在 C++ 中添加两个 C 字符串

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，那么至少使用 C 标准库函数，这样你就有希望得到正确的 null 终止。您要查找的函数是 std::strcat ，它连接两个字符串。

之后，使用 std::string .

您在stringAdd中有一个拼写错误，导致错误

size1=+size2;

这应该是

size1 += size2;

否则，您只是用 size2 的值覆盖size1。话虽如此，在C++您也不允许这样做

char newWord[size1];

数组的大小必须在编译时而不是运行时知道。

stringAdd不正确，并且具有未定义的行为，因为它不返回任何内容。

还有这个 if 语句在函数stringLength

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 while 语句中的条件

while (*strPtr != '')

与 if 语句中的相同。

{

函数可以按以下方式编写

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;
}

总的来说，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。

你把它标记为"c ++"和"c-strings"，这有点像要求一辆由脚力驱动的汽车。

您有 2 个选项：

1. 使用C++字符串
2. 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;
}

为什么你做的事情是错误的：

// 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 != '')
    {
        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
}