C/C++函数调用栈的实现方法_C 语言

本文实例讲述了C/C++函数调用栈的实现方法。可用于实现简单的脚本解释器。分享给大家供大家参考。具体实现方法如下：

头文件声明部分:

复制代码 代码如下:

#pragma once
const int BUFFERSIZE = 1024;
const int growfactor = 2;
 
// this stack is used as call stack.
class TStack{
private:
size_t size;   // the stack length
size_t pos;    // the stack top position   
char *buffer;  // the buffer

private:
void push(void* D, size_t bytecount);  // the implementation of push
void* pop(size_t bytecount);   // the implementation of pop
public:
TStack(size_t _size = BUFFERSIZE, size_t _pos = 0);  // initialize
TStack(const TStack& o);  // copy
TStack& operator=(const TStack& o);  // assignment
void pushInt(int i) { push(&i, sizeof(int)); }  // push an int
void pushLong(long l) { push(&l, sizeof(long)); }  // push a long
void pushfloat(double f) { push(&f, sizeof(f));}  // push  a double
void pushPointer(void* p){ push(p, sizeof(p)); }
// int
int popInt() { return *(int *)pop(sizeof(int));}  // pop an int
long popLong() { return *(long *)pop(sizeof(long)); }  // pop an int   
double* popfloat() { return (double*)pop(sizeof(double)); }  // pop a double
void* popPointer() { return pop(sizeof(void*)) ; }
void clear() { pos = 0; } 
};

 
实现部分：
 

复制代码 代码如下:

#include "stdafx.h"
#include "TStack.h"
#include "new.h"
 
void TStack::push( void* D, size_t bytecount )
{
// if memory is not enough
// if run under multithread envionment,
// a lock or critical section should be added
if (pos + bytecount > size)
{   
  size_t oldsize = size;
       size *= growfactor;  
  char *newbuffer = new char[size];
  memcpy(newbuffer, buffer, oldsize);
  delete buffer;
  buffer = newbuffer;   
}
memcpy(buffer+pos, D, bytecount);
pos += bytecount;
}
 
void* TStack::pop( size_t bytecount )
{
// need synchronization for multithread environment
pos -= bytecount;
return &buffer[pos];
}
 
TStack::TStack( size_t _size , size_t _pos )
:size(_size),
pos(_pos),
buffer(new char[size])
{
}
 
TStack::TStack( const TStack &O )
:size(O.size),
pos(O.pos)
{
   buffer = new char[size];
   if (buffer != NULL)
   {
  memcpy(buffer, O.buffer, size);
   }
}
 
TStack& TStack::operator=( const TStack& O )
{
if (this == &O)
 return *this;     
    this->size = O.size;
this->pos = O.pos;
 
if (buffer != NULL)
{
delete buffer;
}
    buffer = new char[this->size];
if (buffer != NULL)
{
      memcpy(buffer, O.buffer, this->size);
}
return *this;
}

希望本文所述对大家的C++程序设计有所帮助。