ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r14
  lea    D25(%rip),%rdi
  lea    D0(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  lea    0x14(%rsp),%r12
  lea    0x10(%rsp),%r13
  mov    %r12,%rcx
  mov    %rbp,%rsi
  mov    %r13,%rdx
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D11(%rip),%rbp
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D25
data_type: string
name: retrieve_data
arguments: data_tag: D0
data_type: string
name: retrieve_data
arguments: data_tag: D11
data_type: string

tool: D25: %d%d%d

tool: D0:   %d


tool: D11: %d 

assistant: ```c
int main() {
  int i, j, k, l, m, n;
  int *p[6];
  scanf("%d%d%d", &i, &j, &k);
  p[0] = &i;
  printf("  %d\n", &i);
  p[1] = &j;
  printf("  %d\n", &j);
  p[2] = &k;
  printf("  %d\n", &k);
  for (m = 0; m < 3; m++)
    printf("%d ", p[m]);
  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   %r14
  lea    D23(%rip),%rdi
  lea    D17(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  lea    0x14(%rsp),%r12
  lea    0x10(%rsp),%r13
  mov    %r12,%rcx
  mov    %rbp,%rsi
  mov    %r13,%rdx
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D28(%rip),%rbp
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D23
datatype: string
name: retrieve_data
arguments: label: D17
datatype: string
name: retrieve_data
arguments: label: D28
datatype: string

tool: D23: %d%d%d

tool: D17:   %d


tool: D28: %d 

assistant: ```c
int main() {
  int i, j, k, l, m, n;
  int *p[6];
  scanf("%d%d%d", &i, &j, &k);
  p[0] = &i;
  printf("  %d\n", &i);
  p[1] = &j;
  printf("  %d\n", &j);
  p[2] = &k;
  printf("  %d\n", &k);
  for (m = 0; m < 3; m++)
    printf("%d ", p[m]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<setPort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,0x8(%rax)
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,0x4(%rax)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <htons@plt>
  mov    -0x8(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
struct sockaddr *setPort(struct sockaddr_in *addr, uint16_t port) {
  addr->sin_family = AF_INET;
  addr->sin_addr.s_addr = INADDR_ANY;
  addr->sin_port = htons(port);
  return (struct sockaddr *)addr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<setPort>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x8(%rbx)
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x4(%rbx)
  call   <htons@plt>
  mov    %eax,(%rbx)
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
struct sockaddr *setPort(struct sockaddr_in *addr, uint16_t port) {
  addr->sin_family = AF_INET;
  addr->sin_addr.s_addr = INADDR_ANY;
  addr->sin_port = htons(port);
  return (struct sockaddr *)addr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<setPort>:
  endbr64
  mov    D5(%rip),%rax
  push   %r12
  mov    %rdi,%r12
  mov    %esi,%edi
  mov    (%rax),%eax
  mov    %eax,0x8(%r12)
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x4(%r12)
  call   <htons@plt>
  mov    %eax,(%r12)
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
struct sockaddr *setPort(struct sockaddr_in *addr, uint16_t port) {
  addr->sin_family = AF_INET;
  addr->sin_addr.s_addr = INADDR_ANY;
  addr->sin_port = htons(port);
  return (struct sockaddr *)addr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<setPort>:
  endbr64
  mov    D4(%rip),%rax
  mov    D8(%rip),%rdx
  push   %r12
  mov    %rdi,%r12
  mov    %esi,%edi
  mov    (%rax),%eax
  movd   (%rdx),%xmm0
  movd   %eax,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,0x4(%r12)
  call   <htons@plt>
  mov    %eax,(%r12)
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
struct sockaddr *setPort(struct sockaddr_in *addr, uint16_t port) {
  addr->sin_family = AF_INET;
  addr->sin_addr.s_addr = INADDR_ANY;
  addr->sin_port = htons(port);
  return (struct sockaddr *)addr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<md5_hash>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%r13
  mov    %esi,%r12d
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    $0x67452301,%edi
  call   <UINT32_C@plt>
  mov    %eax,0x0(%rbp)
  mov    $0xefcdab89,%edi
  call   <UINT32_C@plt>
  mov    %eax,0x4(%rbp)
  mov    $0x98badcfe,%edi
  call   <UINT32_C@plt>
  mov    %eax,0x8(%rbp)
  mov    $0x10325476,%edi
  call   <UINT32_C@plt>
  mov    %eax,0xc(%rbp)
  cmp    $0x3f,%r12d
  jle    L44
  mov    %r13,%rbx
  lea    -0x40(%r12),%r14d
  shr    $0x6,%r14d
  mov    %r14d,%eax
  shl    $0x8,%rax
  lea    0x100(%r13,%rax,1),%r15
L7:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <md5_compress@plt>
  add    $0x100,%rbx
  cmp    %r15,%rbx
  jne    L7
  lea    0x1(%r14),%eax
  shl    $0x6,%eax
L50:
  mov    %r12d,%ebx
  sub    %eax,%ebx
  cltq
  lea    0x0(%r13,%rax,4),%rsi
  mov    %rsp,%rdi
  mov    %ebx,%edx
  call   <memcpy@plt>
  movslq %ebx,%rax
  movl   $0x80,(%rsp,%rax,4)
  add    $0x1,%ebx
  mov    $0x40,%edx
  sub    %ebx,%edx
  cmp    $0x7,%edx
  jle    L21
  mov    $0x38,%edx
  sub    %ebx,%edx
  movslq %ebx,%rbx
  lea    (%rsp,%rbx,4),%rdi
  mov    $0x0,%esi
  call   <memset@plt>
L40:
  lea    0x0(,%r12,8),%eax
  mov    %eax,0x38(%rsp)
  sar    $0x1d,%r12d
  mov    %r12d,0x3c(%rsp)
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <md5_compress@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L44:
  mov    $0x0,%eax
  jmp    L50
L21:
  mov    %rsp,%r13
  movslq %ebx,%rbx
  lea    0x0(%r13,%rbx,4),%rdi
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <md5_compress@plt>
  mov    $0x38,%edx
  mov    $0x0,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  jmp    L40
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void md5_hash(uint8_t *message, uint32_t len, uint32_t *hash) {
  hash[0] = UINT32_C(0x67452301);
  hash[1] = UINT32_C(0xEFCDAB89);
  hash[2] = UINT32_C(0x98BADCFE);
  hash[3] = UINT32_C(0x10325476);
  int i;
  for (i = 0; i + 64 <= len; i += 64)
    md5_compress(hash, (uint32_t *)(message + i));
  uint32_t block[16];
  uint8_t *byteBlock = (uint8_t *)block;
  int rem = len - i;
  memcpy(byteBlock, message + i, rem);
  byteBlock[rem] = 0x80;
  rem++;
  if (64 - rem >= 8)
    memset(byteBlock + rem, 0, 56 - rem);
  else {
    memset(byteBlock + rem, 0, 64 - rem);
    md5_compress(hash, block);
    memset(block, 0, 56);
  }
  block[14] = len << 3;
  block[15] = len >> 29;
  md5_compress(hash, block);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<md5_hash>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  mov    $0x67452301,%edi
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  call   <UINT32_C@plt>
  mov    $0xefcdab89,%edi
  mov    %eax,0x0(%rbp)
  call   <UINT32_C@plt>
  mov    $0x98badcfe,%edi
  mov    %eax,0x4(%rbp)
  call   <UINT32_C@plt>
  mov    $0x10325476,%edi
  mov    %eax,0x8(%rbp)
  call   <UINT32_C@plt>
  mov    %eax,0xc(%rbp)
  cmp    $0x3f,%r12d
  jle    L37
  lea    -0x40(%r12),%r14d
  mov    %r15,%rbx
  shr    $0x6,%r14d
  mov    %r14d,%eax
  shl    $0x8,%rax
  lea    0x100(%r15,%rax,1),%r13
  nopl   0x0(%rax)
L9:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x100,%rbx
  call   <md5_compress@plt>
  cmp    %rbx,%r13
  jne    L9
  add    $0x1,%r14d
  mov    %r12d,%ebx
  shl    $0x6,%r14d
  movslq %r14d,%rax
  sub    %r14d,%ebx
  lea    (%r15,%rax,4),%r15
L8:
  mov    %rsp,%r13
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rax
  mov    $0x40,%edx
  add    $0x1,%ebx
  shl    $0x2,%rax
  sub    %ebx,%edx
  movl   $0x80,(%rsp,%rax,1)
  lea    0x4(%r13,%rax,1),%rdi
  cmp    $0x7,%edx
  jle    L38
  mov    $0x38,%edx
  xor    %esi,%esi
  sub    %ebx,%edx
  call   <memset@plt>
L10:
  lea    0x0(,%r12,8),%eax
  mov    %r13,%rsi
  sar    $0x1d,%r12d
  mov    %rbp,%rdi
  mov    %eax,0x38(%rsp)
  mov    %r12d,0x3c(%rsp)
  call   <md5_compress@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  xor    %esi,%esi
  call   <memset@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <md5_compress@plt>
  mov    $0x38,%edx
  xor    %esi,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L37:
  mov    %r12d,%ebx
  jmp    L8
L55:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void md5_hash(uint8_t *message, uint32_t len, uint32_t *hash) {
  hash[0] = UINT32_C(0x67452301);
  hash[1] = UINT32_C(0xEFCDAB89);
  hash[2] = UINT32_C(0x98BADCFE);
  hash[3] = UINT32_C(0x10325476);
  int i;
  for (i = 0; i + 64 <= len; i += 64)
    md5_compress(hash, (uint32_t *)(message + i));
  uint32_t block[16];
  uint8_t *byteBlock = (uint8_t *)block;
  int rem = len - i;
  memcpy(byteBlock, message + i, rem);
  byteBlock[rem] = 0x80;
  rem++;
  if (64 - rem >= 8)
    memset(byteBlock + rem, 0, 56 - rem);
  else {
    memset(byteBlock + rem, 0, 64 - rem);
    md5_compress(hash, block);
    memset(block, 0, 56);
  }
  block[14] = len << 3;
  block[15] = len >> 29;
  md5_compress(hash, block);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<md5_hash>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  mov    $0x67452301,%edi
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  call   <UINT32_C@plt>
  mov    $0xefcdab89,%edi
  mov    %eax,0x0(%rbp)
  call   <UINT32_C@plt>
  mov    $0x98badcfe,%edi
  mov    %eax,0x4(%rbp)
  call   <UINT32_C@plt>
  mov    $0x10325476,%edi
  mov    %eax,0x8(%rbp)
  call   <UINT32_C@plt>
  mov    %eax,0xc(%rbp)
  cmp    $0x3f,%r12d
  jle    L53
  lea    -0x40(%r12),%r14d
  mov    %r15,%rbx
  shr    $0x6,%r14d
  mov    %r14d,%eax
  shl    $0x8,%rax
  lea    0x100(%r15,%rax,1),%r13
  nopl   0x0(%rax)
L18:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x100,%rbx
  call   <md5_compress@plt>
  cmp    %rbx,%r13
  jne    L18
  add    $0x1,%r14d
  mov    %r12d,%ebx
  shl    $0x6,%r14d
  movslq %r14d,%rax
  sub    %r14d,%ebx
  lea    (%r15,%rax,4),%r15
L26:
  mov    %rsp,%r13
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rax
  mov    $0x40,%edx
  add    $0x1,%ebx
  shl    $0x2,%rax
  sub    %ebx,%edx
  movl   $0x80,(%rsp,%rax,1)
  lea    0x4(%r13,%rax,1),%rdi
  cmp    $0x7,%edx
  jle    L44
  mov    $0x38,%edx
  xor    %esi,%esi
  sub    %ebx,%edx
  call   <memset@plt>
L32:
  lea    0x0(,%r12,8),%eax
  sar    $0x1d,%r12d
  mov    %r13,%rsi
  mov    %rbp,%rdi
  movd   %eax,%xmm0
  movd   %r12d,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,0x38(%rsp)
  call   <md5_compress@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L44:
  xor    %esi,%esi
  call   <memset@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <md5_compress@plt>
  mov    $0x38,%edx
  xor    %esi,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L53:
  mov    %r12d,%ebx
  jmp    L26
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void md5_hash(uint8_t *message, uint32_t len, uint32_t *hash) {
  hash[0] = UINT32_C(0x67452301);
  hash[1] = UINT32_C(0xEFCDAB89);
  hash[2] = UINT32_C(0x98BADCFE);
  hash[3] = UINT32_C(0x10325476);
  int i;
  for (i = 0; i + 64 <= len; i += 64)
    md5_compress(hash, (uint32_t *)(message + i));
  uint32_t block[16];
  uint8_t *byteBlock = (uint8_t *)block;
  int rem = len - i;
  memcpy(byteBlock, message + i, rem);
  byteBlock[rem] = 0x80;
  rem++;
  if (64 - rem >= 8)
    memset(byteBlock + rem, 0, 56 - rem);
  else {
    memset(byteBlock + rem, 0, 64 - rem);
    md5_compress(hash, block);
    memset(block, 0, 56);
  }
  block[14] = len << 3;
  block[15] = len >> 29;
  md5_compress(hash, block);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rindex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,%rax
  mov    %esi,%edx
  mov    %edx,%ecx
  movq   $0x0,-0x8(%rbp)
  jmp    L11
L27:
  movzbl (%rax),%edx
  cmp    %dl,%cl
  jne    L21
  mov    %rax,-0x8(%rbp)
L21:
  add    $0x1,%rax
L11:
  movzbl (%rax),%edx
  test   %dl,%dl
  jne    L27
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *rindex(register char *s, register char c) {
  char *p;
  for (p = 0; *s; s++)
    if (*s == c)
      p = s;
  return (p);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rindex>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L10
  mov    $0x0,%edx
L2:
  cmp    %al,%sil
  cmove  %rdi,%rdx
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L2
L4:
  mov    %rdx,%rax
  ret
L10:
  mov    $0x0,%edx
  jmp    L4
```

assistant: ```c
char *rindex(register char *s, register char c) {
  char *p;
  for (p = 0; *s; s++)
    if (*s == c)
      p = s;
  return (p);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rindex>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L4
  xchg   %ax,%ax
L8:
  cmp    %al,%sil
  cmove  %rdi,%r8
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L8
L4:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *rindex(register char *s, register char c) {
  char *p;
  for (p = 0; *s; s++)
    if (*s == c)
      p = s;
  return (p);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rindex>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L1
  xchg   %ax,%ax
L11:
  cmp    %sil,%al
  cmove  %rdi,%r8
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L11
L1:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *rindex(register char *s, register char c) {
  char *p;
  for (p = 0; *s; s++)
    if (*s == c)
      p = s;
  return (p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <foo@plt>
  test   %rax,%rax
  je     L9
  mov    $0x0,%eax
  call   <abort@plt>
L9:
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (foo((struct a *)0) != 0)
    abort();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <foo@plt>
  test   %rax,%rax
  jne    L11
L6:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L11:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L6
```

assistant: ```c
int main() {
  if (foo((struct a *)0) != 0)
    abort();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <foo@plt>
  test   %rax,%rax
  jne    L0
L17:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L0:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L17
```

assistant: ```c
int main() {
  if (foo((struct a *)0) != 0)
    abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <foo@plt>
  test   %rax,%rax
  jne    L10
L5:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L10:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L5
```

assistant: ```c
int main() {
  if (foo((struct a *)0) != 0)
    abort();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<eventfd_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  lea    -0x8(%rbp),%rcx
  mov    -0x4(%rbp),%eax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  cmp    $0x3,%rax
  jg     L9
  mov    $0xffffffff,%eax
  jmp    L5
L9:
  mov    $0x0,%eax
L5:
  leave
  ret
```

assistant: ```c
int eventfd_write(int fd, uint64_t val) {
  if (write(fd, &val, sizeof(val)) < (ssize_t)sizeof(val))
    return (-1);
  else
    return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<eventfd_write>:
  endbr64
  sub    $0x18,%rsp
  mov    %esi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x4,%edx
  call   <write@plt>
  cmp    $0x3,%rax
  setle  %al
  movzbl %al,%eax
  neg    %eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int eventfd_write(int fd, uint64_t val) {
  if (write(fd, &val, sizeof(val)) < (ssize_t)sizeof(val))
    return (-1);
  else
    return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<eventfd_write>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%edx
  mov    %esi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  call   <write@plt>
  cmp    $0x3,%rax
  setle  %al
  add    $0x18,%rsp
  movzbl %al,%eax
  neg    %eax
  ret
```

assistant: ```c
int eventfd_write(int fd, uint64_t val) {
  if (write(fd, &val, sizeof(val)) < (ssize_t)sizeof(val))
    return (-1);
  else
    return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<eventfd_write>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%edx
  mov    %esi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  call   <write@plt>
  cmp    $0x3,%rax
  setle  %al
  add    $0x18,%rsp
  movzbl %al,%eax
  neg    %eax
  ret
```

assistant: ```c
int eventfd_write(int fd, uint64_t val) {
  if (write(fd, &val, sizeof(val)) < (ssize_t)sizeof(val))
    return (-1);
  else
    return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  lea    D46(%rip),%rdi
  call   <puts@plt>
  lea    D81(%rip),%rdi
  call   <puts@plt>
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%r13d
  mov    D87(%rip),%rbp
  mov    D62(%rip),%rbx
  lea    D64(%rip),%r15
L0:
  movq   %rbx,%xmm0
  call   <f@plt>
  movq   %xmm0,%r12
  movq   %rbx,%xmm0
  call   <diff@plt>
  movapd %xmm0,%xmm1
  mov    %rbx,0x8(%rsp)
  movq   %r12,%xmm0
  divsd  %xmm1,%xmm0
  movq   %rbx,%xmm4
  subsd  %xmm0,%xmm4
  movq   %xmm4,%rbx
  mov    %rbp,%r14
  movapd %xmm4,%xmm5
  subsd  0x8(%rsp),%xmm5
  movapd %xmm5,%xmm7
  andpd  D73(%rip),%xmm7
  movq   %xmm7,%rbp
  movapd %xmm7,%xmm0
  call   <log@plt>
  movq   %xmm0,%r12
  movq   %r14,%xmm0
  call   <log@plt>
  movq   %r12,%xmm2
  divsd  %xmm0,%xmm2
  movq   %xmm2,%r12
  movsd  0x8(%rsp),%xmm0
  call   <f@plt>
  movapd %xmm0,%xmm1
  mov    %r13d,%esi
  add    $0x1,%r13d
  movq   %r12,%xmm3
  movq   %rbp,%xmm2
  movsd  0x8(%rsp),%xmm0
  mov    %r15,%rdi
  mov    $0x4,%eax
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  movq   %rbx,%xmm3
  subsd  %xmm3,%xmm0
  andpd  D73(%rip),%xmm0
  movsd  D42(%rip),%xmm6
  comisd %xmm0,%xmm6
  jbe    L0
  movq   %rbx,%xmm0
  lea    D89(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D46
type: string
name: parse_data
arguments: label: D81
type: string
name: parse_data
arguments: label: D14
type: string
name: parse_data
arguments: label: D87
type: f64
name: parse_data
arguments: label: D62
type: f64
name: parse_data
arguments: label: D64
type: string
name: parse_data
arguments: label: D73
type: qword[2]
name: parse_data
arguments: label: D42
type: f64
name: parse_data
arguments: label: D89
type: string

tool: D46: Solution for e^x = 2x+1 using newton raphson method

tool: D81: Initial approximation:1

tool: D14: i	Root		f(Root)		Error		Order of convergence

tool: D87: 2

tool: D62: 1

tool: D64: %d	%f	%f	%f	%f


tool: {"D73": [0x7fffffffffffffff,0x0000000000000000]}

tool: D42: 1e-06

tool: D89: 

Final Answer:%f


assistant: ```c
int main() {
  printf("Solution for e^x = 2x+1 using newton raphson method\n");
  printf("Initial approximation:1\n");
  double newRoot, oldRoot = 1;
  double oldError = 2, newError;
  int i = 0;
  printf("i\tRoot\t\tf(Root)\t\tError\t\tOrder of convergence\n");
  while (1) {
    newRoot = oldRoot - (f(oldRoot) / diff(oldRoot));
    newError = fabs(newRoot - oldRoot);
    printf("%d\t%f\t%f\t%f\t%f\n", i++, oldRoot, f(oldRoot), newError, log(newError) / log(oldError));
    if (fabs(oldRoot - newRoot) < 0.000001)
      break;
    oldRoot = newRoot;
    oldError = newError;
  }
  printf("\n\nFinal Answer:%f\n", newRoot);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<peer_start>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x4(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <peer_connect@plt>
  cmp    $0x1,%eax
  jne    L37
  mov    $0x1,%eax
  jmp    L11
L37:
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <peer_listen@plt>
  cmp    $0x1,%eax
  jne    L34
  mov    $0x1,%eax
  jmp    L11
L34:
  mov    $0x0,%eax
L11:
  leave
  ret
```

assistant: ```c
int peer_start(int sockfd, const char *host, const char *port) {
  if ((peer_connect(sockfd, host, port)) == 1) {
    return 1;
  }
  if ((peer_listen(sockfd, port)) == 1) {
    return 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<peer_start>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rdx,%rbp
  call   <peer_connect@plt>
  cmp    $0x1,%eax
  je     L3
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <peer_listen@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int peer_start(int sockfd, const char *host, const char *port) {
  if ((peer_connect(sockfd, host, port)) == 1) {
    return 1;
  }
  if ((peer_listen(sockfd, port)) == 1) {
    return 1;
  }
  return 0;
}
```