試圖使用自己的strcpy並複製一個數組元素不工作？

Question

親愛的朋友， 我是新來的，請檢查我的一段代碼。我的意圖是將名稱複製到結構數組元素中。我是c新手，無法理解發生了什麼......請引導我？

#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 

typedef struct new_st{ 

     const char name[100]; 
     int value; 

    }var1; 

char *myStringcopy(char *dst, const char *src) 
{ 
     char *ptr; 
     ptr = dst; 
     while(*dst++!=NULL) 
     { 
      *dst++=*src++; 
      } 

     return(ptr); 
} 

int main() 
{ 

    char my_str[] = {"HelloWord", "MY var1", "my var2"}; 

    var1 *new_st1; 


    new_st1 = malloc(sizeof(struct new_st)); 


    //trying just first name then i thought of using for loop for rest 


     myStringcopy(my_str, new_st1->name[0]); 

     printf("%s\n",new_st1->name[0]); 



    return 0; 


} 

Answer 1

在這個函數char *myStringcopy(char *dst, const char *src)你的第一個參數是目的地。但是你用源地址作爲第一個參數來調用這個函數。

您遞增目的地址，而條件，並在同時身體*dst++=*src++;

你而條件檢查的內容等於NULL

String數組decalration應該是這樣的char *my_str[] = {"HelloWord", "MY var1", "my var2"};兩次INT函數在循環while(*dst++!=NULL)

Answer 2

坦率地說，你的代碼似乎有很多邏輯錯誤。這是我的修復：

#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 

typedef struct new_st{ 
    char name[100]; //It should not be const because the content inside the array is intended to be modified. 
    int value; 
}var1; 

char *myStringcopy(char *dst, const char *src) 
{ 
     char *ptr; 
     ptr = dst; 
     while(*src!='\0') //src should be copy until a '\0' in src is reach. 
      *dst++=*src++; 
     *dst='\0'; //making dst a null-terminated string 

     return(ptr); 
} 

int main() 
{ 

    const char* my_str[] = {"HelloWord", "MY var1", "my var2"}; //I have no idea why your code in this line even compile. The type of my_str should be char**, or an array of char* . Also, you should use const if the string will not be modified. 
    var1 *new_st1; 

    new_st1 = malloc(sizeof(struct new_st)); 
    myStringcopy(new_st1->name, my_str[0]); //new_st1->name[0] is a char. There is no reason to copy a char. Instead, you should copy a char* . I *guess* that you want to copy stuffs from my_str[n] to new_st1->name 
    printf("%s\n",new_st1->name); //%s accepts a string(i.e. char*) , while new_st1->name[0] is a char. In this case, new_st1->name should be passed as a parameter, instead of new_st1->name[0] 
    free(new_st1); //you should free() it after allocating memory with malloc() to avoid memory leak. 
    return 0; 
} 

這是你想要的嗎？

編輯：現在的解釋。