使用指针表示法反转cstring时出现意外结果

时间:2015-08-09 03:36:28

标签: c++ pointers c-strings

我将两个c字符串(一个包含字符,另一个空为空)传递给一个函数,该函数应该包含包含字符的c-string,并使用从后面读取的字符串填充空c-string。操作简单,但不起作用。我无法弄清楚我做错了什么。

#include <iostream>
#include <cstring>

using namespace std;

void magic(char*, const char*, int);

int main()
{
   char string1[20];
   char string2[] = "This is a test";
   char *str1 = string1; 
   char *str2 = string2;
   int stringsize = strlen(string2);
   magic(str1, str2, stringsize);
   return 0;
}

void magic(char* string1, const char* string2, int stringsize)
{
   cout << string2 << endl; //here it prints as expected
   *(string1 + stringsize) = '\0';
   string1 = string1 + stringsize - 1;
   for(;*string2 != '\0' ;string1--, string2++)
   {
      *string1 = *string2;
   }
   cout << string1 << endl; //blank space is printed - expected "This is a test" reversed
   cout << string2 << endl; //blank space is printed - expected "This is a test" 
}

1 个答案:

答案 0 :(得分:1)

如果您使用非常简单的字符串执行for循环中的步骤,则可以在逻辑中看到问题。让我们说吧,

string1 = "This";

在循环的第一次迭代中,

string1 = "This";
string2 = "T";

在循环的第二次迭代中,

string1 = "his";
string2 = "hT";

在循环的第三次迭代中,

string1 = "is";
string2 = "ihT";

在循环的第四次迭代中,

string1 = "s";
string2 = "sihT";

在循环的第五次迭代中,

string1 = "";
string2 = "?sihT";  // You go past the start of the array.

因此,在for循环结束后,您需要递增string2以确保它指向数组的开头。

void magic(char* string1, const char* string2, int stringsize)
{
   cout << string2 << endl;
   *(string1 + stringsize) = '\0';
   string1 = string1 + stringsize-1;
   for(;*string2 != '\0' ;string1--, string2++)
   {
      *string1 = *string2;
   }

   // Add this.
   string1++;

   cout << string1 << endl; // This is now the reversed string.
   cout << string2 << endl; // This is now the empty string since
                            // you kept incrementing it in the for
                            // loop until you reached the end of the
                            // string.
}