[英](C++) Reversing a string using stacks?
我正在尝试使用堆栈来反转字符串。 它正确地反转了字符串,但是当我达到0时,for循环崩溃。我收到“ 字符串下标超出范围 ”错误。 当前,for循环仅递减为1。如何获取并推送它并显示s1 [0]?
这是主要代码:
#include <cstdlib> // Provides EXIT_SUCCESS
#include <iostream> // Provides cin, cout
#include <stack> // Provides stack
#include <string> // Provides string
using namespace std;
. . .
string reverse(string & s1)
{
stack<char> stk1;
string::size_type i;
// this for loop sets the rest of the characters
for (i = s1.size() - 1; i > 0; i--)
{
stk1.push(s1[i]);
cout << stk1.top();
}
return "The function was a success. Now that's what I call reverse psychology.";
}
这是头文件:
#ifndef MAIN_SAVITCH_STACK1_H
#define MAIN_SAVITCH_STACK1_H
#include <cstdlib> // Provides size_t
namespace main_savitch_7A
{
template <class Item>
class stack
{
public:
// TYPEDEFS AND MEMBER CONSTANT -- See Appendix E if this fails to compile.
typedef std::size_t size_type;
typedef Item value_type;
static const size_type CAPACITY = 30;
// CONSTRUCTOR
stack( ) { used = 0; }
// MODIFICATION MEMBER FUNCTIONS
void push(const Item& entry);
void pop( );
// CONSTANT MEMBER FUNCTIONS
bool empty( ) const { return (used == 0); }
size_type size( ) const { return used; }
Item top( ) const;
private:
Item data[CAPACITY]; // Partially filled array
size_type used; // How much of array is being used
};
}
#include "stack1.template" // Include the implementation.
#endif
这是堆栈实现(模板文件):
#include <cassert> // Provides assert
namespace main_savitch_7A
{
template <class Item>
const typename stack<Item>::size_type stack<Item>::CAPACITY;
template <class Item>
void stack<Item>::push(const Item& entry)
// Library facilities used: cassert
{
assert(size( ) < CAPACITY);
data[used] = entry;
++used;
}
template <class Item>
void stack<Item>::pop( )
// Library facilities used: cassert
{
assert(!empty( ));
--used;
}
template <class Item>
Item stack<Item>::top( ) const
// Library facilities used: cassert
{
assert(!empty( ));
return data[used-1];
}
}
我想将for循环更改为此,但是它不起作用:
// this for loop sets the rest of the characters
for (i = s1.size() - 1; i >= 0; i--) // i > -1 doesn't work either
{
stk1.push(s1[i]);
cout << stk1.top();
}
cout << s1[0] << "\n\n";
return "The function was a success. Now that's what I call reverse psychology.";
}
我是无符号的,因此如果它等于0,则它在减量时会回绕。您需要对其使用带符号的类型,或者检查边界条件而不涉及负数(即,不要将其与-1比较并执行如果为0,则不递减)。
我可以想到以下几种选择。
使用string::size_type
作为循环计数器:
string::size_type i;
for (i = s1.size(); i > 0; i--)
{
stk1.push(s1[i-1]);
cout << stk1.top();
}
要么
将int
用作循环计数器:
int i = 0;
for (i = s1.size()-1; i >= 0; i--)
{
stk1.push(s1[i]);
cout << stk1.top();
}
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