Question

所以在c ++中，正常的cstyle字符串可以像这样启动：

   char cString[] = "my string";

如何使用动态内存做同样的事情？

为什么这不起作用？

   char *charPtr; 

   charPtr = new char("make this the value of the c string");

感谢您的帮助

Answer 1

你不能像这样初始化一个c风格的字符串。首先，你需要一个指向字符串开头的指针。然后需要分配足够的内存来保存该字符串，即字符数+终止'\ 0'。对于strlen和strcpy，您应该包含cstring标头。最后，您需要将字符串复制到内存块。

   char *str; //pointer to string
   str = new char[strlen("The string to copy")+1]; //allocate memory 
   //strlen returns the number of charaters of a string exluding the '\0', so we need to add 1
   //either use a string constant, or another cstyle string, or input the length manually
   strcopy(str, "The string to copy");//strcpy copies the content of the string constant to str
   //be sure to set the last element of the string to '\0'
   str[strlen(str)] = 0; // or '\0'
   //if str has 5 char it returns 4 because the last is \0
   //but the array goes from 0 to 4, so 4 is the last element

并且不要忘记使用delete [] str;如果你摆脱了字符串！

c ++在动态内存中初始化cstyle字符串的默认值

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