2oct(const EC_GROUP *group, const EC_POINT *p, point_conversion_form_t form, unsigned char *buf, size_t len, BN_CTX *ctx) {
  BIGNUM *X;
  BIGNUM *Y;
  int length;
  BN_CTX_start(ctx);
  X = BN_CTX_get(ctx);
  Y = BN_CTX_get(ctx);
  EC_POINT_get_Jprojective_coordinates_GFp(group, p, X, Y, NULL, ctx);
  BN_CTX_end(ctx);
  length = BN_num_bytes(X);
  memset(buf, 0, len);
  length = BN_bn2bin(X, buf + (33 - length));
  if (form == POINT_CONVERSION_UNCOMPRESSED) {
    length = BN_num_bytes(Y);
    length = BN_bn2bin(Y, buf + (65 - length));
    buf[0] = 0x04;
    length = 65;
  } else {
    if (BN_is_odd(Y)) {
      buf[0] = 0x03;
    } else {
      buf[0] = 0x02;
    }
    length = 33;
  }
  return (length);
}
```
user: Translate the below assembly code into C programming language:

```asm
<struct_EC_POINT_point2oct>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    %rsi,%r15
  mov    %rdx,%r14
  mov    %rcx,%rbp
  mov    %r8,0x8(%rsp)
  mov    %r9,%rbx
  mov    %r9,%rdi
  call   <BN_CTX_start@plt>
  mov    %rbx,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,%r12
  mov    %rbx,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,%r13
  mov    %rbx,%r9
  mov    $0x0,%r8d
  mov    %rax,%rcx
  mov    %r12,%rdx
  mov    %r15,%rsi
  mov    (%rsp),%rdi
  call   <EC_POINT_get_Jprojective_coordinates_GFp@plt>
  mov    %rbx,%rdi
  call   <BN_CTX_end@plt>
  mov    %r12,%rdi
  call   <BN_num_bytes@plt>
  mov    %eax,%ebx
  mov    0x8(%rsp),%rdx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    $0x21,%esi
  sub    %ebx,%esi
  movslq %esi,%rsi
  add    %rbp,%rsi
  mov    %r12,%rdi
  call   <BN_bn2bin@plt>
  mov    D6(%rip),%rax
  cmp    %r14,(%rax)
  je     L2
  mov    %r13,%rdi
  call   <BN_is_odd@plt>
  cmp    $0x1,%rax
  sbb    %edx,%edx
  add    $0x3,%edx
  mov    $0x21,%eax
L15:
  mov    %dl,0x0(%rbp)
  cltq
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  mov    %r13,%rdi
  call   <BN_num_bytes@plt>
  mov    $0x41,%esi
  sub    %eax,%esi
  movslq %esi,%rsi
  add    %rbp,%rsi
  mov    %r13,%rdi
  call   <BN_bn2bin@plt>
  mov    $0x4,%edx
  mov    $0x41,%eax
  jmp    L15
```

assistant: ```c
size_t struct_EC_POINT_point2oct(const EC_GROUP *group, const EC_POINT *p, point_conversion_form_t form, unsigned char *buf, size_t len, BN_CTX *ctx) {
  BIGNUM *X;
  BIGNUM *Y;
  int length;
  BN_CTX_start(ctx);
  X = BN_CTX_get(ctx);
  Y = BN_CTX_get(ctx);
  EC_POINT_get_Jprojective_coordinates_GFp(group, p, X, Y, NULL, ctx);
  BN_CTX_end(ctx);
  length = BN_num_bytes(X);
  memset(buf, 0, len);
  length = BN_bn2bin(X, buf + (33 - length));
  if (form == POINT_CONVERSION_UNCOMPRESSED) {
    length = BN_num_bytes(Y);
    length = BN_bn2bin(Y, buf + (65 - length));
    buf[0] = 0x04;
    length = 65;
  } else {
    if (BN_is_odd(Y)) {
      buf[0] = 0x03;
    } else {
      buf[0] = 0x02;
    }
    length = 33;
  }
  return (length);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<struct_EC_POINT_point2oct>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %r9,%rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %r9,%rdi
  mov    %rsi,(%rsp)
  call   <BN_CTX_start@plt>
  mov    %rbp,%rdi
  call   <BN_CTX_get@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <BN_CTX_get@plt>
  mov    (%rsp),%rsi
  mov    %rbp,%r9
  xor    %r8d,%r8d
  mov    0x8(%rsp),%rdi
  mov    %rax,%rcx
  mov    %r12,%rdx
  mov    %rax,%r14
  call   <EC_POINT_get_Jprojective_coordinates_GFp@plt>
  mov    %rbp,%rdi
  call   <BN_CTX_end@plt>
  mov    %r12,%rdi
  call   <BN_num_bytes@plt>
  mov    %r15,%rdx
  xor    %esi,%esi
  mov    %rbx,%rdi
  mov    %eax,%ebp
  call   <memset@plt>
  mov    $0x21,%esi
  mov    %r12,%rdi
  sub    %ebp,%esi
  movslq %esi,%rsi
  add    %rbx,%rsi
  call   <BN_bn2bin@plt>
  mov    D1(%rip),%rax
  mov    %r14,%rdi
  cmp    %r13,(%rax)
  je     L7
  call   <BN_is_odd@plt>
  mov    $0x21,%r8d
  cmp    $0x1,%rax
  sbb    %eax,%eax
  add    $0x3,%eax
L9:
  mov    %al,(%rbx)
  add    $0x18,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  call   <BN_num_bytes@plt>
  mov    $0x41,%esi
  mov    %r14,%rdi
  sub    %eax,%esi
  movslq %esi,%rsi
  add    %rbx,%rsi
  call   <BN_bn2bin@plt>
  mov    $0x4,%eax
  mov    $0x41,%r8d
  jmp    L9
```

assistant: ```c
size_t struct_EC_POINT_point2oct(const EC_GROUP *group, const EC_POINT *p, point_conversion_form_t form, unsigned char *buf, size_t len, BN_CTX *ctx) {
  BIGNUM *X;
  BIGNUM *Y;
  int length;
  BN_CTX_start(ctx);
  X = BN_CTX_get(ctx);
  Y = BN_CTX_get(ctx);
  EC_POINT_get_Jprojective_coordinates_GFp(group, p, X, Y, NULL, ctx);
  BN_CTX_end(ctx);
  length = BN_num_bytes(X);
  memset(buf, 0, len);
  length = BN_bn2bin(X, buf + (33 - length));
  if (form == POINT_CONVERSION_UNCOMPRESSED) {
    length = BN_num_bytes(Y);
    length = BN_bn2bin(Y, buf + (65 - length));
    buf[0] = 0x04;
    length = 65;
  } else {
    if (BN_is_odd(Y)) {
      buf[0] = 0x03;
    } else {
      buf[0] = 0x02;
    }
    length = 33;
  }
  return (length);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<struct_EC_POINT_point2oct>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %r9,%rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %r9,%rdi
  mov    %rsi,(%rsp)
  call   <BN_CTX_start@plt>
  mov    %rbp,%rdi
  call   <BN_CTX_get@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <BN_CTX_get@plt>
  mov    (%rsp),%rsi
  mov    %rbp,%r9
  xor    %r8d,%r8d
  mov    0x8(%rsp),%rdi
  mov    %rax,%rcx
  mov    %r12,%rdx
  mov    %rax,%r14
  call   <EC_POINT_get_Jprojective_coordinates_GFp@plt>
  mov    %rbp,%rdi
  call   <BN_CTX_end@plt>
  mov    %r12,%rdi
  call   <BN_num_bytes@plt>
  mov    %r15,%rdx
  xor    %esi,%esi
  mov    %rbx,%rdi
  mov    %eax,%ebp
  call   <memset@plt>
  mov    $0x21,%esi
  mov    %r12,%rdi
  sub    %ebp,%esi
  movslq %esi,%rsi
  add    %rbx,%rsi
  call   <BN_bn2bin@plt>
  mov    D8(%rip),%rax
  mov    %r14,%rdi
  cmp    %r13,(%rax)
  je     L17
  call   <BN_is_odd@plt>
  mov    $0x21,%r8d
  cmp    $0x1,%rax
  sbb    %eax,%eax
  add    $0x3,%eax
L19:
  mov    %al,(%rbx)
  add    $0x18,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  call   <BN_num_bytes@plt>
  mov    $0x41,%esi
  mov    %r14,%rdi
  sub    %eax,%esi
  movslq %esi,%rsi
  add    %rbx,%rsi
  call   <BN_bn2bin@plt>
  mov    $0x4,%eax
  mov    $0x41,%r8d
  jmp    L19
```

assistant: ```c
size_t struct_EC_POINT_point2oct(const EC_GROUP *group, const EC_POINT *p, point_conversion_form_t form, unsigned char *buf, size_t len, BN_CTX *ctx) {
  BIGNUM *X;
  BIGNUM *Y;
  int length;
  BN_CTX_start(ctx);
  X = BN_CTX_get(ctx);
  Y = BN_CTX_get(ctx);
  EC_POINT_get_Jprojective_coordinates_GFp(group, p, X, Y, NULL, ctx);
  BN_CTX_end(ctx);
  length = BN_num_bytes(X);
  memset(buf, 0, len);
  length = BN_bn2bin(X, buf + (33 - length));
  if (form == POINT_CONVERSION_UNCOMPRESSED) {
    length = BN_num_bytes(Y);
    length = BN_bn2bin(Y, buf + (65 - length));
    buf[0] = 0x04;
    length = 65;
  } else {
    if (BN_is_odd(Y)) {
      buf[0] = 0x03;
    } else {
      buf[0] = 0x02;
    }
    length = 33;
  }
  return (length);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fib_w_goto>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L10
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L11:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  subl   $0x1,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L15
  jmp    L11
L10:
  nop
  jmp    L31
L15:
  nop
L31:
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int fib_w_goto(int n) {
  int val = 1;
  int nval = 1;
  int nmi, t;
  if (val >= n)
    goto done;
  nmi = n - 1;
loop:
  t = val + nval;
  val = nval;
  nval = t;
  nmi--;
  if (nmi)
    goto loop;
done:
  return val;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fib_w_goto>:
  endbr64
  mov    $0x1,%edx
  cmp    $0x1,%edi
  jle    L12
  sub    $0x1,%edi
  mov    $0x1,%eax
  mov    $0x1,%ecx
L8:
  mov    %eax,%edx
  add    %ecx,%eax
  mov    %edx,%ecx
  sub    $0x1,%edi
  jne    L8
L12:
  mov    %edx,%eax
  ret
```

assistant: ```c
int fib_w_goto(int n) {
  int val = 1;
  int nval = 1;
  int nmi, t;
  if (val >= n)
    goto done;
  nmi = n - 1;
loop:
  t = val + nval;
  val = nval;
  nval = t;
  nmi--;
  if (nmi)
    goto loop;
done:
  return val;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fib_w_goto>:
  endbr64
  mov    $0x1,%r8d
  cmp    $0x1,%edi
  jle    L2
  sub    $0x1,%edi
  mov    $0x1,%eax
  mov    $0x1,%edx
  nopl   0x0(%rax)
L15:
  mov    %eax,%r8d
  add    %edx,%eax
  mov    %r8d,%edx
  sub    $0x1,%edi
  jne    L15
L2:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int fib_w_goto(int n) {
  int val = 1;
  int nval = 1;
  int nmi, t;
  if (val >= n)
    goto done;
  nmi = n - 1;
loop:
  t = val + nval;
  val = nval;
  nval = t;
  nmi--;
  if (nmi)
    goto loop;
done:
  return val;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<fib_w_goto>:
  endbr64
  mov    $0x1,%r8d
  cmp    $0x1,%edi
  jle    L4
  sub    $0x1,%edi
  mov    $0x1,%eax
  mov    $0x1,%edx
  nopl   0x0(%rax)
L18:
  mov    %eax,%r8d
  add    %edx,%eax
  mov    %r8d,%edx
  sub    $0x1,%edi
  jne    L18
L4:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int fib_w_goto(int n) {
  int val = 1;
  int nval = 1;
  int nmi, t;
  if (val >= n)
    goto done;
  nmi = n - 1;
loop:
  t = val + nval;
  val = nval;
  nval = t;
  nmi--;
  if (nmi)
    goto loop;
done:
  return val;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L24
L22:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L25
  addq   $0x1,-0x20(%rbp)
L25:
  addq   $0x1,-0x8(%rbp)
  subq   $0x1,-0x28(%rbp)
L24:
  cmpq   $0x0,-0x28(%rbp)
  jne    L22
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strncpy(char *dest, const char *src, size_t count) {
  char *tmp = dest;
  while (count) {
    if ((*tmp = *src) != 0)
      src++;
    tmp++;
    count--;
  }
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L5
  add    %rdi,%rdx
  mov    %rdi,%rcx
L0:
  movzbl (%rsi),%r8d
  mov    %r8b,(%rcx)
  cmp    $0x1,%r8b
  sbb    $0xffffffffffffffff,%rsi
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  jne    L0
L5:
  ret
```

assistant: ```c
char *strncpy(char *dest, const char *src, size_t count) {
  char *tmp = dest;
  while (count) {
    if ((*tmp = *src) != 0)
      src++;
    tmp++;
    count--;
  }
  return dest;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L7
  add    %rdi,%rdx
  mov    %rdi,%rcx
  nopw   0x0(%rax,%rax,1)
L18:
  movzbl (%rsi),%r8d
  cmp    $0x1,%r8b
  mov    %r8b,(%rcx)
  sbb    $0xffffffffffffffff,%rsi
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L18
L7:
  ret
```

assistant: ```c
char *strncpy(char *dest, const char *src, size_t count) {
  char *tmp = dest;
  while (count) {
    if ((*tmp = *src) != 0)
      src++;
    tmp++;
    count--;
  }
  return dest;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L10
  add    %rdi,%rdx
  mov    %rdi,%rcx
  nopw   0x0(%rax,%rax,1)
L0:
  movzbl (%rsi),%r8d
  cmp    $0x1,%r8b
  mov    %r8b,(%rcx)
  sbb    $0xffffffffffffffff,%rsi
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  jne    L0
L10:
  ret
```

assistant: ```c
char *strncpy(char *dest, const char *src, size_t count) {
  char *tmp = dest;
  while (count) {
    if ((*tmp = *src) != 0)
      src++;
    tmp++;
    count--;
  }
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<search_start>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L12:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0x7,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcmp@plt>
  test   %eax,%eax
  jne    L9
  mov    -0x4(%rbp),%eax
  jmp    L19
L9:
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L12
  mov    $0xffffffff,%eax
L19:
  leave
  ret
```

assistant: ```c
int search_start(char *pData, int Size, char *match) {
  int i;
  for (i = 0; i < Size; i++) {
    if (!memcmp(pData + i, match, 7))
      return i;
  }
  return -1;
}
```