Question

在我的代码中，我试图将两个字符串添加到一起，但由于某种原因，我似乎无法为我的stringAdd函数获取正确的返回类型。我希望能够返回一个c字符串。我的实施似乎也没有用。有什么建议吗？

 #include <iostream>
 #include<cstring>

 using namespace std;
 int stringLength(char *); // Function prototype 
 char stringAdd(char *strPtr, char *strPtr2);//Function prototype

int main()
{
   const int SIZE = 51; // Array size 
   char letter; // The character to count 
   char word1[SIZE] = "Happy ";
   char word2[SIZE] = "Birthday";

   cout <<"Your first c-string is: "<<word1<<"Your second c-string is: "<<word2<<"\n";
   cout << "The length of your first c-string is: ";

   cout << stringLength(word1) << " chars long.\n";
   cout << "The length of your second c-string is: ";

   cout << stringLength(word2) << " chars long.\n";

   if (SIZE >= (stringLength(word1) + stringLength(word2) + 1))
   {
      cout << "we are gunna add ur strings";
      stringAdd(word1, word2);
   }
   else
   {
      cout << "String1 is not large enough for both strings.\n";
   }
   return 0;
}

int stringLength(char *strPtr)
{
    int times = 0; // Number of times a char appears in the string 

    // Step through the string each char. 
    while (*strPtr != '\0')
    {
        if (*strPtr != '0') // If the current character doesnt equals the null terminator... 
           times++; // Increments the counter 
        strPtr++; // Goes to the next char in the string. 
    }
    return times;
}

到目前为止，我的代码工作正常但是下面的函数似乎根本不起作用。我不确定如何使用参考

添加两个c字符串

char stringAdd(char *strPtr, char *strPtr2)
{
   int size1;
   int size2;
   size1= stringLength(strPtr);
   int j=size1+1; // counter set to the num of chars in the first c-string
   int i = 0; // counter for to add to the 2nd c-string
   size2= stringLength(strPtr2);
   size1=+size2;
   char newWord[size1];

   for(int i=0;i<size1;i++)
      newWord[i] = *strPtr[i]
   for(int j=0;j<size2;j++)
      newWord[i]= *str
}

Answer 1

首先，使用std::string。

然后，使用std::string。

最后，如果你真的必须手动操作char数组，那么至少要使用C标准库函数，这样你就有希望得到正确的空终止。您正在寻找的功能是std::strcat，它连接两个字符串。

之后，请使用std::string。

Answer 2

stringAdd中的拼写错误会导致错误

size1=+size2;

这应该是

size1 += size2;

否则，您只需使用size1的值覆盖size2。话虽如此，在C ++中你也不允许这样做

char newWord[size1];

必须在编译时而不是运行时知道数组的大小。

Answer 3

函数stringAdd不正确，并且具有未定义的行为，因为它不返回任何内容。

此函数stringLength中的if语句

if (*strPtr != '0') // If the current character doesnt equals the null terminator... 

    times++; // Increments the counter

doea没有多大意义，因为封闭的while语句中的条件

while (*strPtr != '\0')

与if语句中的相同。

{

可以按以下方式编写函数

size_t stringLength( const char *strPtr )
{
    size_t n = 0;

    while ( strPtr[n] ) ++n;

    return n;
}


char * stringAdd( char *strPtr, const char *strPtr2 )
{
    char *p = strPtr + stringLength( strPtr );

    while ( *p++ = *strPtr2++ );

    return strPtr;
}

在主要内容你可以写

if (SIZE >= (stringLength(word1) + stringLength(word2) + 1)) {
    cout << "we are gunna add ur strings" << endl;
    cout << stringAdd(word1, word2) << endl;
}
//...

在这种情况下，word2将附加到word1。

Answer 4

您已将此标记为＆＃34; c ++＆＃34;和＆＃34; c-strings＆＃34;，这有点像要求用脚力驱动的汽车。

您有两个选择：

使用C ++字符串
使用C字符串

对于前者：

#include <iostream>
#include <string>

int main()
{
    std::string word1 = "Happy";
    std::string word2 = "Birthday";
    // ... your other stuff
    std::string result = word1 + " " + word2 + "!";
    std::cout << "Result is " << result << std::endl;
    return 0;
}

对于后者：

#include <iostream> // if you are stuck using c-strings, this is kind of odd
#include <cstring>
#include <memory>

int main()
{
    const char* word1 = "Happy";
    const char* word2 = "Birthday";
    const unsigned int newWordSize = 20; // you only need 16 for this, so 20 is sufficient
    char newWord[newWordSize];
    std::memset(newWord, newWordSize, 0);
    std::strcpy(newWord, word1);
    std::strcat(newWord, " ");
    std::strcat(newWord, word2);
    std::strcat(newWord, "!");
    std::cout << "New Word is " << newWord << std::endl;
    return 0;
}

为什么你做错了：

// NOTE:  If your null-terminators are not set, this breaks as it is an infinite loop.
int stringLength(char *strPtr)
{
    // NOTE:  pointer arithmetic can speed up this function, and make this variable unnecessary
    int times = 0; // Number of times a char appears in the string 

    // Step through the string each char. 
    while (*strPtr != '\0')
    {
        if (*strPtr != '0') // If the current character doesnt equals the null terminator... 
           times++; // Increments the counter 
        strPtr++; // Goes to the next char in the string. 
    }
    return times;
}

char stringAdd(char *strPtr, char *strPtr2) // ERROR:  your return value should probably be char*, and you will need to free that memory later
{
   int size1;
   int size2;
   size1= stringLength(strPtr);
   int j=size1+1; // counter set to the num of chars in the first c-string
   int i = 0; // counter for to add to the 2nd c-string
   size2= stringLength(strPtr2);
   size1=+size2;
   char newWord[size1]; // ERROR:  you cannot allocate a dynamic array this way.

   for(int i=0;i<size1;i++) // ERROR:  you've set size1 = size1 + size2 + 1, and you attempt to access the first word with this new size.  You will access memory outside the bounds of your array
      newWord[i] = *strPtr[i]
   for(int j=0;j<size2;j++)
      newWord[i]= *str
    // ERROR:  You do not set the null-terminator for the new string
    // ERROR:  you do not return anything
}

在c ++中添加两个c字符串

4 个答案: