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

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D83(%rip),%rdi
  call   <printf@plt>
  lea    D75(%rip),%rdi
  call   <printf@plt>
  lea    D60(%rip),%rdi
  call   <printf@plt>
  lea    D89(%rip),%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    D64(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  call   <printf@plt>
  lea    D84(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_label: D83
data_type: string
name: parse_data
arguments: data_label: D75
data_type: string
name: parse_data
arguments: data_label: D60
data_type: string
name: parse_data
arguments: data_label: D89
data_type: string
name: parse_data
arguments: data_label: D1
data_type: string
name: parse_data
arguments: data_label: D64
data_type: string
name: parse_data
arguments: data_label: D20
data_type: string
name: parse_data
arguments: data_label: D84
data_type: string
name: parse_data
arguments: data_label: D24
data_type: string

tool: D83: Usage for SIMD MICRO benchmarks:



tool: D75: 	 -b, --bench NAME 	 name of the benchmark - possible values is simd.


tool: D60: 	 -r, --reps N 		 number of repetitions. Default value is ULONG_MAX.


tool: D89: 	 -o, --op TYPE 		 TYPE of operation.


tool: D1: 				 --> for simd benchmark: "+", "-" and "*". Default is "+".


tool: D64: 	 -d, --dtype DATATYPE 	 DATATYPE to be used - possible values are int, long, float, double. Default is int.


tool: D20: 	 -i, --info 		 Print out system information such as current CPU frequency, core counts, cache size, plus datatype sizes.


tool: D84: 	 -h, --help 		 Displays this help.


tool: D24: 



assistant: ```c
void usage() {
  printf("Usage for SIMD MICRO benchmarks:\n\n");
  printf("\t -b, --bench NAME \t name of the benchmark - possible values is simd.\n");
  printf("\t -r, --reps N \t\t number of repetitions. Default value is ULONG_MAX.\n");
  printf("\t -o, --op TYPE \t\t TYPE of operation.\n");
  printf("\t\t\t\t --> for simd benchmark: \"+\", \"-\" and \"*\". Default is \"+\".\n");
  printf("\t -d, --dtype DATATYPE \t DATATYPE to be used - possible values are int, long, float, double. Default is int.\n");
  printf("\t -i, --info \t\t Print out system information such as current CPU frequency, core counts, cache size, plus datatype sizes.\n");
  printf("\t -h, --help \t\t Displays this help.\n");
  printf("\n\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D88(%rip),%rdi
  call   <printf@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    D69(%rip),%rdi
  call   <printf@plt>
  lea    D61(%rip),%rdi
  call   <printf@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D51(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D36
type: string
name: extract_symbolic_data
arguments: tag: D24
type: string
name: extract_symbolic_data
arguments: tag: D88
type: string
name: extract_symbolic_data
arguments: tag: D35
type: string
name: extract_symbolic_data
arguments: tag: D9
type: string
name: extract_symbolic_data
arguments: tag: D69
type: string
name: extract_symbolic_data
arguments: tag: D61
type: string
name: extract_symbolic_data
arguments: tag: D21
type: string
name: extract_symbolic_data
arguments: tag: D51
type: string

tool: D36: Usage for SIMD MICRO benchmarks:



tool: D24: 	 -b, --bench NAME 	 name of the benchmark - possible values is simd.


tool: D88: 	 -r, --reps N 		 number of repetitions. Default value is ULONG_MAX.


tool: D35: 	 -o, --op TYPE 		 TYPE of operation.


tool: D9: 				 --> for simd benchmark: "+", "-" and "*". Default is "+".


tool: D69: 	 -d, --dtype DATATYPE 	 DATATYPE to be used - possible values are int, long, float, double. Default is int.


tool: D61: 	 -i, --info 		 Print out system information such as current CPU frequency, core counts, cache size, plus datatype sizes.


tool: D21: 	 -h, --help 		 Displays this help.


tool: D51: 



assistant: ```c
void usage() {
  printf("Usage for SIMD MICRO benchmarks:\n\n");
  printf("\t -b, --bench NAME \t name of the benchmark - possible values is simd.\n");
  printf("\t -r, --reps N \t\t number of repetitions. Default value is ULONG_MAX.\n");
  printf("\t -o, --op TYPE \t\t TYPE of operation.\n");
  printf("\t\t\t\t --> for simd benchmark: \"+\", \"-\" and \"*\". Default is \"+\".\n");
  printf("\t -d, --dtype DATATYPE \t DATATYPE to be used - possible values are int, long, float, double. Default is int.\n");
  printf("\t -i, --info \t\t Print out system information such as current CPU frequency, core counts, cache size, plus datatype sizes.\n");
  printf("\t -h, --help \t\t Displays this help.\n");
  printf("\n\n");
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<reverseBytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L7
L26:
  mov    -0x40(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x40(%rbp),%eax
  cltq
  lea    -0x10(%rbp),%rdx
  add    %rdx,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L36
L34:
  subq   $0x1,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    -0x28(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L36:
  lea    -0x10(%rbp),%rax
  cmp    %rax,-0x28(%rbp)
  ja     L34
  mov    -0x40(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
  addl   $0x1,-0x2c(%rbp)
L7:
  mov    -0x2c(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L26
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
void reverseBytes(void *s, int nItems, int nBytes) {
  int i;
  unsigned char tmp[8];
  unsigned char *f, *t;
  unsigned char *src = s;
  for (i = 0; i < nItems; i++) {
    memcpy(tmp, src, nBytes);
    for (f = tmp + nBytes, t = src; f > tmp;) {
      *t++ = *--f;
    }
    src += nBytes;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<reverseBytes>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L38
  mov    %rdi,%rbp
  mov    %esi,%r14d
  mov    %edx,%r13d
  mov    $0x0,%r12d
  mov    %rsp,%rbx
L37:
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  movslq %r13d,%rsi
  lea    (%rbx,%rsi,1),%rax
  cmp    %rbx,%rax
  jbe    L34
  mov    %rbp,%rdx
L26:
  sub    $0x1,%rax
  add    $0x1,%rdx
  movzbl (%rax),%ecx
  mov    %cl,-0x1(%rdx)
  cmp    %rbx,%rax
  jne    L26
L34:
  add    %rsi,%rbp
  add    $0x1,%r12d
  cmp    %r12d,%r14d
  jne    L37
L38:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverseBytes(void *s, int nItems, int nBytes) {
  int i;
  unsigned char tmp[8];
  unsigned char *f, *t;
  unsigned char *src = s;
  for (i = 0; i < nItems; i++) {
    memcpy(tmp, src, nBytes);
    for (f = tmp + nBytes, t = src; f > tmp;) {
      *t++ = *--f;
    }
    src += nBytes;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<reverseBytes>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L44
  movslq %edx,%r13
  lea    0x10(%rsp),%r14
  mov    %rdi,%r15
  mov    %edx,%r12d
  lea    (%r14,%r13,1),%rbx
  xor    %ebp,%ebp
  xchg   %ax,%ax
L18:
  mov    %r12d,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %r15,%rdx
  mov    %rbx,%rax
  cmp    %r14,%rbx
  jbe    L27
  nopl   0x0(%rax)
L0:
  movzbl -0x1(%rax),%ecx
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %cl,-0x1(%rdx)
  cmp    %r14,%rax
  jne    L0
L27:
  add    %r13,%r15
  add    $0x1,%ebp
  cmp    %ebp,0xc(%rsp)
  jne    L18
L44:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverseBytes(void *s, int nItems, int nBytes) {
  int i;
  unsigned char tmp[8];
  unsigned char *f, *t;
  unsigned char *src = s;
  for (i = 0; i < nItems; i++) {
    memcpy(tmp, src, nBytes);
    for (f = tmp + nBytes, t = src; f > tmp;) {
      *t++ = *--f;
    }
    src += nBytes;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<reverseBytes>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L26
  movslq %edx,%r13
  lea    0x10(%rsp),%r14
  mov    %rdi,%r15
  mov    %edx,%r12d
  lea    (%r14,%r13,1),%rbx
  xor    %ebp,%ebp
  xchg   %ax,%ax
L8:
  mov    %r12d,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  cmp    %r14,%rbx
  jbe    L40
  mov    %r15,%rdx
  mov    %rbx,%rax
  nopl   0x0(%rax)
L21:
  movzbl -0x1(%rax),%ecx
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %cl,-0x1(%rdx)
  cmp    %r14,%rax
  jne    L21
L40:
  add    %r13,%r15
  add    $0x1,%ebp
  cmp    %ebp,0xc(%rsp)
  jne    L8
L26:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverseBytes(void *s, int nItems, int nBytes) {
  int i;
  unsigned char tmp[8];
  unsigned char *f, *t;
  unsigned char *src = s;
  for (i = 0; i < nItems; i++) {
    memcpy(tmp, src, nBytes);
    for (f = tmp + nBytes, t = src; f > tmp;) {
      *t++ = *--f;
    }
    src += nBytes;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ResetISR>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xe000ed88,%eax
  mov    (%rax),%edx
  mov    $0xe000ed88,%eax
  or     $0xf00000,%edx
  mov    %edx,(%rax)
  mov    $0x0,%eax
  call   <__iar_program_start@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ResetISR(void) {
  (*((volatile uint32_t *)(0xE000ED88))) = (((*((volatile uint32_t *)(0xE000ED88))) & ~(0x00300000 | 0x00C00000)) | 0x00300000 | 0x00C00000);
  __iar_program_start();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ResetISR>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xe000ed88,%edx
  mov    (%rdx),%eax
  or     $0xf00000,%eax
  mov    %eax,(%rdx)
  mov    $0x0,%eax
  call   <__iar_program_start@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void ResetISR(void) {
  (*((volatile uint32_t *)(0xE000ED88))) = (((*((volatile uint32_t *)(0xE000ED88))) & ~(0x00300000 | 0x00C00000)) | 0x00300000 | 0x00C00000);
  __iar_program_start();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ResetISR>:
  endbr64
  mov    $0xe000ed88,%edx
  mov    (%rdx),%eax
  or     $0xf00000,%eax
  mov    %eax,(%rdx)
  xor    %eax,%eax
  jmp    1050 <__iar_program_start@plt>
```

assistant: ```c
void ResetISR(void) {
  (*((volatile uint32_t *)(0xE000ED88))) = (((*((volatile uint32_t *)(0xE000ED88))) & ~(0x00300000 | 0x00C00000)) | 0x00300000 | 0x00C00000);
  __iar_program_start();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ResetISR>:
  endbr64
  mov    $0xe000ed88,%edx
  mov    (%rdx),%eax
  or     $0xf00000,%eax
  mov    %eax,(%rdx)
  xor    %eax,%eax
  jmp    1050 <__iar_program_start@plt>
```

assistant: ```c
void ResetISR(void) {
  (*((volatile uint32_t *)(0xE000ED88))) = (((*((volatile uint32_t *)(0xE000ED88))) & ~(0x00300000 | 0x00C00000)) | 0x00300000 | 0x00C00000);
  __iar_program_start();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D15(%rip),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L3:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  jl     L3
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D15
type_infered: string
name: parse_data
arguments: tag: D5
type_infered: string

tool: D15: hello, world!

tool: D5: %s


assistant: ```c
int main(int argc, char **argv) {
  int i;
  char *str = "hello, world!", ch;
  for (i = 0; i < strlen(str); i++)
    ch = str[i];
  printf("%s\n", str);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  lea    D3(%rip),%rbp
  jmp    L16
L4:
  add    $0x1,%ebx
L16:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L4
  lea    D3(%rip),%rsi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_infered: string
name: parse_data
arguments: data_tag: D16
type_infered: string

tool: D3: hello, world!

tool: D16: %s


assistant: ```c
int main(int argc, char **argv) {
  int i;
  char *str = "hello, world!", ch;
  for (i = 0; i < strlen(str); i++)
    ch = str[i];
  printf("%s\n", str);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L6
  nopl   (%rax)
L11:
  add    $0x1,%ebx
L6:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L11
  mov    %rbp,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
type_infered: string
name: retrieve_data
arguments: label: D3
type_infered: string

tool: D5: hello, world!

tool: D3: %s


assistant: ```c
int main(int argc, char **argv) {
  int i;
  char *str = "hello, world!", ch;
  for (i = 0; i < strlen(str); i++)
    ch = str[i];
  printf("%s\n", str);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D16(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L16
  nopl   (%rax)
L15:
  add    $0x1,%ebx
L16:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L15
  mov    %rbp,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D16
type_guessed: string
name: read_data
arguments: label: D8
type_guessed: string

tool: D16: hello, world!

tool: D8: %s


assistant: ```c
int main(int argc, char **argv) {
  int i;
  char *str = "hello, world!", ch;
  for (i = 0; i < strlen(str); i++)
    ch = str[i];
  printf("%s\n", str);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<decchk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L78
L42:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  cmpb   $0x2f,-0xd(%rbp)
  jle    L48
  cmpb   $0x39,-0xd(%rbp)
  jg     L48
  movsbl -0xd(%rbp),%eax
  lea    0x2(%rax),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,-0x4(%rbp)
  and    $0x7,%eax
  movslq %eax,%rcx
  movslq %edx,%rax
  shl    $0x3,%rax
  add    %rcx,%rax
  lea    0x0(,%rax,4),%rdx
  lea    D33(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  movslq %eax,%rcx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  lea    0x0(,%rax,4),%rdx
  lea    D12(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  mov    %eax,-0x8(%rbp)
L48:
  addl   $0x1,-0xc(%rbp)
L78:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L42
  movl   $0x0,-0xc(%rbp)
  jmp    L24
L35:
  mov    -0x4(%rbp),%eax
  and    $0x7,%eax
  movslq %eax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  lea    D33(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  movslq %eax,%rcx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  lea    0x0(,%rax,4),%rdx
  lea    D12(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  test   %eax,%eax
  je     L34
  addl   $0x1,-0xc(%rbp)
L24:
  cmpl   $0x9,-0xc(%rbp)
  jle    L35
  jmp    L9
L34:
  nop
L9:
  mov    -0xc(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %dl,(%rax)
  cmpl   $0x0,-0x8(%rbp)
  sete   %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int decchk(char string[], int n, char *ch) {
  char c;
  int j, k = 0, m = 0;
  static int ip[10][8] = {0, 1, 5, 8, 9, 4, 2, 7, 1, 5, 8, 9, 4, 2, 7, 0, 2, 7, 0, 1, 5, 8, 9, 4, 3, 6, 3, 6, 3, 6, 3, 6, 4, 2, 7, 0, 1, 5, 8, 9, 5, 8, 9, 4, 2, 7, 0, 1, 6, 3, 6, 3, 6, 3, 6, 3, 7, 0, 1, 5, 8, 9, 4, 2, 8, 9, 4, 2, 7, 0, 1, 5, 9, 4, 2, 7, 0, 1, 5, 8};
  static int ij[10][10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 0, 6, 7, 8, 9, 5, 2, 3, 4, 0, 1, 7, 8, 9, 5, 6, 3, 4, 0, 1, 2, 8, 9, 5, 6, 7, 4, 0, 1, 2, 3, 9, 5, 6, 7, 8, 5, 9, 8, 7, 6, 0, 4, 3, 2, 1, 6, 5, 9, 8, 7, 1, 0, 4, 3, 2, 7, 6, 5, 9, 8, 2, 1, 0, 4, 3, 8, 7, 6, 5, 9, 3, 2, 1, 0, 4, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0};
  for (j = 0; j < n; j++) {
    c = string[j];
    if (c >= 48 && c <= 57)
      k = ij[k][ip[(c + 2) % 10][7 & m++]];
  }
  for (j = 0; j <= 9; j++)
    if (!ij[k][ip[j][m & 7]])
      break;
  *ch = j + 48;
  return k == 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<decchk>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdx,%r11
  test   %esi,%esi
  jle    L25
  mov    %rdi,%rdx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rdi
  mov    $0x0,%esi
  mov    $0x0,%ecx
  lea    D25(%rip),%r10
  lea    D29(%rip),%r9
  jmp    L13
L14:
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  je     L11
L13:
  movzbl (%rdx),%eax
  lea    -0x30(%rax),%r8d
  cmp    $0x9,%r8b
  ja     L14
  mov    %esi,%r8d
  and    $0x7,%r8d
  movsbl %al,%eax
  lea    0x2(%rax),%ebx
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%ebp
  sar    $0x1f,%ebp
  sub    %ebp,%eax
  lea    (%rax,%rax,4),%ebp
  add    %ebp,%ebp
  mov    %ebx,%eax
  sub    %ebp,%eax
  cltq
  lea    (%r8,%rax,8),%rax
  movslq (%r9,%rax,4),%rax
  movslq %ecx,%rcx
  lea    (%rcx,%rcx,4),%rcx
  lea    (%rax,%rcx,2),%rax
  mov    (%r10,%rax,4),%ecx
  lea    0x1(%rsi),%esi
  jmp    L14
L25:
  mov    $0x0,%esi
  mov    $0x0,%ecx
L11:
  and    $0x7,%esi
  lea    D29(%rip),%rax
  lea    (%rax,%rsi,4),%rsi
  mov    $0x0,%edx
  lea    D25(%rip),%r8
  movslq %ecx,%rax
  lea    (%rax,%rax,4),%rdi
  add    %rdi,%rdi
L53:
  movslq (%rsi),%rax
  add    %rdi,%rax
  cmpl   $0x0,(%r8,%rax,4)
  je     L48
  add    $0x1,%edx
  add    $0x20,%rsi
  cmp    $0xa,%edx
  jne    L53
L48:
  add    $0x30,%edx
  mov    %dl,(%r11)
  test   %ecx,%ecx
  sete   %al
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D25
type_infered: string
name: read_data
arguments: tag: D29
type_infered: string

tool: 

tool: 

assistant: ```c
int decchk(char string[], int n, char *ch) {
  char c;
  int j, k = 0, m = 0;
  static int ip[10][8] = {0, 1, 5, 8, 9, 4, 2, 7, 1, 5, 8, 9, 4, 2, 7, 0, 2, 7, 0, 1, 5, 8, 9, 4, 3, 6, 3, 6, 3, 6, 3, 6, 4, 2, 7, 0, 1, 5, 8, 9, 5, 8, 9, 4, 2, 7, 0, 1, 6, 3, 6, 3, 6, 3, 6, 3, 7, 0, 1, 5, 8, 9, 4, 2, 8, 9, 4, 2, 7, 0, 1, 5, 9, 4, 2, 7, 0, 1, 5, 8};
  static int ij[10][10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 0, 6, 7, 8, 9, 5, 2, 3, 4, 0, 1, 7, 8, 9, 5, 6, 3, 4, 0, 1, 2, 8, 9, 5, 6, 7, 4, 0, 1, 2, 3, 9, 5, 6, 7, 8, 5, 9, 8, 7, 6, 0, 4, 3, 2, 1, 6, 5, 9, 8, 7, 1, 0, 4, 3, 2, 7, 6, 5, 9, 8, 2, 1, 0, 4, 3, 8, 7, 6, 5, 9, 3, 2, 1, 0, 4, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0};
  for (j = 0; j < n; j++) {
    c = string[j];
    if (c >= 48 && c <= 57)
      k = ij[k][ip[(c + 2) % 10][7 & m++]];
  }
  for (j = 0; j <= 9; j++)
    if (!ij[k][ip[j][m & 7]])
      break;
  *ch = j + 48;
  return k == 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<decchk>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  test   %esi,%esi
  jle    L65
  lea    -0x1(%rsi),%eax
  xor    %ecx,%ecx
  xor    %esi,%esi
  mov    $0xcccccccd,%ebx
  lea    0x1(%rdi,%rax,1),%r8
  lea    D18(%rip),%r11
  lea    D21(%rip),%r10
L52:
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L8
  add    $0x2,%eax
  mov    %esi,%r9d
  add    $0x1,%esi
  movsbl %al,%edx
  and    $0x7,%r9d
  mov    %rdx,%rax
  imul   %rbx,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  lea    (%rcx,%rcx,4),%rdx
  cltq
  lea    (%r9,%rax,8),%rax
  movslq (%r10,%rax,4),%rax
  lea    (%rax,%rdx,2),%rax
  movslq (%r11,%rax,4),%rcx
L8:
  add    $0x1,%rdi
  cmp    %rdi,%r8
  jne    L52
  xor    %r8d,%r8d
  test   %ecx,%ecx
  movslq %ecx,%rax
  sete   %r8b
  and    $0x7,%esi
  lea    (%rax,%rax,4),%rax
  movslq (%r10,%rsi,4),%rdx
  lea    (%rdx,%rax,2),%rax
  mov    (%r11,%rax,4),%eax
L42:
  lea    (%rcx,%rcx,4),%rcx
  lea    (%r10,%rsi,4),%rsi
  xor    %edx,%edx
  lea    D18(%rip),%rdi
  add    %rcx,%rcx
  jmp    L53
  nopl   0x0(%rax)
L20:
  movslq (%rsi),%rax
  add    %rcx,%rax
  mov    (%rdi,%rax,4),%eax
L53:
  test   %eax,%eax
  je     L0
  add    $0x1,%edx
  add    $0x20,%rsi
  cmp    $0xa,%edx
  jne    L20
  mov    $0x3a,%edx
  mov    %r8d,%eax
  pop    %rbx
  mov    %dl,0x0(%rbp)
  pop    %rbp
  ret
  nop
L0:
  add    $0x30,%edx
  mov    %r8d,%eax
  pop    %rbx
  mov    %dl,0x0(%rbp)
  pop    %rbp
  ret
  nopl   0x0(%rax)
L65:
  xor    %eax,%eax
  mov    $0x1,%r8d
  xor    %ecx,%ecx
  xor    %esi,%esi
  lea    D21(%rip),%r10
  jmp    L42
```

tool_calls: name: parse_data
arguments: tag: D18
type_infered: string
name: parse_data
arguments: tag: D21
type_infered: string

tool: 

tool: 

assistant: ```c
int decchk(char string[], int n, char *ch) {
  char c;
  int j, k = 0, m = 0;
  static int ip[10][8] = {0, 1, 5, 8, 9, 4, 2, 7, 1, 5, 8, 9, 4, 2, 7, 0, 2, 7, 0, 1, 5, 8, 9, 4, 3, 6, 3, 6, 3, 6, 3, 6, 4, 2, 7, 0, 1, 5, 8, 9, 5, 8, 9, 4, 2, 7, 0, 1, 6, 3, 6, 3, 6, 3, 6, 3, 7, 0, 1, 5, 8, 9, 4, 2, 8, 9, 4, 2, 7, 0, 1, 5, 9, 4, 2, 7, 0, 1, 5, 8};
  static int ij[10][10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 0, 6, 7, 8, 9, 5, 2, 3, 4, 0, 1, 7, 8, 9, 5, 6, 3, 4, 0, 1, 2, 8, 9, 5, 6, 7, 4, 0, 1, 2, 3, 9, 5, 6, 7, 8, 5, 9, 8, 7, 6, 0, 4, 3, 2, 1, 6, 5, 9, 8, 7, 1, 0, 4, 3, 2, 7, 6, 5, 9, 8, 2, 1, 0, 4, 3, 8, 7, 6, 5, 9, 3, 2, 1, 0, 4, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0};
  for (j = 0; j < n; j++) {
    c = string[j];
    if (c >= 48 && c <= 57)
      k = ij[k][ip[(c + 2) % 10][7 & m++]];
  }
  for (j = 0; j <= 9; j++)
    if (!ij[k][ip[j][m & 7]])
      break;
  *ch = j + 48;
  return k == 0;
}
```