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

```asm
<barChartDataLoadForLocus>:
  endbr64
  push   %r12
  mov    %rsi,%rcx
  mov    %rdx,%r8
  mov    $0x200,%esi
  push   %rbp
  lea    D0(%rip),%rdx
  mov    %rdi,%rbp
  sub    $0x218,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <sqlSafef@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <barChartDataLoadByQuery@plt>
  mov    0x208(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x218,%rsp
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
datatype: string

tool: D0: SELECT * FROM %s WHERE locus='%s'

assistant: ```c
struct barChartData *barChartDataLoadForLocus(struct sqlConnection *conn, char *table, char *locus) {
  char query[512];
  sqlSafef(query, sizeof(query), "SELECT * FROM %s WHERE locus='%s'", table, locus);
  return barChartDataLoadByQuery(conn, query);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<negate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  neg    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int negate(int x) {
  int result;
L:
  result = -x;
  return (result);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<negate>:
  endbr64
  mov    %edi,%eax
  neg    %eax
  ret
```

assistant: ```c
int negate(int x) {
  int result;
L:
  result = -x;
  return (result);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<negate>:
  endbr64
  mov    %edi,%eax
  neg    %eax
  ret
```

assistant: ```c
int negate(int x) {
  int result;
L:
  result = -x;
  return (result);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<c_main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <read_burst@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int c_main() {
  read_burst();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<c_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <read_burst@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int c_main() {
  read_burst();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<c_main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <read_burst@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int c_main() {
  read_burst();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<c_main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <read_burst@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int c_main() {
  read_burst();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_cp_hash_value>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L4:
  movsbl -0x9(%rbp),%eax
  add    %eax,-0x8(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x0,-0x9(%rbp)
  jne    L4
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffffae4c415d,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x2f,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int get_cp_hash_value(char *name) {
  int hash = 0;
  int index = 0;
  char c;
  while ((c = name[index++]))
    hash += (int)c;
  hash %= 47;
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_cp_hash_value>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L27
  add    $0x1,%rdi
  mov    $0x0,%edx
L7:
  movsbl %al,%eax
  add    %eax,%edx
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L7
L3:
  movslq %edx,%rax
  imul   $0xffffffffae4c415d,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x2f,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  ret
L27:
  mov    $0x0,%edx
  jmp    L3
```

assistant: ```c
int get_cp_hash_value(char *name) {
  int hash = 0;
  int index = 0;
  char c;
  while ((c = name[index++]))
    hash += (int)c;
  hash %= 47;
  return hash;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_cp_hash_value>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L4
  add    $0x1,%rdi
  xor    %edx,%edx
  nopl   0x0(%rax)
L0:
  add    $0x1,%rdi
  add    %eax,%edx
  movsbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L0
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0xffffffffae4c415d,%rax,%rax
  sar    $0x1f,%ecx
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x5,%eax
  sub    %ecx,%eax
  imul   $0x2f,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int get_cp_hash_value(char *name) {
  int hash = 0;
  int index = 0;
  char c;
  while ((c = name[index++]))
    hash += (int)c;
  hash %= 47;
  return hash;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_cp_hash_value>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L14
  add    $0x1,%rdi
  xor    %edx,%edx
  nopl   0x0(%rax)
L2:
  add    $0x1,%rdi
  add    %eax,%edx
  movsbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L2
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0xffffffffae4c415d,%rax,%rax
  sar    $0x1f,%ecx
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x5,%eax
  sub    %ecx,%eax
  imul   $0x2f,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  xor    %eax,%eax
  ret
```

assistant: ```c
int get_cp_hash_value(char *name) {
  int hash = 0;
  int index = 0;
  char c;
  while ((c = name[index++]))
    hash += (int)c;
  hash %= 47;
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<examples>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rdi
  mov    -0x8(%rbp),%rsi
  mov    -0x8(%rbp),%rcx
  mov    -0x8(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  push   -0x8(%rbp)
  push   -0x8(%rbp)
  push   -0x8(%rbp)
  mov    %rdi,%r9
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  add    $0x20,%rsp
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
type: string

tool: D3: 
Request activation for Office2013 using V4 protocol from 192.168.1.5:1688
	%s -l O15 -4 192.168.1.5
	%s -l O15 -4 192.168.1.5:1688

Request activation for Windows Server 2012 using V4 protocol from localhost:1688
	%s -4 -l Windows -k 8665cb71-468c-4aa3-a337-cb9bc9d5eaac
	%s -4 -l 2012
	%s -4 -l 2012 [::1]:1688
	%s -4 -l 12 127.0.0.2:1688

Send 100,000 requests to localhost:1688
	%s -n 100000 -l Office2010

Request Activation for Windows 8 from 10.0.0.1:4711 and pretend to be Steve Ballmer
	%s -l Windows8 -w steveb1.redmond.microsoft.com 10.0.0.1:4711



assistant: ```c
__attribute__((noreturn)) void examples(const char *const programName) {
  printf("\nRequest activation for Office2013 using V4 protocol from 192.168.1.5:1688\n"
         "\t%s -l O15 -4 192.168.1.5\n"
         "\t%s -l O15 -4 192.168.1.5:1688\n\n"
         "Request activation for Windows Server 2012 using V4 protocol from localhost:1688\n"
         "\t%s -4 -l Windows -k 8665cb71-468c-4aa3-a337-cb9bc9d5eaac\n"
         "\t%s -4 -l 2012\n"
         "\t%s -4 -l 2012 [::1]:1688\n"
         "\t%s -4 -l 12 127.0.0.2:1688\n\n"
         "Send 100,000 requests to localhost:1688\n"
         "\t%s -n 100000 -l Office2010\n\n"
         "Request Activation for Windows 8 from 10.0.0.1:4711 and pretend to be Steve Ballmer\n"
         "\t%s -l Windows8 -w steveb1.redmond.microsoft.com 10.0.0.1:4711\n\n",
         programName, programName, programName, programName, programName, programName, programName, programName);
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<examples>:
  endbr64
  push   %rax
  pop    %rax
  sub    $0x10,%rsp
  mov    %rdi,%rsi
  push   %rdi
  push   %rdi
  push   %rdi
  mov    %rdi,%r9
  mov    %rdi,%r8
  mov    %rdi,%rcx
  mov    %rdi,%rdx
  lea    D8(%rip),%rdi
  call   <printf@plt>
  add    $0x20,%rsp
  mov    $0x0,%edi
  call   <exit@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D8
type_infered: string

tool: D8: 
Request activation for Office2013 using V4 protocol from 192.168.1.5:1688
	%s -l O15 -4 192.168.1.5
	%s -l O15 -4 192.168.1.5:1688

Request activation for Windows Server 2012 using V4 protocol from localhost:1688
	%s -4 -l Windows -k 8665cb71-468c-4aa3-a337-cb9bc9d5eaac
	%s -4 -l 2012
	%s -4 -l 2012 [::1]:1688
	%s -4 -l 12 127.0.0.2:1688

Send 100,000 requests to localhost:1688
	%s -n 100000 -l Office2010

Request Activation for Windows 8 from 10.0.0.1:4711 and pretend to be Steve Ballmer
	%s -l Windows8 -w steveb1.redmond.microsoft.com 10.0.0.1:4711



assistant: ```c
__attribute__((noreturn)) void examples(const char *const programName) {
  printf("\nRequest activation for Office2013 using V4 protocol from 192.168.1.5:1688\n"
         "\t%s -l O15 -4 192.168.1.5\n"
         "\t%s -l O15 -4 192.168.1.5:1688\n\n"
         "Request activation for Windows Server 2012 using V4 protocol from localhost:1688\n"
         "\t%s -4 -l Windows -k 8665cb71-468c-4aa3-a337-cb9bc9d5eaac\n"
         "\t%s -4 -l 2012\n"
         "\t%s -4 -l 2012 [::1]:1688\n"
         "\t%s -4 -l 12 127.0.0.2:1688\n\n"
         "Send 100,000 requests to localhost:1688\n"
         "\t%s -n 100000 -l Office2010\n\n"
         "Request Activation for Windows 8 from 10.0.0.1:4711 and pretend to be Steve Ballmer\n"
         "\t%s -l Windows8 -w steveb1.redmond.microsoft.com 10.0.0.1:4711\n\n",
         programName, programName, programName, programName, programName, programName, programName, programName);
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<examples>:
  endbr64
  push   %rax
  pop    %rax
  sub    $0x10,%rsp
  mov    %rdi,%rsi
  mov    %rdi,%r9
  mov    %rdi,%r8
  push   %rdi
  mov    %rdi,%rcx
  mov    %rdi,%rdx
  push   %rdi
  push   %rdi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  add    $0x20,%rsp
  xor    %edi,%edi
  call   <exit@plt>
```

tool_calls: name: parse_data
arguments: data_label: D0
datatype: string

tool: D0: 
Request activation for Office2013 using V4 protocol from 192.168.1.5:1688
	%s -l O15 -4 192.168.1.5
	%s -l O15 -4 192.168.1.5:1688

Request activation for Windows Server 2012 using V4 protocol from localhost:1688
	%s -4 -l Windows -k 8665cb71-468c-4aa3-a337-cb9bc9d5eaac
	%s -4 -l 2012
	%s -4 -l 2012 [::1]:1688
	%s -4 -l 12 127.0.0.2:1688

Send 100,000 requests to localhost:1688
	%s -n 100000 -l Office2010

Request Activation for Windows 8 from 10.0.0.1:4711 and pretend to be Steve Ballmer
	%s -l Windows8 -w steveb1.redmond.microsoft.com 10.0.0.1:4711



assistant: ```c
__attribute__((noreturn)) void examples(const char *const programName) {
  printf("\nRequest activation for Office2013 using V4 protocol from 192.168.1.5:1688\n"
         "\t%s -l O15 -4 192.168.1.5\n"
         "\t%s -l O15 -4 192.168.1.5:1688\n\n"
         "Request activation for Windows Server 2012 using V4 protocol from localhost:1688\n"
         "\t%s -4 -l Windows -k 8665cb71-468c-4aa3-a337-cb9bc9d5eaac\n"
         "\t%s -4 -l 2012\n"
         "\t%s -4 -l 2012 [::1]:1688\n"
         "\t%s -4 -l 12 127.0.0.2:1688\n\n"
         "Send 100,000 requests to localhost:1688\n"
         "\t%s -n 100000 -l Office2010\n\n"
         "Request Activation for Windows 8 from 10.0.0.1:4711 and pretend to be Steve Ballmer\n"
         "\t%s -l Windows8 -w steveb1.redmond.microsoft.com 10.0.0.1:4711\n\n",
         programName, programName, programName, programName, programName, programName, programName, programName);
  exit(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
<examples>:
  endbr64
  push   %rax
  pop    %rax
  sub    $0x10,%rsp
  mov    %rdi,%rsi
  mov    %rdi,%r9
  mov    %rdi,%r8
  push   %rdi
  mov    %rdi,%rcx
  mov    %rdi,%rdx
  push   %rdi
  push   %rdi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  add    $0x20,%rsp
  xor    %edi,%edi
  call   <exit@plt>
```

tool_calls: name: parse_data
arguments: data_label: D1
type: string

tool: D1: 
Request activation for Office2013 using V4 protocol from 192.168.1.5:1688
	%s -l O15 -4 192.168.1.5
	%s -l O15 -4 192.168.1.5:1688

Request activation for Windows Server 2012 using V4 protocol from localhost:1688
	%s -4 -l Windows -k 8665cb71-468c-4aa3-a337-cb9bc9d5eaac
	%s -4 -l 2012
	%s -4 -l 2012 [::1]:1688
	%s -4 -l 12 127.0.0.2:1688

Send 100,000 requests to localhost:1688
	%s -n 100000 -l Office2010

Request Activation for Windows 8 from 10.0.0.1:4711 and pretend to be Steve Ballmer
	%s -l Windows8 -w steveb1.redmond.microsoft.com 10.0.0.1:4711



assistant: ```c
__attribute__((noreturn)) void examples(const char *const programName) {
  printf("\nRequest activation for Office2013 using V4 protocol from 192.168.1.5:1688\n"
         "\t%s -l O15 -4 192.168.1.5\n"
         "\t%s -l O15 -4 192.168.1.5:1688\n\n"
         "Request activation for Windows Server 2012 using V4 protocol from localhost:1688\n"
         "\t%s -4 -l Windows -k 8665cb71-468c-4aa3-a337-cb9bc9d5eaac\n"
         "\t%s -4 -l 2012\n"
         "\t%s -4 -l 2012 [::1]:1688\n"
         "\t%s -4 -l 12 127.0.0.2:1688\n\n"
         "Send 100,000 requests to localhost:1688\n"
         "\t%s -n 100000 -l Office2010\n\n"
         "Request Activation for Windows 8 from 10.0.0.1:4711 and pretend to be Steve Ballmer\n"
         "\t%s -l Windows8 -w steveb1.redmond.microsoft.com 10.0.0.1:4711\n\n",
         programName, programName, programName, programName, programName, programName, programName, programName);
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<AllocKey>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %ebx,%eax
  add    $0x4,%eax
  mov    %eax,%edi
  call   <Alloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rsi
  mov    -0x2c(%rbp),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rsi,%r8
  lea    D8(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
type: string

tool: D8: %s
%c
%s

assistant: ```c
char *AllocKey(char *normal, int type, char *abbrev) {
  char *tmp = Alloc((unsigned)(4 + strlen(normal) + strlen(abbrev)));
  (void)sprintf(tmp, "%s\n%c\n%s", normal, type, abbrev);
  return tmp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<AllocKey>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r13d
  mov    %rdx,%r12
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x4(%rbx,%rax,1),%edi
  call   <Alloc@plt>
  mov    %rax,%rbx
  mov    %r12,%r8
  mov    %r13d,%ecx
  mov    %rbp,%rdx
  lea    D0(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
type_infered: string

tool: D0: %s
%c
%s

assistant: ```c
char *AllocKey(char *normal, int type, char *abbrev) {
  char *tmp = Alloc((unsigned)(4 + strlen(normal) + strlen(abbrev)));
  (void)sprintf(tmp, "%s\n%c\n%s", normal, type, abbrev);
  return tmp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<AllocKey>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x4(%r12,%rax,1),%edi
  call   <Alloc@plt>
  mov    %rbx,%r8
  mov    %r14d,%ecx
  mov    %r13,%rdx
  lea    D4(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D4
data_type: string

tool: D4: %s
%c
%s

assistant: ```c
char *AllocKey(char *normal, int type, char *abbrev) {
  char *tmp = Alloc((unsigned)(4 + strlen(normal) + strlen(abbrev)));
  (void)sprintf(tmp, "%s\n%c\n%s", normal, type, abbrev);
  return tmp;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<AllocKey>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x4(%r12,%rax,1),%edi
  call   <Alloc@plt>
  mov    %rbx,%r8
  mov    %r14d,%ecx
  mov    %r13,%rdx
  lea    D1(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: label: D1
type_infered: string

tool: D1: %s
%c
%s

assistant: ```c
char *AllocKey(char *normal, int type, char *abbrev) {
  char *tmp = Alloc((unsigned)(4 + strlen(normal) + strlen(abbrev)));
  (void)sprintf(tmp, "%s\n%c\n%s", normal, type, abbrev);
  return tmp;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <test_assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int FORWARD = 0;
  char *a = "baba";
  char *(*f)(const char *s, int c) = strchr;
  if (FORWARD) {
    f = strrchr;
  }
  char *x = f(a, 'b');
  test_assert(x == a);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <test_assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int FORWARD = 0;
  char *a = "baba";
  char *(*f)(const char *s, int c) = strchr;
  if (FORWARD) {
    f = strrchr;
  }
  char *x = f(a, 'b');
  test_assert(x == a);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <test_assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int FORWARD = 0;
  char *a = "baba";
  char *(*f)(const char *s, int c) = strchr;
  if (FORWARD) {
    f = strrchr;
  }
  char *x = f(a, 'b');
  test_assert(x == a);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<I2CMasterInitExpClk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <I2CMasterEnable@plt>
  cmpl   $0x1,-0x1c(%rbp)
  jne    L26
  movl   $0x61a80,-0x8(%rbp)
  jmp    L19
L26:
  movl   $0x186a0,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  lea    -0x1(%rax),%ecx
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  mov    %eax,%esi
  mov    %ecx,%eax
  cltd
  idiv   %esi
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0xc,%eax
  cltq
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  mov    %rax,(%rdx)
  mov    -0x14(%rbp),%eax
  add    $0xfc0,%eax
  cltq
  mov    (%rax),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L22
  mov    -0x18(%rbp),%eax
  add    $0x137477f,%eax
  movslq %eax,%rdx
  imul   $0x6944d07d,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x17,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  or     $0x80,%al
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    $0xc,%eax
  cltq
  mov    %rax,%rcx
  movslq %edx,%rax
  mov    %rax,(%rcx)
L22:
  nop
  leave
  ret
```

assistant: ```c
void I2CMasterInitExpClk(uint32_t ui32Base, uint32_t ui32I2CClk, bool bFast) {
  uint32_t ui32SCLFreq;
  uint32_t ui32TPR;
  ;
  I2CMasterEnable(ui32Base);
  if (bFast == 1) {
    ui32SCLFreq = 400000;
  } else {
    ui32SCLFreq = 100000;
  }
  ui32TPR = ((ui32I2CClk + (2 * 10 * ui32SCLFreq) - 1) / (2 * 10 * ui32SCLFreq)) - 1;
  (*((volatile unsigned long *)(ui32Base + 0x0000000C))) = ui32TPR;
  if ((*((volatile unsigned long *)(ui32Base + 0x00000FC0))) & 0x00000001) {
    ui32TPR = ((ui32I2CClk + (2 * 3 * 3400000) - 1) / (2 * 3 * 3400000)) - 1;
    (*((volatile unsigned long *)(ui32Base + 0x0000000C))) = 0x00000080 | ui32TPR;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<I2CMasterInitExpClk>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edx,%r12d
  call   <I2CMasterEnable@plt>
  cmp    $0x1,%r12d
  mov    $0x61a80,%eax
  mov    $0x186a0,%edx
  cmovne %edx,%eax
  lea    (%rax,%rax,4),%esi
  shl    $0x2,%esi
  lea    0xc(%rbx),%ecx
  movslq %ecx,%rcx
  lea    -0x1(%rbp,%rsi,1),%eax
  cltd
  idiv   %esi
  sub    $0x1,%eax
  cltq
  mov    %rax,(%rcx)
  add    $0xfc0,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%rax
  test   $0x1,%al
  je     L2
  add    $0x137477f,%ebp
  movslq %ebp,%rax
  imul   $0x6944d07d,%rax,%rax
  sar    $0x37,%rax
  sar    $0x1f,%ebp
  sub    %ebp,%eax
  sub    $0x1,%eax
  or     $0x80,%al
  cltq
  mov    %rax,(%rcx)
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void I2CMasterInitExpClk(uint32_t ui32Base, uint32_t ui32I2CClk, bool bFast) {
  uint32_t ui32SCLFreq;
  uint32_t ui32TPR;
  ;
  I2CMasterEnable(ui32Base);
  if (bFast == 1) {
    ui32SCLFreq = 400000;
  } else {
    ui32SCLFreq = 100000;
  }
  ui32TPR = ((ui32I2CClk + (2 * 10 * ui32SCLFreq) - 1) / (2 * 10 * ui32SCLFreq)) - 1;
  (*((volatile unsigned long *)(ui32Base + 0x0000000C))) = ui32TPR;
  if ((*((volatile unsigned long *)(ui32Base + 0x00000FC0))) & 0x00000001) {
    ui32TPR = ((ui32I2CClk + (2 * 3 * 3400000) - 1) / (2 * 3 * 3400000)) - 1;
    (*((volatile unsigned long *)(ui32Base + 0x0000000C))) = 0x00000080 | ui32TPR;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<I2CMasterInitExpClk>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  call   <I2CMasterEnable@plt>
  cmp    $0x1,%r12d
  mov    $0x1e8480,%eax
  lea    0xc(%rbx),%ecx
  mov    $0x7a1200,%esi
  movslq %ecx,%rcx
  cmovne %eax,%esi
  add    $0xfc0,%ebx
  movslq %ebx,%rbx
  lea    -0x1(%rsi,%rbp,1),%eax
  cltd
  idiv   %esi
  sub    $0x1,%eax
  cltq
  mov    %rax,(%rcx)
  mov    (%rbx),%rax
  test   $0x1,%al
  je     L2
  add    $0x137477f,%ebp
  movslq %ebp,%rax
  sar    $0x1f,%ebp
  imul   $0x6944d07d,%rax,%rax
  sar    $0x37,%rax
  sub    %ebp,%eax
  sub    $0x1,%eax
  or     $0x80,%al
  cltq
  mov    %rax,(%rcx)
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void I2CMasterInitExpClk(uint32_t ui32Base, uint32_t ui32I2CClk, bool bFast) {
  uint32_t ui32SCLFreq;
  uint32_t ui32TPR;
  ;
  I2CMasterEnable(ui32Base);
  if (bFast == 1) {
    ui32SCLFreq = 400000;
  } else {
    ui32SCLFreq = 100000;
  }
  ui32TPR = ((ui32I2CClk + (2 * 10 * ui32SCLFreq) - 1) / (2 * 10 * ui32SCLFreq)) - 1;
  (*((volatile unsigned long *)(ui32Base + 0x0000000C))) = ui32TPR;
  if ((*((volatile unsigned long *)(ui32Base + 0x00000FC0))) & 0x00000001) {
    ui32TPR = ((ui32I2CClk + (2 * 3 * 3400000) - 1) / (2 * 3 * 3400000)) - 1;
    (*((volatile unsigned long *)(ui32Base + 0x0000000C))) = 0x00000080 | ui32TPR;
  }
}
```