C语言 如何为 const char* 数组赋值并打印到屏幕

How to assign values to const char* array and print to screen

提问by jshapy8

Although I am using C++, as a requirement I need to use const char* arrays instead of string or char arrays. Since I am new, I am required to learn about how to use const char*. I have declared my const char* array as const char* str[5]. At a later point in the program I need to populate each on of the 5 elements with values.

尽管我使用的是 C++，但作为要求，我需要使用 const char* 数组而不是字符串或字符数组。因为我是新手，所以我需要学习如何使用 const char*。我已将 const char* 数组声明为const char* str[5]. 在程序的稍后点，我需要用值填充 5 个元素中的每一个。

However, if I try to assign a value like this:

但是，如果我尝试分配这样的值：

const char* str[5];
char value[5];
value[0] = "hello";
str[0] = value[0];

It will not compile. What is the proper way to add a char array to char to a const char* array and then print the array? Any help would be appreciated. Thanks.

它不会编译。将 char 数组添加到 char 到 const char* 数组然后打印数组的正确方法是什么？任何帮助，将不胜感激。谢谢。

回答by Bill Lynch

1. The string "hello"is made up of 6 characters, not 5.

   {'h', 'e', 'l', 'l', 'o', '
   char value[6] = "hello";
   '}
   

2. If you assign the string when you declare value, your code can look similar to what your currently have:

   char value[6];
   strncpy(value, "hello", sizeof(value));
   

3. If you want to do it in two separate lines, you should use strncpy().

   const char * str[5];
   char value[6] = "hello";
   str[0] = value;
   

4. To place a pointer to valuein the list of strings named str:

   {'h', 'e', 'l', 'l', 'o', '
   char value[6] = "hello";
   '}
   

   Note that this leaves str[1]through str[4]with unspecified values.

1. 字符串"hello"由 6 个字符组成，而不是 5 个。

   char value[6];
   strncpy(value, "hello", sizeof(value));
   

2. 如果您在声明时分配字符串value，则您的代码可能与您当前拥有的类似：

   const char * str[5];
   char value[6] = "hello";
   str[0] = value;
   

3. 如果您想在两个单独的行中执行此操作，则应使用strncpy().

   int main(void) {
     const char* str[5];
   
     str[0] = "He" "llo";  // He llo are string literals that concat into 1
   
     char Planet[] = "Earth";  
     str[1] = Planet;  // OK to assign a char * to const char *, but not visa-versa
   
     const char Greet[] = "How";
     str[2] = Greet;
   
     char buf[4];
     buf[0] = 'R'; // For a character array to qualify as a string, need a null character.  
     buf[1] = 0;   // '
   Hello  
   Earth  
   How  
   R  
   ' same _value_ as 0  
     str[3] = buf; // array buf converts to address of first element: char *
   
     str[4] = NULL; // Do not want to print this.
   
     for (int i = 0; str[i]; i++)
       puts(str[i]);
     return 0;
   }
   

4. value在名为 的字符串列表中放置一个指针str：

   value[0] = "hello";
   

   请注意，这使得str[1]通过str[4]与未定值。

回答by chux - Reinstate Monica

Various methods to populate const char* str[5]later point in the program.

const char* str[5]在程序中稍后填充的各种方法。

char value[6];
strncpy(value,"hello",sizeof value);

.

.

char value[]="hello";

回答by ameyCU

str[0]=value;

How can this hold "hello". It can hold only one character.

这怎么能保持“你好”。它只能容纳一个字符。

Also value[5]is not enough for this . There will be no space for '\0'and program will exhibit UB.

value[5]对此也还不够。将没有空间，'\0'并且程序将展示UB。

Thus either use value[6]-

因此要么使用value[6]-

##代码##

Or declare like this -

或者这样声明——

##代码##

And after that just make the pointer point to this array. Something like this-

然后让指针指向这个数组。像这样的东西——

##代码##